How to check if all elements of 1 list are in the *same quantity* and in any order, in the list2?

How to check if all elements of 1 list are in the same quantity and in any order, in the list2?

I know its a very common question at first, but I haven't found one that specific. (If you do, please tell me.) And all ways I found didnt work for me.
I need to check if all elements of list 1 appears in the same amount in the list2.

Ex :

#If list1 = [2,2,2,6]    

# and list2 =[2,6,2,5,2,4]     

#then all list1 are in list2.

#If list2 = [2,6] then all list1 are not in list2.

i'm trying this way :

list1 = [6,2]



import itertools



for i in itertools.product((2,4,5,1), repeat=3) :

    asd = i[0] + i[1]

    asd2= i[1] + i[2]



    list2 = [asd, asd2]

    if all(elem in list2  for elem in list1):

        print (i,list2)

It works when the elements are not repeated in the list1, like [1,2]. But when they are repeated, all repeated elements is beeing counted as only 1 : [2,2,2] its beeing understanded as [2]. Or so i think.

edited 4 hours ago

petezurich

3,76581936

asked 5 hours ago

Vitor Oliveira

475

1

After reading both questions, it does not look like a duplicate to me. This question cares about quantity, but not order. The other one cares about order, but not quantity.

– gilch
4 hours ago

add a comment |

Ex :

#If list1 = [2,2,2,6]    

# and list2 =[2,6,2,5,2,4]     

#then all list1 are in list2.

#If list2 = [2,6] then all list1 are not in list2.

i'm trying this way :

list1 = [6,2]



import itertools



for i in itertools.product((2,4,5,1), repeat=3) :

    asd = i[0] + i[1]

    asd2= i[1] + i[2]



    list2 = [asd, asd2]

    if all(elem in list2  for elem in list1):

        print (i,list2)

edited 4 hours ago

petezurich

3,76581936

asked 5 hours ago

Vitor Oliveira

475

1

After reading both questions, it does not look like a duplicate to me. This question cares about quantity, but not order. The other one cares about order, but not quantity.

– gilch
4 hours ago

add a comment |

Ex :

#If list1 = [2,2,2,6]    

# and list2 =[2,6,2,5,2,4]     

#then all list1 are in list2.

#If list2 = [2,6] then all list1 are not in list2.

i'm trying this way :

list1 = [6,2]



import itertools



for i in itertools.product((2,4,5,1), repeat=3) :

    asd = i[0] + i[1]

    asd2= i[1] + i[2]



    list2 = [asd, asd2]

    if all(elem in list2  for elem in list1):

        print (i,list2)

edited 4 hours ago

petezurich

3,76581936

asked 5 hours ago

Vitor Oliveira

475

Ex :

#If list1 = [2,2,2,6]    

# and list2 =[2,6,2,5,2,4]     

#then all list1 are in list2.

#If list2 = [2,6] then all list1 are not in list2.

i'm trying this way :

list1 = [6,2]



import itertools



for i in itertools.product((2,4,5,1), repeat=3) :

    asd = i[0] + i[1]

    asd2= i[1] + i[2]



    list2 = [asd, asd2]

    if all(elem in list2  for elem in list1):

        print (i,list2)

python python-3.x

edited 4 hours ago

petezurich

3,76581936

asked 5 hours ago

Vitor Oliveira

475

edited 4 hours ago

petezurich

3,76581936

asked 5 hours ago

Vitor Oliveira

475

edited 4 hours ago

petezurich

3,76581936

edited 4 hours ago

petezurich

3,76581936

edited 4 hours ago

petezurich

3,76581936

asked 5 hours ago

Vitor Oliveira

475

asked 5 hours ago

Vitor Oliveira

475

asked 5 hours ago

Vitor Oliveira

475

1

After reading both questions, it does not look like a duplicate to me. This question cares about quantity, but not order. The other one cares about order, but not quantity.

– gilch
4 hours ago

add a comment |

1

After reading both questions, it does not look like a duplicate to me. This question cares about quantity, but not order. The other one cares about order, but not quantity.

– gilch
4 hours ago

After reading both questions, it does not look like a duplicate to me. This question cares about quantity, but not order. The other one cares about order, but not quantity.

– gilch
4 hours ago

add a comment |

2 Answers
2

active

oldest

votes

Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.

from collections import Counter



def a_all_in_b(a, b):

    """True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""

    return Counter(a).items() <= Counter(b).items()

Note that Counter only works on hashable elements because it's a subclass of dict.

edited 4 hours ago

answered 4 hours ago

gilch

4,3101716

1

that's slick @gilch

– modesitt
4 hours ago

Does this work for something like a_all_in_b([1], [1, 1])?

– Tomothy32
4 hours ago

@Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".

– gilch
4 hours ago

@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.

– Tomothy32
4 hours ago

1

Also it should be <=, not <.

– user2357112
4 hours ago

|
show 3 more comments

Modify this answer to Checking if list is a sublist to check for equality of occurences:

from collections import Counter 



list1 = [2,2,2,6]    

list2 =[2,6,2,5,2,4]



def same_amount(a,b):

    c1 = Counter(a)

    c2 = Counter(b)



    for key,value in c1.items():

        if c2[key] != value:

            return False

    return True





print(same_amount(list1,list2))

print(same_amount(list1 + [2],list2))

Output:

True

False

There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.

answered 4 hours ago

Patrick Artner

26k62544

It works too. Thanks !

– Vitor Oliveira
3 hours ago

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.

from collections import Counter



def a_all_in_b(a, b):

    """True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""

    return Counter(a).items() <= Counter(b).items()

Note that Counter only works on hashable elements because it's a subclass of dict.

edited 4 hours ago

answered 4 hours ago

gilch

4,3101716

1

that's slick @gilch

– modesitt
4 hours ago

Does this work for something like a_all_in_b([1], [1, 1])?

– Tomothy32
4 hours ago

@Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".

– gilch
4 hours ago

@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.

– Tomothy32
4 hours ago

1

Also it should be <=, not <.

– user2357112
4 hours ago

|
show 3 more comments

Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.

from collections import Counter



def a_all_in_b(a, b):

    """True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""

    return Counter(a).items() <= Counter(b).items()

Note that Counter only works on hashable elements because it's a subclass of dict.

edited 4 hours ago

answered 4 hours ago

gilch

4,3101716

1

that's slick @gilch

– modesitt
4 hours ago

Does this work for something like a_all_in_b([1], [1, 1])?

– Tomothy32
4 hours ago

@Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".

– gilch
4 hours ago

@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.

– Tomothy32
4 hours ago

1

Also it should be <=, not <.

– user2357112
4 hours ago

|
show 3 more comments

Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.

from collections import Counter



def a_all_in_b(a, b):

    """True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""

    return Counter(a).items() <= Counter(b).items()

Note that Counter only works on hashable elements because it's a subclass of dict.

edited 4 hours ago

answered 4 hours ago

gilch

4,3101716

Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.

from collections import Counter



def a_all_in_b(a, b):

    """True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""

    return Counter(a).items() <= Counter(b).items()

Note that Counter only works on hashable elements because it's a subclass of dict.

edited 4 hours ago

answered 4 hours ago

gilch

4,3101716

edited 4 hours ago

answered 4 hours ago

gilch

4,3101716

answered 4 hours ago

gilch

4,3101716

answered 4 hours ago

gilch

4,3101716

1

that's slick @gilch

– modesitt
4 hours ago

Does this work for something like a_all_in_b([1], [1, 1])?

– Tomothy32
4 hours ago

@Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".

– gilch
4 hours ago

@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.

– Tomothy32
4 hours ago

1

Also it should be <=, not <.

– user2357112
4 hours ago

|
show 3 more comments

1

that's slick @gilch

– modesitt
4 hours ago

Does this work for something like a_all_in_b([1], [1, 1])?

– Tomothy32
4 hours ago

@Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".

– gilch
4 hours ago

@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.

– Tomothy32
4 hours ago

1

Also it should be <=, not <.

– user2357112
4 hours ago

that's slick @gilch

– modesitt
4 hours ago

Does this work for something like a_all_in_b([1], [1, 1])?

– Tomothy32
4 hours ago

@Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".

– gilch
4 hours ago

@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.

– Tomothy32
4 hours ago

Also it should be <=, not <.

– user2357112
4 hours ago

|
show 3 more comments

Modify this answer to Checking if list is a sublist to check for equality of occurences:

from collections import Counter 



list1 = [2,2,2,6]    

list2 =[2,6,2,5,2,4]



def same_amount(a,b):

    c1 = Counter(a)

    c2 = Counter(b)



    for key,value in c1.items():

        if c2[key] != value:

            return False

    return True





print(same_amount(list1,list2))

print(same_amount(list1 + [2],list2))

Output:

True

False

There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.

answered 4 hours ago

Patrick Artner

26k62544

It works too. Thanks !

– Vitor Oliveira
3 hours ago

add a comment |

Modify this answer to Checking if list is a sublist to check for equality of occurences:

from collections import Counter 



list1 = [2,2,2,6]    

list2 =[2,6,2,5,2,4]



def same_amount(a,b):

    c1 = Counter(a)

    c2 = Counter(b)



    for key,value in c1.items():

        if c2[key] != value:

            return False

    return True





print(same_amount(list1,list2))

print(same_amount(list1 + [2],list2))

Output:

True

False

There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.

answered 4 hours ago

Patrick Artner

26k62544

It works too. Thanks !

– Vitor Oliveira
3 hours ago

add a comment |

Modify this answer to Checking if list is a sublist to check for equality of occurences:

from collections import Counter 



list1 = [2,2,2,6]    

list2 =[2,6,2,5,2,4]



def same_amount(a,b):

    c1 = Counter(a)

    c2 = Counter(b)



    for key,value in c1.items():

        if c2[key] != value:

            return False

    return True





print(same_amount(list1,list2))

print(same_amount(list1 + [2],list2))

Output:

True

False

There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.

answered 4 hours ago

Patrick Artner

26k62544

Modify this answer to Checking if list is a sublist to check for equality of occurences:

from collections import Counter 



list1 = [2,2,2,6]    

list2 =[2,6,2,5,2,4]



def same_amount(a,b):

    c1 = Counter(a)

    c2 = Counter(b)



    for key,value in c1.items():

        if c2[key] != value:

            return False

    return True





print(same_amount(list1,list2))

print(same_amount(list1 + [2],list2))

Output:

True

False

There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.

answered 4 hours ago

Patrick Artner

26k62544

answered 4 hours ago

Patrick Artner

26k62544

answered 4 hours ago

Patrick Artner

26k62544

answered 4 hours ago

Patrick Artner

26k62544

It works too. Thanks !

– Vitor Oliveira
3 hours ago

add a comment |

It works too. Thanks !

– Vitor Oliveira
3 hours ago

It works too. Thanks !

– Vitor Oliveira
3 hours ago

add a comment |

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