Sort array by month and year
I need help with sorting an array by month and year to display on chart respectively.
Array1: ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'....];
Desired Output : [....'Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
I also have another array with values for each month according to the array above respectively
Array2: ['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
,'Jan19_value'....];
Array2: ['55','2','3','0','21','132'....]; //real values
Now when the monthyear array sorts i want the data in this array to move to a new position according to the monthyear position. like this:
Desired Array1: [....'Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
Desired Array2: [....'Apr18_value','May18_value','Jun18_value','Jul18_value','Jan19_value','Mar19_value'];
So i can pick the data later like this:
var label = array[4];
var value = array2[4];
Any help will be appreciated.
javascript html arrays sorting
add a comment |
I need help with sorting an array by month and year to display on chart respectively.
Array1: ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'....];
Desired Output : [....'Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
I also have another array with values for each month according to the array above respectively
Array2: ['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
,'Jan19_value'....];
Array2: ['55','2','3','0','21','132'....]; //real values
Now when the monthyear array sorts i want the data in this array to move to a new position according to the monthyear position. like this:
Desired Array1: [....'Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
Desired Array2: [....'Apr18_value','May18_value','Jun18_value','Jul18_value','Jan19_value','Mar19_value'];
So i can pick the data later like this:
var label = array[4];
var value = array2[4];
Any help will be appreciated.
javascript html arrays sorting
10
What have you tried so far?
– Anurag Srivastava
yesterday
add a comment |
I need help with sorting an array by month and year to display on chart respectively.
Array1: ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'....];
Desired Output : [....'Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
I also have another array with values for each month according to the array above respectively
Array2: ['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
,'Jan19_value'....];
Array2: ['55','2','3','0','21','132'....]; //real values
Now when the monthyear array sorts i want the data in this array to move to a new position according to the monthyear position. like this:
Desired Array1: [....'Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
Desired Array2: [....'Apr18_value','May18_value','Jun18_value','Jul18_value','Jan19_value','Mar19_value'];
So i can pick the data later like this:
var label = array[4];
var value = array2[4];
Any help will be appreciated.
javascript html arrays sorting
I need help with sorting an array by month and year to display on chart respectively.
Array1: ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'....];
Desired Output : [....'Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
I also have another array with values for each month according to the array above respectively
Array2: ['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
,'Jan19_value'....];
Array2: ['55','2','3','0','21','132'....]; //real values
Now when the monthyear array sorts i want the data in this array to move to a new position according to the monthyear position. like this:
Desired Array1: [....'Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
Desired Array2: [....'Apr18_value','May18_value','Jun18_value','Jul18_value','Jan19_value','Mar19_value'];
So i can pick the data later like this:
var label = array[4];
var value = array2[4];
Any help will be appreciated.
javascript html arrays sorting
javascript html arrays sorting
edited yesterday
isanae
2,51711437
2,51711437
asked yesterday
Amir QureshiAmir Qureshi
222112
222112
10
What have you tried so far?
– Anurag Srivastava
yesterday
add a comment |
10
What have you tried so far?
– Anurag Srivastava
yesterday
10
10
What have you tried so far?
– Anurag Srivastava
yesterday
What have you tried so far?
– Anurag Srivastava
yesterday
add a comment |
9 Answers
9
active
oldest
votes
You could get the date as sortable string and sort this sting.
For getting more than one array sorted by one signature array, you could take sorting with map, where you sort an array of indices, indicating the final sorting and then reassign all arrays with this sorting.
The getD
function returns a formatted string by taking an index of array0
for sorting. Inside of the fuinction the string is destructed into month and year parts and replaced by its ISO 8601 representation. The callback of the replacement function takes the matched items, retuns an array with a formatted year and the month of an object with the month names and the related month numbers. Then this array is joined and returned.
The sorting takes place with a comparsion with String#localeCompare
.
var array1 = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'],
array2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'],
array3 = ['55', '2', '3', '0', '21', '132'],
indices = Object
.keys(array1)
.sort(function (a, b) {
function getD(i) {
var months = { Jan: '01', Feb: '02', Mar: '03', Apr: '04', May: '05', Jun: '06', Jul: '07', Aug: '08', Sep: '09', Oct: '10', Nov: '11', Dec: '12' },
s = array1[i];
return s.replace(/^(...)(.+)$/, (_, m, y) => [y.padStart(4, '0'), months[m]].join('-'));
}
return getD(a).localeCompare(getD(b));
}),
result = [array1, array3].map(a => indices.map(i => a[i]));
result.forEach(a => console.log(...a));
Are you sure the sorting is correct format? i think you are sortedMMDD
Format its aMMYY
– prasanth
yesterday
@prasanth, thank you for the hint. i thought, the number is a day.
– Nina Scholz
yesterday
if i only want to perform this with array 1 and array 3?
– Amir Qureshi
yesterday
@AmirQureshi, then you need oly the wanted array in the array for getting a new sort. please see edit.
– Nina Scholz
yesterday
do you want to get new arrays?
– Nina Scholz
yesterday
|
show 1 more comment
You need change the string for to new Date(dateString)
format like
new Date(Month Date Year)
Updated regex Pattern for Both Array
https://regex101.com/r/h1cm1z/2/
Updated Sort Second array based on first array sorting index
var arr1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
var arr2 =['55','2','3','0','21','132'];
function datesort(arr){
return arr.concat().sort((a,b)=>{
a = a.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
b = b.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
return new Date(a) - new Date(b)
})
}
var after_arr1 =new datesort(arr1);
var after_arr2 = arr1.reduce(function(a,b,c){
var ind = after_arr1.indexOf(b);
a[ind] = arr2[c]
return a
},);
console.log(after_arr1.join(','));
console.log(after_arr2.join(','))
Thank you , it works as required. What if i have another array with values for each month respectively according to the monthyear array. how do i sort it when the monthyear moves to a new position. do you understand what i am saying?
– Amir Qureshi
yesterday
i could added the format of date .you could change your array value like this.If you have any new position .Kindly change regex pattern and reproduce the same date format.Or post your new position. i will give the matching pattern
– prasanth
yesterday
Or where did you get this array? You could format array with in pattern on the place of array create.or with something added delimiters likeMM_YY,YY-MM
– prasanth
yesterday
i have updated the question please have a look .
– Amir Qureshi
yesterday
1
you need to sort array2 based on array1 sorted index position right?
– prasanth
yesterday
|
show 8 more comments
To order properly you need to know the order of the months. This is not alphabetical so you can use an array with the order of the months then look them up.
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
Is already properly sorted and can be used in conjunction with .indexOf() to get the position of a month.
const myArr = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const myArr2 = ['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
,'Jan19_value'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0,3))
let yearA = a.slice(3,6)
let monthB = monthOrder.indexOf(b.slice(0,3))
let yearB = b.slice(3,6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
let sortedMonths = myArr.sort(sortYearMonth)
let sortedMonths2 = myArr2.sort(sortYearMonth)
console.log(sortedMonths )
console.log(sortedMonths2 )
Update: values in same position
Updated version links the two arrays together then sorts the first while keeping the relative position to the second.
Idea: Link two arrays with a temporary Object then extract the key/value pair using Object.entries
. Then sorting the array based on the first value of the pair, which is the value of array1. Then it returns the key/value pair in the right order you can extract the values into two arrays again by using .map()
I added a run with the string based example and the real values examples below
const myArr = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
const myArr2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'];
const myArr3 = ['55','2','3','0','21','132'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0, 3))
let yearA = a.slice(3, 6)
let monthB = monthOrder.indexOf(b.slice(0, 3))
let yearB = b.slice(3, 6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
function sortByFirst(myArr, myArr2) {
let keyValue = myArr.reduce((links, item, i) => {
links[item] = myArr2[i];
return links
}, {})
let entries = Object.entries(keyValue)
return entries.sort((a, b) => sortYearMonth(a[0], b[0]))
}
let sortedEntries = sortByFirst(myArr, myArr2)
let sortedMonths = sortedEntries.map(i => i[0])
let sortedValues = sortedEntries.map(i => i[1])
let sortedEntries2 = sortByFirst(myArr, myArr3)
let sortedMonths2 = sortedEntries2.map(i => i[0])
let sortedValues2 = sortedEntries2.map(i => i[1])
console.log(sortedMonths)
console.log(sortedValues)
console.log(sortedMonths2)
console.log(sortedValues2)
add a comment |
const input = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
//const output : ['Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
//Can be done easily by using momentjs, darte-fns, but here i will do it natively
const t = {
Jan: 1,
Feb: 2,
Mar: 3,
Apr: 4,
May: 5,
Jun: 6,
Jul: 7,
Aug: 8,
Sep: 9,
Oct: 10,
Nov: 11,
Dec: 12
}
const giveConcatString = (a, t) => {
const monthPart = a.substr(0, 3)
const yearPart = a.substr(3)
return `${yearPart}${t[monthPart]}`
}
const sortedArray = input.sort((a, b) => {
const concatString = giveConcatString(a, t)
const concatStringB = giveConcatString(b, t)
return concatString <= concatStringB ? -1 : 1
})
console.log(sortedArray)
This may help you solve this problem. Did it natively.
add a comment |
You can try this raw snippet.
var MY = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
var s = "JanFebMarAprMayJunJulAugSepOctNovDec";
// converting to raw date format
for (i in MY) {
num = MY[i].match(/d+/g);
letr = MY[i].match(/[a-zA-Z]+/g);
MY[i] = (s.indexOf(letr) / 3 + 1) + '-' + '20' + num;
}
// sorting logic
var sorted = MY.sort(function(a, b) {
a = a.split("-");
b = b.split("-")
return new Date(a[1], a[0], 1) - new Date(b[1], b[0], 1)
});
// converting back to original array after sorting
res = ;
for (i in sorted) {
var a = sorted[i].split('-');
res[i] = (s.substr((parseInt(a[0]) - 1) * 3, 3)) + '-' + a[1].substr(2, 2);
}
console.log(res);
add a comment |
I used lodash for sorting and substring
method to divide date into month and day parts
const data = ['Mar20','Mar21', 'Mar01', 'Mar19','Apr18','Jun18','Jul18','May18','Jan19']
const monthMap = ["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"];
const sorted = _.sortBy(data, date => monthMap.indexOf(date.substring(0,3)), date => date.substring(3,5))
console.log(sorted)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
add a comment |
let mth = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
mth.sort((itemA, itemB)=>{
let mthAstr = itemA.slice(0,3)+" 01 "+itemA.slice(3,5);
let mthA = new Date(mthAstr);
let mthBstr = itemB.slice(0,3)+" 01 "+itemB.slice(3,5);
let mthB = new Date(mthBstr);
if (mthA < mthB)
return -1;
if (mthA > mthB)
return 1;
return 0;
});
add a comment |
You can try something like that :
var Array = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
Array.sort(function(a, b) {
a = [a.slice(0,3), ' 20', a.slice(3)].join('');
b = [b.slice(0,3), ' 20', b.slice(3)].join('')
return new Date() - new Date(b);
});
console.log(Array);
New contributor
add a comment |
I would suggest that your data structure is unfortunate. Working with shared indices makes it much more difficult than a data structure that combines them, such as an array of objects, [{label: 'Mar19', value: 55}, ...]
.
If your data is coming from an upstream solution you cannot control, you can still manage this in your own work, converting before you use it. (And if you really have to converting back to pass to others.)
A common name for a function combining two arrays is zip
-- think of it acting like a zipper. This version uses one that takes a function to say how the paired elements should be combined. (Elsewhere such a function might be called zipWith
.)
Here sortArraysByDate
calls zip
passing a function that turns 'Mar19'
and 55
into {label: 'Mar19, value: 55, month: 'Mar', year: 19}
using dateFields
to extract the month and year from that label, then sorts these using the straightforward dateSort
const months = {Jan: 1, Feb: 2, Mar: 3, Apr: 4, May: 5, Jun: 6,
Jul: 7, Aug: 8, Sep: 9, Oct: 10, Nov: 11, Dec: 12}
const dateFields = (d) => ({
month: d.slice(0, 3),
year: Number(d.slice(3))
})
const dateSort = ({month: m1, year: y1}, {month: m2, year: y2}) =>
(y1 - y2) || (months[m1] - months[m2])
const zip = (fn, a1, a2) => a1.map((a, i) => fn(a, a2[i]))
const sortArraysByDate = (a1, a2) =>
zip((label, value) => ({label, value, ...dateFields(label)}), a1, a2)
.sort(dateSort)
.map(({label, value}) => ({label, value}))
const Array1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const Array2 = ['55','2','3','0','21','132']; //real values
const result = sortArraysByDate(Array1, Array2)
console.log(result)
The map
call in sortArraysByDate
is quite possibly not necessary. Without it, the resulting data includes extra year
and month
fields; that might not be an issue.
If you really need those updated arrays in the original format, you can just map
the result:
const newArray1 = result.map(o => o.label)
const newArray2 = result.map(o => o.value)
But I would urge you not to do this unless it's absolutely necessary. This structure is really useful. The paired arrays are much less so.
Also, if you need to combine more than two arrays, you could write a slightly more sophisticated version of zip
:
const zip = (fn, ...as) => as[0].map((_, i) => fn(...as.map(a => a[i])))
This takes a function on n
arguments, and n
arrays, and yields a new array containing the result of calling that function respectively on the successive items in each of the arrays.
add a comment |
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9 Answers
9
active
oldest
votes
9 Answers
9
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could get the date as sortable string and sort this sting.
For getting more than one array sorted by one signature array, you could take sorting with map, where you sort an array of indices, indicating the final sorting and then reassign all arrays with this sorting.
The getD
function returns a formatted string by taking an index of array0
for sorting. Inside of the fuinction the string is destructed into month and year parts and replaced by its ISO 8601 representation. The callback of the replacement function takes the matched items, retuns an array with a formatted year and the month of an object with the month names and the related month numbers. Then this array is joined and returned.
The sorting takes place with a comparsion with String#localeCompare
.
var array1 = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'],
array2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'],
array3 = ['55', '2', '3', '0', '21', '132'],
indices = Object
.keys(array1)
.sort(function (a, b) {
function getD(i) {
var months = { Jan: '01', Feb: '02', Mar: '03', Apr: '04', May: '05', Jun: '06', Jul: '07', Aug: '08', Sep: '09', Oct: '10', Nov: '11', Dec: '12' },
s = array1[i];
return s.replace(/^(...)(.+)$/, (_, m, y) => [y.padStart(4, '0'), months[m]].join('-'));
}
return getD(a).localeCompare(getD(b));
}),
result = [array1, array3].map(a => indices.map(i => a[i]));
result.forEach(a => console.log(...a));
Are you sure the sorting is correct format? i think you are sortedMMDD
Format its aMMYY
– prasanth
yesterday
@prasanth, thank you for the hint. i thought, the number is a day.
– Nina Scholz
yesterday
if i only want to perform this with array 1 and array 3?
– Amir Qureshi
yesterday
@AmirQureshi, then you need oly the wanted array in the array for getting a new sort. please see edit.
– Nina Scholz
yesterday
do you want to get new arrays?
– Nina Scholz
yesterday
|
show 1 more comment
You could get the date as sortable string and sort this sting.
For getting more than one array sorted by one signature array, you could take sorting with map, where you sort an array of indices, indicating the final sorting and then reassign all arrays with this sorting.
The getD
function returns a formatted string by taking an index of array0
for sorting. Inside of the fuinction the string is destructed into month and year parts and replaced by its ISO 8601 representation. The callback of the replacement function takes the matched items, retuns an array with a formatted year and the month of an object with the month names and the related month numbers. Then this array is joined and returned.
The sorting takes place with a comparsion with String#localeCompare
.
var array1 = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'],
array2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'],
array3 = ['55', '2', '3', '0', '21', '132'],
indices = Object
.keys(array1)
.sort(function (a, b) {
function getD(i) {
var months = { Jan: '01', Feb: '02', Mar: '03', Apr: '04', May: '05', Jun: '06', Jul: '07', Aug: '08', Sep: '09', Oct: '10', Nov: '11', Dec: '12' },
s = array1[i];
return s.replace(/^(...)(.+)$/, (_, m, y) => [y.padStart(4, '0'), months[m]].join('-'));
}
return getD(a).localeCompare(getD(b));
}),
result = [array1, array3].map(a => indices.map(i => a[i]));
result.forEach(a => console.log(...a));
Are you sure the sorting is correct format? i think you are sortedMMDD
Format its aMMYY
– prasanth
yesterday
@prasanth, thank you for the hint. i thought, the number is a day.
– Nina Scholz
yesterday
if i only want to perform this with array 1 and array 3?
– Amir Qureshi
yesterday
@AmirQureshi, then you need oly the wanted array in the array for getting a new sort. please see edit.
– Nina Scholz
yesterday
do you want to get new arrays?
– Nina Scholz
yesterday
|
show 1 more comment
You could get the date as sortable string and sort this sting.
For getting more than one array sorted by one signature array, you could take sorting with map, where you sort an array of indices, indicating the final sorting and then reassign all arrays with this sorting.
The getD
function returns a formatted string by taking an index of array0
for sorting. Inside of the fuinction the string is destructed into month and year parts and replaced by its ISO 8601 representation. The callback of the replacement function takes the matched items, retuns an array with a formatted year and the month of an object with the month names and the related month numbers. Then this array is joined and returned.
The sorting takes place with a comparsion with String#localeCompare
.
var array1 = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'],
array2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'],
array3 = ['55', '2', '3', '0', '21', '132'],
indices = Object
.keys(array1)
.sort(function (a, b) {
function getD(i) {
var months = { Jan: '01', Feb: '02', Mar: '03', Apr: '04', May: '05', Jun: '06', Jul: '07', Aug: '08', Sep: '09', Oct: '10', Nov: '11', Dec: '12' },
s = array1[i];
return s.replace(/^(...)(.+)$/, (_, m, y) => [y.padStart(4, '0'), months[m]].join('-'));
}
return getD(a).localeCompare(getD(b));
}),
result = [array1, array3].map(a => indices.map(i => a[i]));
result.forEach(a => console.log(...a));
You could get the date as sortable string and sort this sting.
For getting more than one array sorted by one signature array, you could take sorting with map, where you sort an array of indices, indicating the final sorting and then reassign all arrays with this sorting.
The getD
function returns a formatted string by taking an index of array0
for sorting. Inside of the fuinction the string is destructed into month and year parts and replaced by its ISO 8601 representation. The callback of the replacement function takes the matched items, retuns an array with a formatted year and the month of an object with the month names and the related month numbers. Then this array is joined and returned.
The sorting takes place with a comparsion with String#localeCompare
.
var array1 = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'],
array2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'],
array3 = ['55', '2', '3', '0', '21', '132'],
indices = Object
.keys(array1)
.sort(function (a, b) {
function getD(i) {
var months = { Jan: '01', Feb: '02', Mar: '03', Apr: '04', May: '05', Jun: '06', Jul: '07', Aug: '08', Sep: '09', Oct: '10', Nov: '11', Dec: '12' },
s = array1[i];
return s.replace(/^(...)(.+)$/, (_, m, y) => [y.padStart(4, '0'), months[m]].join('-'));
}
return getD(a).localeCompare(getD(b));
}),
result = [array1, array3].map(a => indices.map(i => a[i]));
result.forEach(a => console.log(...a));
var array1 = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'],
array2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'],
array3 = ['55', '2', '3', '0', '21', '132'],
indices = Object
.keys(array1)
.sort(function (a, b) {
function getD(i) {
var months = { Jan: '01', Feb: '02', Mar: '03', Apr: '04', May: '05', Jun: '06', Jul: '07', Aug: '08', Sep: '09', Oct: '10', Nov: '11', Dec: '12' },
s = array1[i];
return s.replace(/^(...)(.+)$/, (_, m, y) => [y.padStart(4, '0'), months[m]].join('-'));
}
return getD(a).localeCompare(getD(b));
}),
result = [array1, array3].map(a => indices.map(i => a[i]));
result.forEach(a => console.log(...a));
var array1 = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'],
array2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'],
array3 = ['55', '2', '3', '0', '21', '132'],
indices = Object
.keys(array1)
.sort(function (a, b) {
function getD(i) {
var months = { Jan: '01', Feb: '02', Mar: '03', Apr: '04', May: '05', Jun: '06', Jul: '07', Aug: '08', Sep: '09', Oct: '10', Nov: '11', Dec: '12' },
s = array1[i];
return s.replace(/^(...)(.+)$/, (_, m, y) => [y.padStart(4, '0'), months[m]].join('-'));
}
return getD(a).localeCompare(getD(b));
}),
result = [array1, array3].map(a => indices.map(i => a[i]));
result.forEach(a => console.log(...a));
edited yesterday
answered yesterday
Nina ScholzNina Scholz
190k1599173
190k1599173
Are you sure the sorting is correct format? i think you are sortedMMDD
Format its aMMYY
– prasanth
yesterday
@prasanth, thank you for the hint. i thought, the number is a day.
– Nina Scholz
yesterday
if i only want to perform this with array 1 and array 3?
– Amir Qureshi
yesterday
@AmirQureshi, then you need oly the wanted array in the array for getting a new sort. please see edit.
– Nina Scholz
yesterday
do you want to get new arrays?
– Nina Scholz
yesterday
|
show 1 more comment
Are you sure the sorting is correct format? i think you are sortedMMDD
Format its aMMYY
– prasanth
yesterday
@prasanth, thank you for the hint. i thought, the number is a day.
– Nina Scholz
yesterday
if i only want to perform this with array 1 and array 3?
– Amir Qureshi
yesterday
@AmirQureshi, then you need oly the wanted array in the array for getting a new sort. please see edit.
– Nina Scholz
yesterday
do you want to get new arrays?
– Nina Scholz
yesterday
Are you sure the sorting is correct format? i think you are sorted
MMDD
Format its a MMYY
– prasanth
yesterday
Are you sure the sorting is correct format? i think you are sorted
MMDD
Format its a MMYY
– prasanth
yesterday
@prasanth, thank you for the hint. i thought, the number is a day.
– Nina Scholz
yesterday
@prasanth, thank you for the hint. i thought, the number is a day.
– Nina Scholz
yesterday
if i only want to perform this with array 1 and array 3?
– Amir Qureshi
yesterday
if i only want to perform this with array 1 and array 3?
– Amir Qureshi
yesterday
@AmirQureshi, then you need oly the wanted array in the array for getting a new sort. please see edit.
– Nina Scholz
yesterday
@AmirQureshi, then you need oly the wanted array in the array for getting a new sort. please see edit.
– Nina Scholz
yesterday
do you want to get new arrays?
– Nina Scholz
yesterday
do you want to get new arrays?
– Nina Scholz
yesterday
|
show 1 more comment
You need change the string for to new Date(dateString)
format like
new Date(Month Date Year)
Updated regex Pattern for Both Array
https://regex101.com/r/h1cm1z/2/
Updated Sort Second array based on first array sorting index
var arr1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
var arr2 =['55','2','3','0','21','132'];
function datesort(arr){
return arr.concat().sort((a,b)=>{
a = a.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
b = b.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
return new Date(a) - new Date(b)
})
}
var after_arr1 =new datesort(arr1);
var after_arr2 = arr1.reduce(function(a,b,c){
var ind = after_arr1.indexOf(b);
a[ind] = arr2[c]
return a
},);
console.log(after_arr1.join(','));
console.log(after_arr2.join(','))
Thank you , it works as required. What if i have another array with values for each month respectively according to the monthyear array. how do i sort it when the monthyear moves to a new position. do you understand what i am saying?
– Amir Qureshi
yesterday
i could added the format of date .you could change your array value like this.If you have any new position .Kindly change regex pattern and reproduce the same date format.Or post your new position. i will give the matching pattern
– prasanth
yesterday
Or where did you get this array? You could format array with in pattern on the place of array create.or with something added delimiters likeMM_YY,YY-MM
– prasanth
yesterday
i have updated the question please have a look .
– Amir Qureshi
yesterday
1
you need to sort array2 based on array1 sorted index position right?
– prasanth
yesterday
|
show 8 more comments
You need change the string for to new Date(dateString)
format like
new Date(Month Date Year)
Updated regex Pattern for Both Array
https://regex101.com/r/h1cm1z/2/
Updated Sort Second array based on first array sorting index
var arr1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
var arr2 =['55','2','3','0','21','132'];
function datesort(arr){
return arr.concat().sort((a,b)=>{
a = a.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
b = b.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
return new Date(a) - new Date(b)
})
}
var after_arr1 =new datesort(arr1);
var after_arr2 = arr1.reduce(function(a,b,c){
var ind = after_arr1.indexOf(b);
a[ind] = arr2[c]
return a
},);
console.log(after_arr1.join(','));
console.log(after_arr2.join(','))
Thank you , it works as required. What if i have another array with values for each month respectively according to the monthyear array. how do i sort it when the monthyear moves to a new position. do you understand what i am saying?
– Amir Qureshi
yesterday
i could added the format of date .you could change your array value like this.If you have any new position .Kindly change regex pattern and reproduce the same date format.Or post your new position. i will give the matching pattern
– prasanth
yesterday
Or where did you get this array? You could format array with in pattern on the place of array create.or with something added delimiters likeMM_YY,YY-MM
– prasanth
yesterday
i have updated the question please have a look .
– Amir Qureshi
yesterday
1
you need to sort array2 based on array1 sorted index position right?
– prasanth
yesterday
|
show 8 more comments
You need change the string for to new Date(dateString)
format like
new Date(Month Date Year)
Updated regex Pattern for Both Array
https://regex101.com/r/h1cm1z/2/
Updated Sort Second array based on first array sorting index
var arr1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
var arr2 =['55','2','3','0','21','132'];
function datesort(arr){
return arr.concat().sort((a,b)=>{
a = a.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
b = b.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
return new Date(a) - new Date(b)
})
}
var after_arr1 =new datesort(arr1);
var after_arr2 = arr1.reduce(function(a,b,c){
var ind = after_arr1.indexOf(b);
a[ind] = arr2[c]
return a
},);
console.log(after_arr1.join(','));
console.log(after_arr2.join(','))
You need change the string for to new Date(dateString)
format like
new Date(Month Date Year)
Updated regex Pattern for Both Array
https://regex101.com/r/h1cm1z/2/
Updated Sort Second array based on first array sorting index
var arr1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
var arr2 =['55','2','3','0','21','132'];
function datesort(arr){
return arr.concat().sort((a,b)=>{
a = a.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
b = b.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
return new Date(a) - new Date(b)
})
}
var after_arr1 =new datesort(arr1);
var after_arr2 = arr1.reduce(function(a,b,c){
var ind = after_arr1.indexOf(b);
a[ind] = arr2[c]
return a
},);
console.log(after_arr1.join(','));
console.log(after_arr2.join(','))
var arr1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
var arr2 =['55','2','3','0','21','132'];
function datesort(arr){
return arr.concat().sort((a,b)=>{
a = a.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
b = b.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
return new Date(a) - new Date(b)
})
}
var after_arr1 =new datesort(arr1);
var after_arr2 = arr1.reduce(function(a,b,c){
var ind = after_arr1.indexOf(b);
a[ind] = arr2[c]
return a
},);
console.log(after_arr1.join(','));
console.log(after_arr2.join(','))
var arr1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
var arr2 =['55','2','3','0','21','132'];
function datesort(arr){
return arr.concat().sort((a,b)=>{
a = a.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
b = b.replace(/(d+)(.*)/g,' 1 $1'); // Month 1 YEAR
return new Date(a) - new Date(b)
})
}
var after_arr1 =new datesort(arr1);
var after_arr2 = arr1.reduce(function(a,b,c){
var ind = after_arr1.indexOf(b);
a[ind] = arr2[c]
return a
},);
console.log(after_arr1.join(','));
console.log(after_arr2.join(','))
edited yesterday
answered yesterday
prasanthprasanth
13.7k21335
13.7k21335
Thank you , it works as required. What if i have another array with values for each month respectively according to the monthyear array. how do i sort it when the monthyear moves to a new position. do you understand what i am saying?
– Amir Qureshi
yesterday
i could added the format of date .you could change your array value like this.If you have any new position .Kindly change regex pattern and reproduce the same date format.Or post your new position. i will give the matching pattern
– prasanth
yesterday
Or where did you get this array? You could format array with in pattern on the place of array create.or with something added delimiters likeMM_YY,YY-MM
– prasanth
yesterday
i have updated the question please have a look .
– Amir Qureshi
yesterday
1
you need to sort array2 based on array1 sorted index position right?
– prasanth
yesterday
|
show 8 more comments
Thank you , it works as required. What if i have another array with values for each month respectively according to the monthyear array. how do i sort it when the monthyear moves to a new position. do you understand what i am saying?
– Amir Qureshi
yesterday
i could added the format of date .you could change your array value like this.If you have any new position .Kindly change regex pattern and reproduce the same date format.Or post your new position. i will give the matching pattern
– prasanth
yesterday
Or where did you get this array? You could format array with in pattern on the place of array create.or with something added delimiters likeMM_YY,YY-MM
– prasanth
yesterday
i have updated the question please have a look .
– Amir Qureshi
yesterday
1
you need to sort array2 based on array1 sorted index position right?
– prasanth
yesterday
Thank you , it works as required. What if i have another array with values for each month respectively according to the monthyear array. how do i sort it when the monthyear moves to a new position. do you understand what i am saying?
– Amir Qureshi
yesterday
Thank you , it works as required. What if i have another array with values for each month respectively according to the monthyear array. how do i sort it when the monthyear moves to a new position. do you understand what i am saying?
– Amir Qureshi
yesterday
i could added the format of date .you could change your array value like this.If you have any new position .Kindly change regex pattern and reproduce the same date format.Or post your new position. i will give the matching pattern
– prasanth
yesterday
i could added the format of date .you could change your array value like this.If you have any new position .Kindly change regex pattern and reproduce the same date format.Or post your new position. i will give the matching pattern
– prasanth
yesterday
Or where did you get this array? You could format array with in pattern on the place of array create.or with something added delimiters like
MM_YY,YY-MM
– prasanth
yesterday
Or where did you get this array? You could format array with in pattern on the place of array create.or with something added delimiters like
MM_YY,YY-MM
– prasanth
yesterday
i have updated the question please have a look .
– Amir Qureshi
yesterday
i have updated the question please have a look .
– Amir Qureshi
yesterday
1
1
you need to sort array2 based on array1 sorted index position right?
– prasanth
yesterday
you need to sort array2 based on array1 sorted index position right?
– prasanth
yesterday
|
show 8 more comments
To order properly you need to know the order of the months. This is not alphabetical so you can use an array with the order of the months then look them up.
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
Is already properly sorted and can be used in conjunction with .indexOf() to get the position of a month.
const myArr = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const myArr2 = ['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
,'Jan19_value'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0,3))
let yearA = a.slice(3,6)
let monthB = monthOrder.indexOf(b.slice(0,3))
let yearB = b.slice(3,6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
let sortedMonths = myArr.sort(sortYearMonth)
let sortedMonths2 = myArr2.sort(sortYearMonth)
console.log(sortedMonths )
console.log(sortedMonths2 )
Update: values in same position
Updated version links the two arrays together then sorts the first while keeping the relative position to the second.
Idea: Link two arrays with a temporary Object then extract the key/value pair using Object.entries
. Then sorting the array based on the first value of the pair, which is the value of array1. Then it returns the key/value pair in the right order you can extract the values into two arrays again by using .map()
I added a run with the string based example and the real values examples below
const myArr = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
const myArr2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'];
const myArr3 = ['55','2','3','0','21','132'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0, 3))
let yearA = a.slice(3, 6)
let monthB = monthOrder.indexOf(b.slice(0, 3))
let yearB = b.slice(3, 6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
function sortByFirst(myArr, myArr2) {
let keyValue = myArr.reduce((links, item, i) => {
links[item] = myArr2[i];
return links
}, {})
let entries = Object.entries(keyValue)
return entries.sort((a, b) => sortYearMonth(a[0], b[0]))
}
let sortedEntries = sortByFirst(myArr, myArr2)
let sortedMonths = sortedEntries.map(i => i[0])
let sortedValues = sortedEntries.map(i => i[1])
let sortedEntries2 = sortByFirst(myArr, myArr3)
let sortedMonths2 = sortedEntries2.map(i => i[0])
let sortedValues2 = sortedEntries2.map(i => i[1])
console.log(sortedMonths)
console.log(sortedValues)
console.log(sortedMonths2)
console.log(sortedValues2)
add a comment |
To order properly you need to know the order of the months. This is not alphabetical so you can use an array with the order of the months then look them up.
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
Is already properly sorted and can be used in conjunction with .indexOf() to get the position of a month.
const myArr = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const myArr2 = ['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
,'Jan19_value'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0,3))
let yearA = a.slice(3,6)
let monthB = monthOrder.indexOf(b.slice(0,3))
let yearB = b.slice(3,6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
let sortedMonths = myArr.sort(sortYearMonth)
let sortedMonths2 = myArr2.sort(sortYearMonth)
console.log(sortedMonths )
console.log(sortedMonths2 )
Update: values in same position
Updated version links the two arrays together then sorts the first while keeping the relative position to the second.
Idea: Link two arrays with a temporary Object then extract the key/value pair using Object.entries
. Then sorting the array based on the first value of the pair, which is the value of array1. Then it returns the key/value pair in the right order you can extract the values into two arrays again by using .map()
I added a run with the string based example and the real values examples below
const myArr = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
const myArr2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'];
const myArr3 = ['55','2','3','0','21','132'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0, 3))
let yearA = a.slice(3, 6)
let monthB = monthOrder.indexOf(b.slice(0, 3))
let yearB = b.slice(3, 6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
function sortByFirst(myArr, myArr2) {
let keyValue = myArr.reduce((links, item, i) => {
links[item] = myArr2[i];
return links
}, {})
let entries = Object.entries(keyValue)
return entries.sort((a, b) => sortYearMonth(a[0], b[0]))
}
let sortedEntries = sortByFirst(myArr, myArr2)
let sortedMonths = sortedEntries.map(i => i[0])
let sortedValues = sortedEntries.map(i => i[1])
let sortedEntries2 = sortByFirst(myArr, myArr3)
let sortedMonths2 = sortedEntries2.map(i => i[0])
let sortedValues2 = sortedEntries2.map(i => i[1])
console.log(sortedMonths)
console.log(sortedValues)
console.log(sortedMonths2)
console.log(sortedValues2)
add a comment |
To order properly you need to know the order of the months. This is not alphabetical so you can use an array with the order of the months then look them up.
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
Is already properly sorted and can be used in conjunction with .indexOf() to get the position of a month.
const myArr = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const myArr2 = ['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
,'Jan19_value'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0,3))
let yearA = a.slice(3,6)
let monthB = monthOrder.indexOf(b.slice(0,3))
let yearB = b.slice(3,6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
let sortedMonths = myArr.sort(sortYearMonth)
let sortedMonths2 = myArr2.sort(sortYearMonth)
console.log(sortedMonths )
console.log(sortedMonths2 )
Update: values in same position
Updated version links the two arrays together then sorts the first while keeping the relative position to the second.
Idea: Link two arrays with a temporary Object then extract the key/value pair using Object.entries
. Then sorting the array based on the first value of the pair, which is the value of array1. Then it returns the key/value pair in the right order you can extract the values into two arrays again by using .map()
I added a run with the string based example and the real values examples below
const myArr = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
const myArr2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'];
const myArr3 = ['55','2','3','0','21','132'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0, 3))
let yearA = a.slice(3, 6)
let monthB = monthOrder.indexOf(b.slice(0, 3))
let yearB = b.slice(3, 6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
function sortByFirst(myArr, myArr2) {
let keyValue = myArr.reduce((links, item, i) => {
links[item] = myArr2[i];
return links
}, {})
let entries = Object.entries(keyValue)
return entries.sort((a, b) => sortYearMonth(a[0], b[0]))
}
let sortedEntries = sortByFirst(myArr, myArr2)
let sortedMonths = sortedEntries.map(i => i[0])
let sortedValues = sortedEntries.map(i => i[1])
let sortedEntries2 = sortByFirst(myArr, myArr3)
let sortedMonths2 = sortedEntries2.map(i => i[0])
let sortedValues2 = sortedEntries2.map(i => i[1])
console.log(sortedMonths)
console.log(sortedValues)
console.log(sortedMonths2)
console.log(sortedValues2)
To order properly you need to know the order of the months. This is not alphabetical so you can use an array with the order of the months then look them up.
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
Is already properly sorted and can be used in conjunction with .indexOf() to get the position of a month.
const myArr = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const myArr2 = ['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
,'Jan19_value'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0,3))
let yearA = a.slice(3,6)
let monthB = monthOrder.indexOf(b.slice(0,3))
let yearB = b.slice(3,6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
let sortedMonths = myArr.sort(sortYearMonth)
let sortedMonths2 = myArr2.sort(sortYearMonth)
console.log(sortedMonths )
console.log(sortedMonths2 )
Update: values in same position
Updated version links the two arrays together then sorts the first while keeping the relative position to the second.
Idea: Link two arrays with a temporary Object then extract the key/value pair using Object.entries
. Then sorting the array based on the first value of the pair, which is the value of array1. Then it returns the key/value pair in the right order you can extract the values into two arrays again by using .map()
I added a run with the string based example and the real values examples below
const myArr = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
const myArr2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'];
const myArr3 = ['55','2','3','0','21','132'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0, 3))
let yearA = a.slice(3, 6)
let monthB = monthOrder.indexOf(b.slice(0, 3))
let yearB = b.slice(3, 6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
function sortByFirst(myArr, myArr2) {
let keyValue = myArr.reduce((links, item, i) => {
links[item] = myArr2[i];
return links
}, {})
let entries = Object.entries(keyValue)
return entries.sort((a, b) => sortYearMonth(a[0], b[0]))
}
let sortedEntries = sortByFirst(myArr, myArr2)
let sortedMonths = sortedEntries.map(i => i[0])
let sortedValues = sortedEntries.map(i => i[1])
let sortedEntries2 = sortByFirst(myArr, myArr3)
let sortedMonths2 = sortedEntries2.map(i => i[0])
let sortedValues2 = sortedEntries2.map(i => i[1])
console.log(sortedMonths)
console.log(sortedValues)
console.log(sortedMonths2)
console.log(sortedValues2)
const myArr = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const myArr2 = ['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
,'Jan19_value'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0,3))
let yearA = a.slice(3,6)
let monthB = monthOrder.indexOf(b.slice(0,3))
let yearB = b.slice(3,6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
let sortedMonths = myArr.sort(sortYearMonth)
let sortedMonths2 = myArr2.sort(sortYearMonth)
console.log(sortedMonths )
console.log(sortedMonths2 )
const myArr = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const myArr2 = ['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
,'Jan19_value'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0,3))
let yearA = a.slice(3,6)
let monthB = monthOrder.indexOf(b.slice(0,3))
let yearB = b.slice(3,6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
let sortedMonths = myArr.sort(sortYearMonth)
let sortedMonths2 = myArr2.sort(sortYearMonth)
console.log(sortedMonths )
console.log(sortedMonths2 )
const myArr = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
const myArr2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'];
const myArr3 = ['55','2','3','0','21','132'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0, 3))
let yearA = a.slice(3, 6)
let monthB = monthOrder.indexOf(b.slice(0, 3))
let yearB = b.slice(3, 6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
function sortByFirst(myArr, myArr2) {
let keyValue = myArr.reduce((links, item, i) => {
links[item] = myArr2[i];
return links
}, {})
let entries = Object.entries(keyValue)
return entries.sort((a, b) => sortYearMonth(a[0], b[0]))
}
let sortedEntries = sortByFirst(myArr, myArr2)
let sortedMonths = sortedEntries.map(i => i[0])
let sortedValues = sortedEntries.map(i => i[1])
let sortedEntries2 = sortByFirst(myArr, myArr3)
let sortedMonths2 = sortedEntries2.map(i => i[0])
let sortedValues2 = sortedEntries2.map(i => i[1])
console.log(sortedMonths)
console.log(sortedValues)
console.log(sortedMonths2)
console.log(sortedValues2)
const myArr = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
const myArr2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'];
const myArr3 = ['55','2','3','0','21','132'];
const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
let sortYearMonth = (a, b) => {
let monthA = monthOrder.indexOf(a.slice(0, 3))
let yearA = a.slice(3, 6)
let monthB = monthOrder.indexOf(b.slice(0, 3))
let yearB = b.slice(3, 6)
return (`${yearA}-${monthA}` < `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` > `${yearB}-${monthB}`) ? 1 : 0
}
function sortByFirst(myArr, myArr2) {
let keyValue = myArr.reduce((links, item, i) => {
links[item] = myArr2[i];
return links
}, {})
let entries = Object.entries(keyValue)
return entries.sort((a, b) => sortYearMonth(a[0], b[0]))
}
let sortedEntries = sortByFirst(myArr, myArr2)
let sortedMonths = sortedEntries.map(i => i[0])
let sortedValues = sortedEntries.map(i => i[1])
let sortedEntries2 = sortByFirst(myArr, myArr3)
let sortedMonths2 = sortedEntries2.map(i => i[0])
let sortedValues2 = sortedEntries2.map(i => i[1])
console.log(sortedMonths)
console.log(sortedValues)
console.log(sortedMonths2)
console.log(sortedValues2)
edited yesterday
answered yesterday
Jordan MaduroJordan Maduro
47827
47827
add a comment |
add a comment |
const input = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
//const output : ['Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
//Can be done easily by using momentjs, darte-fns, but here i will do it natively
const t = {
Jan: 1,
Feb: 2,
Mar: 3,
Apr: 4,
May: 5,
Jun: 6,
Jul: 7,
Aug: 8,
Sep: 9,
Oct: 10,
Nov: 11,
Dec: 12
}
const giveConcatString = (a, t) => {
const monthPart = a.substr(0, 3)
const yearPart = a.substr(3)
return `${yearPart}${t[monthPart]}`
}
const sortedArray = input.sort((a, b) => {
const concatString = giveConcatString(a, t)
const concatStringB = giveConcatString(b, t)
return concatString <= concatStringB ? -1 : 1
})
console.log(sortedArray)
This may help you solve this problem. Did it natively.
add a comment |
const input = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
//const output : ['Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
//Can be done easily by using momentjs, darte-fns, but here i will do it natively
const t = {
Jan: 1,
Feb: 2,
Mar: 3,
Apr: 4,
May: 5,
Jun: 6,
Jul: 7,
Aug: 8,
Sep: 9,
Oct: 10,
Nov: 11,
Dec: 12
}
const giveConcatString = (a, t) => {
const monthPart = a.substr(0, 3)
const yearPart = a.substr(3)
return `${yearPart}${t[monthPart]}`
}
const sortedArray = input.sort((a, b) => {
const concatString = giveConcatString(a, t)
const concatStringB = giveConcatString(b, t)
return concatString <= concatStringB ? -1 : 1
})
console.log(sortedArray)
This may help you solve this problem. Did it natively.
add a comment |
const input = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
//const output : ['Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
//Can be done easily by using momentjs, darte-fns, but here i will do it natively
const t = {
Jan: 1,
Feb: 2,
Mar: 3,
Apr: 4,
May: 5,
Jun: 6,
Jul: 7,
Aug: 8,
Sep: 9,
Oct: 10,
Nov: 11,
Dec: 12
}
const giveConcatString = (a, t) => {
const monthPart = a.substr(0, 3)
const yearPart = a.substr(3)
return `${yearPart}${t[monthPart]}`
}
const sortedArray = input.sort((a, b) => {
const concatString = giveConcatString(a, t)
const concatStringB = giveConcatString(b, t)
return concatString <= concatStringB ? -1 : 1
})
console.log(sortedArray)
This may help you solve this problem. Did it natively.
const input = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
//const output : ['Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
//Can be done easily by using momentjs, darte-fns, but here i will do it natively
const t = {
Jan: 1,
Feb: 2,
Mar: 3,
Apr: 4,
May: 5,
Jun: 6,
Jul: 7,
Aug: 8,
Sep: 9,
Oct: 10,
Nov: 11,
Dec: 12
}
const giveConcatString = (a, t) => {
const monthPart = a.substr(0, 3)
const yearPart = a.substr(3)
return `${yearPart}${t[monthPart]}`
}
const sortedArray = input.sort((a, b) => {
const concatString = giveConcatString(a, t)
const concatStringB = giveConcatString(b, t)
return concatString <= concatStringB ? -1 : 1
})
console.log(sortedArray)
This may help you solve this problem. Did it natively.
const input = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
//const output : ['Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
//Can be done easily by using momentjs, darte-fns, but here i will do it natively
const t = {
Jan: 1,
Feb: 2,
Mar: 3,
Apr: 4,
May: 5,
Jun: 6,
Jul: 7,
Aug: 8,
Sep: 9,
Oct: 10,
Nov: 11,
Dec: 12
}
const giveConcatString = (a, t) => {
const monthPart = a.substr(0, 3)
const yearPart = a.substr(3)
return `${yearPart}${t[monthPart]}`
}
const sortedArray = input.sort((a, b) => {
const concatString = giveConcatString(a, t)
const concatStringB = giveConcatString(b, t)
return concatString <= concatStringB ? -1 : 1
})
console.log(sortedArray)
const input = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
//const output : ['Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
//Can be done easily by using momentjs, darte-fns, but here i will do it natively
const t = {
Jan: 1,
Feb: 2,
Mar: 3,
Apr: 4,
May: 5,
Jun: 6,
Jul: 7,
Aug: 8,
Sep: 9,
Oct: 10,
Nov: 11,
Dec: 12
}
const giveConcatString = (a, t) => {
const monthPart = a.substr(0, 3)
const yearPart = a.substr(3)
return `${yearPart}${t[monthPart]}`
}
const sortedArray = input.sort((a, b) => {
const concatString = giveConcatString(a, t)
const concatStringB = giveConcatString(b, t)
return concatString <= concatStringB ? -1 : 1
})
console.log(sortedArray)
edited yesterday
answered yesterday
simbathesailorsimbathesailor
2,23411018
2,23411018
add a comment |
add a comment |
You can try this raw snippet.
var MY = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
var s = "JanFebMarAprMayJunJulAugSepOctNovDec";
// converting to raw date format
for (i in MY) {
num = MY[i].match(/d+/g);
letr = MY[i].match(/[a-zA-Z]+/g);
MY[i] = (s.indexOf(letr) / 3 + 1) + '-' + '20' + num;
}
// sorting logic
var sorted = MY.sort(function(a, b) {
a = a.split("-");
b = b.split("-")
return new Date(a[1], a[0], 1) - new Date(b[1], b[0], 1)
});
// converting back to original array after sorting
res = ;
for (i in sorted) {
var a = sorted[i].split('-');
res[i] = (s.substr((parseInt(a[0]) - 1) * 3, 3)) + '-' + a[1].substr(2, 2);
}
console.log(res);
add a comment |
You can try this raw snippet.
var MY = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
var s = "JanFebMarAprMayJunJulAugSepOctNovDec";
// converting to raw date format
for (i in MY) {
num = MY[i].match(/d+/g);
letr = MY[i].match(/[a-zA-Z]+/g);
MY[i] = (s.indexOf(letr) / 3 + 1) + '-' + '20' + num;
}
// sorting logic
var sorted = MY.sort(function(a, b) {
a = a.split("-");
b = b.split("-")
return new Date(a[1], a[0], 1) - new Date(b[1], b[0], 1)
});
// converting back to original array after sorting
res = ;
for (i in sorted) {
var a = sorted[i].split('-');
res[i] = (s.substr((parseInt(a[0]) - 1) * 3, 3)) + '-' + a[1].substr(2, 2);
}
console.log(res);
add a comment |
You can try this raw snippet.
var MY = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
var s = "JanFebMarAprMayJunJulAugSepOctNovDec";
// converting to raw date format
for (i in MY) {
num = MY[i].match(/d+/g);
letr = MY[i].match(/[a-zA-Z]+/g);
MY[i] = (s.indexOf(letr) / 3 + 1) + '-' + '20' + num;
}
// sorting logic
var sorted = MY.sort(function(a, b) {
a = a.split("-");
b = b.split("-")
return new Date(a[1], a[0], 1) - new Date(b[1], b[0], 1)
});
// converting back to original array after sorting
res = ;
for (i in sorted) {
var a = sorted[i].split('-');
res[i] = (s.substr((parseInt(a[0]) - 1) * 3, 3)) + '-' + a[1].substr(2, 2);
}
console.log(res);
You can try this raw snippet.
var MY = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
var s = "JanFebMarAprMayJunJulAugSepOctNovDec";
// converting to raw date format
for (i in MY) {
num = MY[i].match(/d+/g);
letr = MY[i].match(/[a-zA-Z]+/g);
MY[i] = (s.indexOf(letr) / 3 + 1) + '-' + '20' + num;
}
// sorting logic
var sorted = MY.sort(function(a, b) {
a = a.split("-");
b = b.split("-")
return new Date(a[1], a[0], 1) - new Date(b[1], b[0], 1)
});
// converting back to original array after sorting
res = ;
for (i in sorted) {
var a = sorted[i].split('-');
res[i] = (s.substr((parseInt(a[0]) - 1) * 3, 3)) + '-' + a[1].substr(2, 2);
}
console.log(res);
var MY = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
var s = "JanFebMarAprMayJunJulAugSepOctNovDec";
// converting to raw date format
for (i in MY) {
num = MY[i].match(/d+/g);
letr = MY[i].match(/[a-zA-Z]+/g);
MY[i] = (s.indexOf(letr) / 3 + 1) + '-' + '20' + num;
}
// sorting logic
var sorted = MY.sort(function(a, b) {
a = a.split("-");
b = b.split("-")
return new Date(a[1], a[0], 1) - new Date(b[1], b[0], 1)
});
// converting back to original array after sorting
res = ;
for (i in sorted) {
var a = sorted[i].split('-');
res[i] = (s.substr((parseInt(a[0]) - 1) * 3, 3)) + '-' + a[1].substr(2, 2);
}
console.log(res);
var MY = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
var s = "JanFebMarAprMayJunJulAugSepOctNovDec";
// converting to raw date format
for (i in MY) {
num = MY[i].match(/d+/g);
letr = MY[i].match(/[a-zA-Z]+/g);
MY[i] = (s.indexOf(letr) / 3 + 1) + '-' + '20' + num;
}
// sorting logic
var sorted = MY.sort(function(a, b) {
a = a.split("-");
b = b.split("-")
return new Date(a[1], a[0], 1) - new Date(b[1], b[0], 1)
});
// converting back to original array after sorting
res = ;
for (i in sorted) {
var a = sorted[i].split('-');
res[i] = (s.substr((parseInt(a[0]) - 1) * 3, 3)) + '-' + a[1].substr(2, 2);
}
console.log(res);
answered yesterday
Rahul MeshramRahul Meshram
7,32041942
7,32041942
add a comment |
add a comment |
I used lodash for sorting and substring
method to divide date into month and day parts
const data = ['Mar20','Mar21', 'Mar01', 'Mar19','Apr18','Jun18','Jul18','May18','Jan19']
const monthMap = ["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"];
const sorted = _.sortBy(data, date => monthMap.indexOf(date.substring(0,3)), date => date.substring(3,5))
console.log(sorted)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
add a comment |
I used lodash for sorting and substring
method to divide date into month and day parts
const data = ['Mar20','Mar21', 'Mar01', 'Mar19','Apr18','Jun18','Jul18','May18','Jan19']
const monthMap = ["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"];
const sorted = _.sortBy(data, date => monthMap.indexOf(date.substring(0,3)), date => date.substring(3,5))
console.log(sorted)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
add a comment |
I used lodash for sorting and substring
method to divide date into month and day parts
const data = ['Mar20','Mar21', 'Mar01', 'Mar19','Apr18','Jun18','Jul18','May18','Jan19']
const monthMap = ["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"];
const sorted = _.sortBy(data, date => monthMap.indexOf(date.substring(0,3)), date => date.substring(3,5))
console.log(sorted)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
I used lodash for sorting and substring
method to divide date into month and day parts
const data = ['Mar20','Mar21', 'Mar01', 'Mar19','Apr18','Jun18','Jul18','May18','Jan19']
const monthMap = ["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"];
const sorted = _.sortBy(data, date => monthMap.indexOf(date.substring(0,3)), date => date.substring(3,5))
console.log(sorted)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
const data = ['Mar20','Mar21', 'Mar01', 'Mar19','Apr18','Jun18','Jul18','May18','Jan19']
const monthMap = ["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"];
const sorted = _.sortBy(data, date => monthMap.indexOf(date.substring(0,3)), date => date.substring(3,5))
console.log(sorted)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
const data = ['Mar20','Mar21', 'Mar01', 'Mar19','Apr18','Jun18','Jul18','May18','Jan19']
const monthMap = ["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"];
const sorted = _.sortBy(data, date => monthMap.indexOf(date.substring(0,3)), date => date.substring(3,5))
console.log(sorted)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
answered yesterday
Krzysztof KrzeszewskiKrzysztof Krzeszewski
7641110
7641110
add a comment |
add a comment |
let mth = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
mth.sort((itemA, itemB)=>{
let mthAstr = itemA.slice(0,3)+" 01 "+itemA.slice(3,5);
let mthA = new Date(mthAstr);
let mthBstr = itemB.slice(0,3)+" 01 "+itemB.slice(3,5);
let mthB = new Date(mthBstr);
if (mthA < mthB)
return -1;
if (mthA > mthB)
return 1;
return 0;
});
add a comment |
let mth = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
mth.sort((itemA, itemB)=>{
let mthAstr = itemA.slice(0,3)+" 01 "+itemA.slice(3,5);
let mthA = new Date(mthAstr);
let mthBstr = itemB.slice(0,3)+" 01 "+itemB.slice(3,5);
let mthB = new Date(mthBstr);
if (mthA < mthB)
return -1;
if (mthA > mthB)
return 1;
return 0;
});
add a comment |
let mth = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
mth.sort((itemA, itemB)=>{
let mthAstr = itemA.slice(0,3)+" 01 "+itemA.slice(3,5);
let mthA = new Date(mthAstr);
let mthBstr = itemB.slice(0,3)+" 01 "+itemB.slice(3,5);
let mthB = new Date(mthBstr);
if (mthA < mthB)
return -1;
if (mthA > mthB)
return 1;
return 0;
});
let mth = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
mth.sort((itemA, itemB)=>{
let mthAstr = itemA.slice(0,3)+" 01 "+itemA.slice(3,5);
let mthA = new Date(mthAstr);
let mthBstr = itemB.slice(0,3)+" 01 "+itemB.slice(3,5);
let mthB = new Date(mthBstr);
if (mthA < mthB)
return -1;
if (mthA > mthB)
return 1;
return 0;
});
answered yesterday
EdperEdper
7,21711842
7,21711842
add a comment |
add a comment |
You can try something like that :
var Array = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
Array.sort(function(a, b) {
a = [a.slice(0,3), ' 20', a.slice(3)].join('');
b = [b.slice(0,3), ' 20', b.slice(3)].join('')
return new Date() - new Date(b);
});
console.log(Array);
New contributor
add a comment |
You can try something like that :
var Array = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
Array.sort(function(a, b) {
a = [a.slice(0,3), ' 20', a.slice(3)].join('');
b = [b.slice(0,3), ' 20', b.slice(3)].join('')
return new Date() - new Date(b);
});
console.log(Array);
New contributor
add a comment |
You can try something like that :
var Array = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
Array.sort(function(a, b) {
a = [a.slice(0,3), ' 20', a.slice(3)].join('');
b = [b.slice(0,3), ' 20', b.slice(3)].join('')
return new Date() - new Date(b);
});
console.log(Array);
New contributor
You can try something like that :
var Array = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
Array.sort(function(a, b) {
a = [a.slice(0,3), ' 20', a.slice(3)].join('');
b = [b.slice(0,3), ' 20', b.slice(3)].join('')
return new Date() - new Date(b);
});
console.log(Array);
New contributor
edited yesterday
Enzo BLANCHON
653216
653216
New contributor
answered yesterday
KAMLESH NEHRAKAMLESH NEHRA
212
212
New contributor
New contributor
add a comment |
add a comment |
I would suggest that your data structure is unfortunate. Working with shared indices makes it much more difficult than a data structure that combines them, such as an array of objects, [{label: 'Mar19', value: 55}, ...]
.
If your data is coming from an upstream solution you cannot control, you can still manage this in your own work, converting before you use it. (And if you really have to converting back to pass to others.)
A common name for a function combining two arrays is zip
-- think of it acting like a zipper. This version uses one that takes a function to say how the paired elements should be combined. (Elsewhere such a function might be called zipWith
.)
Here sortArraysByDate
calls zip
passing a function that turns 'Mar19'
and 55
into {label: 'Mar19, value: 55, month: 'Mar', year: 19}
using dateFields
to extract the month and year from that label, then sorts these using the straightforward dateSort
const months = {Jan: 1, Feb: 2, Mar: 3, Apr: 4, May: 5, Jun: 6,
Jul: 7, Aug: 8, Sep: 9, Oct: 10, Nov: 11, Dec: 12}
const dateFields = (d) => ({
month: d.slice(0, 3),
year: Number(d.slice(3))
})
const dateSort = ({month: m1, year: y1}, {month: m2, year: y2}) =>
(y1 - y2) || (months[m1] - months[m2])
const zip = (fn, a1, a2) => a1.map((a, i) => fn(a, a2[i]))
const sortArraysByDate = (a1, a2) =>
zip((label, value) => ({label, value, ...dateFields(label)}), a1, a2)
.sort(dateSort)
.map(({label, value}) => ({label, value}))
const Array1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const Array2 = ['55','2','3','0','21','132']; //real values
const result = sortArraysByDate(Array1, Array2)
console.log(result)
The map
call in sortArraysByDate
is quite possibly not necessary. Without it, the resulting data includes extra year
and month
fields; that might not be an issue.
If you really need those updated arrays in the original format, you can just map
the result:
const newArray1 = result.map(o => o.label)
const newArray2 = result.map(o => o.value)
But I would urge you not to do this unless it's absolutely necessary. This structure is really useful. The paired arrays are much less so.
Also, if you need to combine more than two arrays, you could write a slightly more sophisticated version of zip
:
const zip = (fn, ...as) => as[0].map((_, i) => fn(...as.map(a => a[i])))
This takes a function on n
arguments, and n
arrays, and yields a new array containing the result of calling that function respectively on the successive items in each of the arrays.
add a comment |
I would suggest that your data structure is unfortunate. Working with shared indices makes it much more difficult than a data structure that combines them, such as an array of objects, [{label: 'Mar19', value: 55}, ...]
.
If your data is coming from an upstream solution you cannot control, you can still manage this in your own work, converting before you use it. (And if you really have to converting back to pass to others.)
A common name for a function combining two arrays is zip
-- think of it acting like a zipper. This version uses one that takes a function to say how the paired elements should be combined. (Elsewhere such a function might be called zipWith
.)
Here sortArraysByDate
calls zip
passing a function that turns 'Mar19'
and 55
into {label: 'Mar19, value: 55, month: 'Mar', year: 19}
using dateFields
to extract the month and year from that label, then sorts these using the straightforward dateSort
const months = {Jan: 1, Feb: 2, Mar: 3, Apr: 4, May: 5, Jun: 6,
Jul: 7, Aug: 8, Sep: 9, Oct: 10, Nov: 11, Dec: 12}
const dateFields = (d) => ({
month: d.slice(0, 3),
year: Number(d.slice(3))
})
const dateSort = ({month: m1, year: y1}, {month: m2, year: y2}) =>
(y1 - y2) || (months[m1] - months[m2])
const zip = (fn, a1, a2) => a1.map((a, i) => fn(a, a2[i]))
const sortArraysByDate = (a1, a2) =>
zip((label, value) => ({label, value, ...dateFields(label)}), a1, a2)
.sort(dateSort)
.map(({label, value}) => ({label, value}))
const Array1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const Array2 = ['55','2','3','0','21','132']; //real values
const result = sortArraysByDate(Array1, Array2)
console.log(result)
The map
call in sortArraysByDate
is quite possibly not necessary. Without it, the resulting data includes extra year
and month
fields; that might not be an issue.
If you really need those updated arrays in the original format, you can just map
the result:
const newArray1 = result.map(o => o.label)
const newArray2 = result.map(o => o.value)
But I would urge you not to do this unless it's absolutely necessary. This structure is really useful. The paired arrays are much less so.
Also, if you need to combine more than two arrays, you could write a slightly more sophisticated version of zip
:
const zip = (fn, ...as) => as[0].map((_, i) => fn(...as.map(a => a[i])))
This takes a function on n
arguments, and n
arrays, and yields a new array containing the result of calling that function respectively on the successive items in each of the arrays.
add a comment |
I would suggest that your data structure is unfortunate. Working with shared indices makes it much more difficult than a data structure that combines them, such as an array of objects, [{label: 'Mar19', value: 55}, ...]
.
If your data is coming from an upstream solution you cannot control, you can still manage this in your own work, converting before you use it. (And if you really have to converting back to pass to others.)
A common name for a function combining two arrays is zip
-- think of it acting like a zipper. This version uses one that takes a function to say how the paired elements should be combined. (Elsewhere such a function might be called zipWith
.)
Here sortArraysByDate
calls zip
passing a function that turns 'Mar19'
and 55
into {label: 'Mar19, value: 55, month: 'Mar', year: 19}
using dateFields
to extract the month and year from that label, then sorts these using the straightforward dateSort
const months = {Jan: 1, Feb: 2, Mar: 3, Apr: 4, May: 5, Jun: 6,
Jul: 7, Aug: 8, Sep: 9, Oct: 10, Nov: 11, Dec: 12}
const dateFields = (d) => ({
month: d.slice(0, 3),
year: Number(d.slice(3))
})
const dateSort = ({month: m1, year: y1}, {month: m2, year: y2}) =>
(y1 - y2) || (months[m1] - months[m2])
const zip = (fn, a1, a2) => a1.map((a, i) => fn(a, a2[i]))
const sortArraysByDate = (a1, a2) =>
zip((label, value) => ({label, value, ...dateFields(label)}), a1, a2)
.sort(dateSort)
.map(({label, value}) => ({label, value}))
const Array1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const Array2 = ['55','2','3','0','21','132']; //real values
const result = sortArraysByDate(Array1, Array2)
console.log(result)
The map
call in sortArraysByDate
is quite possibly not necessary. Without it, the resulting data includes extra year
and month
fields; that might not be an issue.
If you really need those updated arrays in the original format, you can just map
the result:
const newArray1 = result.map(o => o.label)
const newArray2 = result.map(o => o.value)
But I would urge you not to do this unless it's absolutely necessary. This structure is really useful. The paired arrays are much less so.
Also, if you need to combine more than two arrays, you could write a slightly more sophisticated version of zip
:
const zip = (fn, ...as) => as[0].map((_, i) => fn(...as.map(a => a[i])))
This takes a function on n
arguments, and n
arrays, and yields a new array containing the result of calling that function respectively on the successive items in each of the arrays.
I would suggest that your data structure is unfortunate. Working with shared indices makes it much more difficult than a data structure that combines them, such as an array of objects, [{label: 'Mar19', value: 55}, ...]
.
If your data is coming from an upstream solution you cannot control, you can still manage this in your own work, converting before you use it. (And if you really have to converting back to pass to others.)
A common name for a function combining two arrays is zip
-- think of it acting like a zipper. This version uses one that takes a function to say how the paired elements should be combined. (Elsewhere such a function might be called zipWith
.)
Here sortArraysByDate
calls zip
passing a function that turns 'Mar19'
and 55
into {label: 'Mar19, value: 55, month: 'Mar', year: 19}
using dateFields
to extract the month and year from that label, then sorts these using the straightforward dateSort
const months = {Jan: 1, Feb: 2, Mar: 3, Apr: 4, May: 5, Jun: 6,
Jul: 7, Aug: 8, Sep: 9, Oct: 10, Nov: 11, Dec: 12}
const dateFields = (d) => ({
month: d.slice(0, 3),
year: Number(d.slice(3))
})
const dateSort = ({month: m1, year: y1}, {month: m2, year: y2}) =>
(y1 - y2) || (months[m1] - months[m2])
const zip = (fn, a1, a2) => a1.map((a, i) => fn(a, a2[i]))
const sortArraysByDate = (a1, a2) =>
zip((label, value) => ({label, value, ...dateFields(label)}), a1, a2)
.sort(dateSort)
.map(({label, value}) => ({label, value}))
const Array1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const Array2 = ['55','2','3','0','21','132']; //real values
const result = sortArraysByDate(Array1, Array2)
console.log(result)
The map
call in sortArraysByDate
is quite possibly not necessary. Without it, the resulting data includes extra year
and month
fields; that might not be an issue.
If you really need those updated arrays in the original format, you can just map
the result:
const newArray1 = result.map(o => o.label)
const newArray2 = result.map(o => o.value)
But I would urge you not to do this unless it's absolutely necessary. This structure is really useful. The paired arrays are much less so.
Also, if you need to combine more than two arrays, you could write a slightly more sophisticated version of zip
:
const zip = (fn, ...as) => as[0].map((_, i) => fn(...as.map(a => a[i])))
This takes a function on n
arguments, and n
arrays, and yields a new array containing the result of calling that function respectively on the successive items in each of the arrays.
const months = {Jan: 1, Feb: 2, Mar: 3, Apr: 4, May: 5, Jun: 6,
Jul: 7, Aug: 8, Sep: 9, Oct: 10, Nov: 11, Dec: 12}
const dateFields = (d) => ({
month: d.slice(0, 3),
year: Number(d.slice(3))
})
const dateSort = ({month: m1, year: y1}, {month: m2, year: y2}) =>
(y1 - y2) || (months[m1] - months[m2])
const zip = (fn, a1, a2) => a1.map((a, i) => fn(a, a2[i]))
const sortArraysByDate = (a1, a2) =>
zip((label, value) => ({label, value, ...dateFields(label)}), a1, a2)
.sort(dateSort)
.map(({label, value}) => ({label, value}))
const Array1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const Array2 = ['55','2','3','0','21','132']; //real values
const result = sortArraysByDate(Array1, Array2)
console.log(result)
const months = {Jan: 1, Feb: 2, Mar: 3, Apr: 4, May: 5, Jun: 6,
Jul: 7, Aug: 8, Sep: 9, Oct: 10, Nov: 11, Dec: 12}
const dateFields = (d) => ({
month: d.slice(0, 3),
year: Number(d.slice(3))
})
const dateSort = ({month: m1, year: y1}, {month: m2, year: y2}) =>
(y1 - y2) || (months[m1] - months[m2])
const zip = (fn, a1, a2) => a1.map((a, i) => fn(a, a2[i]))
const sortArraysByDate = (a1, a2) =>
zip((label, value) => ({label, value, ...dateFields(label)}), a1, a2)
.sort(dateSort)
.map(({label, value}) => ({label, value}))
const Array1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const Array2 = ['55','2','3','0','21','132']; //real values
const result = sortArraysByDate(Array1, Array2)
console.log(result)
edited yesterday
answered yesterday
Scott SauyetScott Sauyet
20.8k22757
20.8k22757
add a comment |
add a comment |
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10
What have you tried so far?
– Anurag Srivastava
yesterday