100 items 100 baskets divisor association analysis problem
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I have the following exercise question:
Suppose there are 100 items, numbered 1 to 100, and also 100 baskets, also numbered 1 to 100. Item i is in basket b if and only if i divides b with no remainder. Thus, item 1 is in all the baskets, item 2 is in all fifty of the even-numbered baskets, etc. for example Basket 12 consists of items {1, 2, 3, 4, 6, 12}
Given this, I'm trying to solve the following 3 questions:
(a) If the support threshold is 5, which items are frequent?
(b) what is the confidence of the following association rules?
(1) {5, 7} → 2.
(2) {2, 3, 4}→ 5.
So I have tried to approach this question but I'm unsure if I'm doing it correctly as I was just introduced to this concept recently.
For question a, the conclusion I reached was everything up to 20 would be the answer, because those are the only numbers that can appear more than 5 times in the 100 baskets.
For question b1, I determine there are 50 baskets for 2, 10 baskets for 5 given 2, and 7 baskets of 7 given 2, the only common basket here I could find was 70, which is one basket which means the confidence factors is 1 in 50 which is 2%?
for b2, I did it similarly, noting the 20 baskets for 5, and finding the elements of 2 given 5, 3 given 5, and 4 given 5, and finding the common element in those, I only found the number 60 which would be 1/20 which is a confidence factor of 5%?
I just want to know if I'm doing this correctly, and if I'm even approaching it in the right way, if not, which way I should be approaching these types of questions?
data-mining market-basket-analysis
asked 3 hours ago
deshawndeshawn
1
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$begingroup$
I have the following exercise question:
Suppose there are 100 items, numbered 1 to 100, and also 100 baskets, also numbered 1 to 100. Item i is in basket b if and only if i divides b with no remainder. Thus, item 1 is in all the baskets, item 2 is in all fifty of the even-numbered baskets, etc. for example Basket 12 consists of items {1, 2, 3, 4, 6, 12}
Given this, I'm trying to solve the following 3 questions:
(a) If the support threshold is 5, which items are frequent?
(b) what is the confidence of the following association rules?
(1) {5, 7} → 2.
(2) {2, 3, 4}→ 5.
So I have tried to approach this question but I'm unsure if I'm doing it correctly as I was just introduced to this concept recently.
For question a, the conclusion I reached was everything up to 20 would be the answer, because those are the only numbers that can appear more than 5 times in the 100 baskets.
For question b1, I determine there are 50 baskets for 2, 10 baskets for 5 given 2, and 7 baskets of 7 given 2, the only common basket here I could find was 70, which is one basket which means the confidence factors is 1 in 50 which is 2%?
for b2, I did it similarly, noting the 20 baskets for 5, and finding the elements of 2 given 5, 3 given 5, and 4 given 5, and finding the common element in those, I only found the number 60 which would be 1/20 which is a confidence factor of 5%?
I just want to know if I'm doing this correctly, and if I'm even approaching it in the right way, if not, which way I should be approaching these types of questions?
data-mining market-basket-analysis
asked 3 hours ago
deshawndeshawn
1
New contributor
deshawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
I have the following exercise question:
Suppose there are 100 items, numbered 1 to 100, and also 100 baskets, also numbered 1 to 100. Item i is in basket b if and only if i divides b with no remainder. Thus, item 1 is in all the baskets, item 2 is in all fifty of the even-numbered baskets, etc. for example Basket 12 consists of items {1, 2, 3, 4, 6, 12}
Given this, I'm trying to solve the following 3 questions:
(a) If the support threshold is 5, which items are frequent?
(b) what is the confidence of the following association rules?
(1) {5, 7} → 2.
(2) {2, 3, 4}→ 5.
So I have tried to approach this question but I'm unsure if I'm doing it correctly as I was just introduced to this concept recently.
For question a, the conclusion I reached was everything up to 20 would be the answer, because those are the only numbers that can appear more than 5 times in the 100 baskets.
For question b1, I determine there are 50 baskets for 2, 10 baskets for 5 given 2, and 7 baskets of 7 given 2, the only common basket here I could find was 70, which is one basket which means the confidence factors is 1 in 50 which is 2%?
for b2, I did it similarly, noting the 20 baskets for 5, and finding the elements of 2 given 5, 3 given 5, and 4 given 5, and finding the common element in those, I only found the number 60 which would be 1/20 which is a confidence factor of 5%?
I just want to know if I'm doing this correctly, and if I'm even approaching it in the right way, if not, which way I should be approaching these types of questions?
data-mining market-basket-analysis
asked 3 hours ago
deshawndeshawn
1
New contributor
deshawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have the following exercise question:
Suppose there are 100 items, numbered 1 to 100, and also 100 baskets, also numbered 1 to 100. Item i is in basket b if and only if i divides b with no remainder. Thus, item 1 is in all the baskets, item 2 is in all fifty of the even-numbered baskets, etc. for example Basket 12 consists of items {1, 2, 3, 4, 6, 12}
Given this, I'm trying to solve the following 3 questions:
(a) If the support threshold is 5, which items are frequent?
(b) what is the confidence of the following association rules?
(1) {5, 7} → 2.
(2) {2, 3, 4}→ 5.
So I have tried to approach this question but I'm unsure if I'm doing it correctly as I was just introduced to this concept recently.
For question a, the conclusion I reached was everything up to 20 would be the answer, because those are the only numbers that can appear more than 5 times in the 100 baskets.
For question b1, I determine there are 50 baskets for 2, 10 baskets for 5 given 2, and 7 baskets of 7 given 2, the only common basket here I could find was 70, which is one basket which means the confidence factors is 1 in 50 which is 2%?
for b2, I did it similarly, noting the 20 baskets for 5, and finding the elements of 2 given 5, 3 given 5, and 4 given 5, and finding the common element in those, I only found the number 60 which would be 1/20 which is a confidence factor of 5%?
I just want to know if I'm doing this correctly, and if I'm even approaching it in the right way, if not, which way I should be approaching these types of questions?
data-mining market-basket-analysis
data-mining market-basket-analysis
asked 3 hours ago
deshawndeshawn
1
New contributor
deshawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
deshawndeshawn
1
New contributor
deshawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
deshawndeshawn
1
New contributor
deshawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
deshawndeshawn
1
asked 3 hours ago
deshawndeshawn
1
1
New contributor
deshawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
deshawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
deshawn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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