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How do I identify clusters that match on categorical data?
The 2019 Stack Overflow Developer Survey Results Are InHow do I cluster data that is a mix of text & categorical data?How to create clusters of position data?Looking for an algorithm that correctly clusters visually separable clustersRecognize a grammar in a sequence of fuzzy tokensFeeding R agnes object into cutreeSample selection through clusteringHow can I handle missing categorical data that has significance?How to identify clusters after multiple runs?Differences between applying KMeans over PCA and applying PCA over KMeansAlgorithm for purely categorical data
$begingroup$
I am seeking some directions for a proper path to research the solve for this problem:
My company made all our employees take a "StrengthFinders" test, which results in every employee being assigned their top five (ordered) "strengths" from a possible list of 34 strengths. We have 500 employees. I am supposed to identify all the employees that match each other for the same 5 strengths (order not important), and also for employees that match each other for 4 out of 5 strengths (again, order doesn't matter). I could potentially have multiple groups matching on different sets of strengths, e.g.:
Group 1: Billy, Sally, Michael have strengths A, H, I, K, Z
Group 2: Bobby and Suzy have strengths A, B, L, S, W
For the case where strengths match for 4 out of 5, I might have the same people from Group 1 above, plus Joe, whose strengths are A, H, M, K, Z; and
Seth, whose strengths are A, H, G, K, Z. I would expect more groupings for the case of 4 out of 5 than the 5 out of 5 case.
The strengths are categorical in nature, so what I've read so far has largely revolved around clustering of continuous numerical variables.
I am looking for an algorithmic way to identify clusters and the members of those clusters for this situation. I think I could do this brute force by repeatedly sorting data in Excel, but I'm confident that a better way must exist, and I ask you to point me in that direction. Thank you.
clustering categorical-data
asked Mar 28 at 16:10
wackojacko1997wackojacko1997
83
$endgroup$
add a comment |
$begingroup$
I am seeking some directions for a proper path to research the solve for this problem:
My company made all our employees take a "StrengthFinders" test, which results in every employee being assigned their top five (ordered) "strengths" from a possible list of 34 strengths. We have 500 employees. I am supposed to identify all the employees that match each other for the same 5 strengths (order not important), and also for employees that match each other for 4 out of 5 strengths (again, order doesn't matter). I could potentially have multiple groups matching on different sets of strengths, e.g.:
Group 1: Billy, Sally, Michael have strengths A, H, I, K, Z
Group 2: Bobby and Suzy have strengths A, B, L, S, W
For the case where strengths match for 4 out of 5, I might have the same people from Group 1 above, plus Joe, whose strengths are A, H, M, K, Z; and
Seth, whose strengths are A, H, G, K, Z. I would expect more groupings for the case of 4 out of 5 than the 5 out of 5 case.
The strengths are categorical in nature, so what I've read so far has largely revolved around clustering of continuous numerical variables.
I am looking for an algorithmic way to identify clusters and the members of those clusters for this situation. I think I could do this brute force by repeatedly sorting data in Excel, but I'm confident that a better way must exist, and I ask you to point me in that direction. Thank you.
clustering categorical-data
asked Mar 28 at 16:10
wackojacko1997wackojacko1997
83
$endgroup$
add a comment |
$begingroup$
I am seeking some directions for a proper path to research the solve for this problem:
My company made all our employees take a "StrengthFinders" test, which results in every employee being assigned their top five (ordered) "strengths" from a possible list of 34 strengths. We have 500 employees. I am supposed to identify all the employees that match each other for the same 5 strengths (order not important), and also for employees that match each other for 4 out of 5 strengths (again, order doesn't matter). I could potentially have multiple groups matching on different sets of strengths, e.g.:
Group 1: Billy, Sally, Michael have strengths A, H, I, K, Z
Group 2: Bobby and Suzy have strengths A, B, L, S, W
For the case where strengths match for 4 out of 5, I might have the same people from Group 1 above, plus Joe, whose strengths are A, H, M, K, Z; and
Seth, whose strengths are A, H, G, K, Z. I would expect more groupings for the case of 4 out of 5 than the 5 out of 5 case.
The strengths are categorical in nature, so what I've read so far has largely revolved around clustering of continuous numerical variables.
I am looking for an algorithmic way to identify clusters and the members of those clusters for this situation. I think I could do this brute force by repeatedly sorting data in Excel, but I'm confident that a better way must exist, and I ask you to point me in that direction. Thank you.
clustering categorical-data
asked Mar 28 at 16:10
wackojacko1997wackojacko1997
83
$endgroup$
I am seeking some directions for a proper path to research the solve for this problem:
My company made all our employees take a "StrengthFinders" test, which results in every employee being assigned their top five (ordered) "strengths" from a possible list of 34 strengths. We have 500 employees. I am supposed to identify all the employees that match each other for the same 5 strengths (order not important), and also for employees that match each other for 4 out of 5 strengths (again, order doesn't matter). I could potentially have multiple groups matching on different sets of strengths, e.g.:
Group 1: Billy, Sally, Michael have strengths A, H, I, K, Z
Group 2: Bobby and Suzy have strengths A, B, L, S, W
For the case where strengths match for 4 out of 5, I might have the same people from Group 1 above, plus Joe, whose strengths are A, H, M, K, Z; and
Seth, whose strengths are A, H, G, K, Z. I would expect more groupings for the case of 4 out of 5 than the 5 out of 5 case.
The strengths are categorical in nature, so what I've read so far has largely revolved around clustering of continuous numerical variables.
I am looking for an algorithmic way to identify clusters and the members of those clusters for this situation. I think I could do this brute force by repeatedly sorting data in Excel, but I'm confident that a better way must exist, and I ask you to point me in that direction. Thank you.
clustering categorical-data
clustering categorical-data
asked Mar 28 at 16:10
wackojacko1997wackojacko1997
83
asked Mar 28 at 16:10
wackojacko1997wackojacko1997
83
asked Mar 28 at 16:10
wackojacko1997wackojacko1997
83
asked Mar 28 at 16:10
wackojacko1997wackojacko1997
83
asked Mar 28 at 16:10
wackojacko1997wackojacko1997
83
83
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You have just 500 data points...
Excel of course is the worst possible tool though.
Anyway, build a dictionary. Put everybody in there 6 times: 1 with all five strengths, and 5 times with one strength omitted. Then you can easily identify the largest groups, and you can also perform various completion operations easily: if you have identified a group with strengths A B C D E, you can add all that have ABCD etc. using the dictionary.
answered Mar 28 at 19:30
Anony-MousseAnony-Mousse
5,165625
$endgroup$
$begingroup$
@wackojacko1997 I think this is the solution. Noting that if each key is a string, strengths need to be sorted alphabetically to place ABCD and CABD in the same group.
$endgroup$
– Esmailian
Mar 30 at 22:25
$begingroup$
While I need to think about the coding with the Dictionary a little bit, this answer does makes sense to me. When I look at @QuantifiedMe's answer (which I perceive as essentially the same thing, but using prime numbers), I think I can use that even without coding (directly in Excel). I'm inclined to mark this the answer, though, as the more general approach.
$endgroup$
– wackojacko1997
Apr 3 at 20:20
$begingroup$
Yes, he is suggesting the same thing, using a prime factor coding instead of a string coding.
$endgroup$
– Anony-Mousse
Apr 3 at 21:16
$begingroup$
Okay, thank you. I am accepting this answer as the general case then.
$endgroup$
– wackojacko1997
Apr 4 at 1:18
add a comment |
$begingroup$
Assign each of the 34 traits a unique prime number.
Compute the product of the 5 prime numbers of each person.
Compare every person's value to find a match.
To find 4 matching traits out of 5, make the product from 4 of the 5 traits. You'll find 5 unique combinations. 1*2*3*4 , 1*2*3*5, 1*2*4*5, 2*3*4*5, and 1*3*4*5. Compare the values again to find the 4th degree matches.
edited Apr 2 at 1:18
Stephen Rauch♦
1,52551330
answered Apr 1 at 18:55
QuantifiedMeQuantifiedMe
111
$endgroup$
$begingroup$
I like this approach for the simplicity and the ease of employing it.
$endgroup$
– wackojacko1997
Apr 3 at 20:21
add a comment |
$begingroup$
You can try k-modes or ROCK which are specifically made to work with categorical values. I don't have experience with them myself but you can look at:
Implementations:
- K-Modes
- ROCK
answered Mar 28 at 17:24
Simon LarssonSimon Larsson
734114
$endgroup$
add a comment |
$begingroup$
If I were you, I would approach this as an Association Mining problem. You most likely will have to pre-process your data for this type of analysis, but it shouldn't be too difficult.
Here is an example in R
answered Mar 28 at 20:45
Rajat S. SubediRajat S. Subedi
1
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have just 500 data points...
Excel of course is the worst possible tool though.
Anyway, build a dictionary. Put everybody in there 6 times: 1 with all five strengths, and 5 times with one strength omitted. Then you can easily identify the largest groups, and you can also perform various completion operations easily: if you have identified a group with strengths A B C D E, you can add all that have ABCD etc. using the dictionary.
answered Mar 28 at 19:30
Anony-MousseAnony-Mousse
5,165625
$endgroup$
$begingroup$
@wackojacko1997 I think this is the solution. Noting that if each key is a string, strengths need to be sorted alphabetically to place ABCD and CABD in the same group.
$endgroup$
– Esmailian
Mar 30 at 22:25
$begingroup$
While I need to think about the coding with the Dictionary a little bit, this answer does makes sense to me. When I look at @QuantifiedMe's answer (which I perceive as essentially the same thing, but using prime numbers), I think I can use that even without coding (directly in Excel). I'm inclined to mark this the answer, though, as the more general approach.
$endgroup$
– wackojacko1997
Apr 3 at 20:20
$begingroup$
Yes, he is suggesting the same thing, using a prime factor coding instead of a string coding.
$endgroup$
– Anony-Mousse
Apr 3 at 21:16
$begingroup$
Okay, thank you. I am accepting this answer as the general case then.
$endgroup$
– wackojacko1997
Apr 4 at 1:18
add a comment |
$begingroup$
You have just 500 data points...
Excel of course is the worst possible tool though.
Anyway, build a dictionary. Put everybody in there 6 times: 1 with all five strengths, and 5 times with one strength omitted. Then you can easily identify the largest groups, and you can also perform various completion operations easily: if you have identified a group with strengths A B C D E, you can add all that have ABCD etc. using the dictionary.
answered Mar 28 at 19:30
Anony-MousseAnony-Mousse
5,165625
$endgroup$
$begingroup$
@wackojacko1997 I think this is the solution. Noting that if each key is a string, strengths need to be sorted alphabetically to place ABCD and CABD in the same group.
$endgroup$
– Esmailian
Mar 30 at 22:25
$begingroup$
While I need to think about the coding with the Dictionary a little bit, this answer does makes sense to me. When I look at @QuantifiedMe's answer (which I perceive as essentially the same thing, but using prime numbers), I think I can use that even without coding (directly in Excel). I'm inclined to mark this the answer, though, as the more general approach.
$endgroup$
– wackojacko1997
Apr 3 at 20:20
$begingroup$
Yes, he is suggesting the same thing, using a prime factor coding instead of a string coding.
$endgroup$
– Anony-Mousse
Apr 3 at 21:16
$begingroup$
Okay, thank you. I am accepting this answer as the general case then.
$endgroup$
– wackojacko1997
Apr 4 at 1:18
add a comment |
$begingroup$
You have just 500 data points...
Excel of course is the worst possible tool though.
Anyway, build a dictionary. Put everybody in there 6 times: 1 with all five strengths, and 5 times with one strength omitted. Then you can easily identify the largest groups, and you can also perform various completion operations easily: if you have identified a group with strengths A B C D E, you can add all that have ABCD etc. using the dictionary.
answered Mar 28 at 19:30
Anony-MousseAnony-Mousse
5,165625
$endgroup$
You have just 500 data points...
Excel of course is the worst possible tool though.
Anyway, build a dictionary. Put everybody in there 6 times: 1 with all five strengths, and 5 times with one strength omitted. Then you can easily identify the largest groups, and you can also perform various completion operations easily: if you have identified a group with strengths A B C D E, you can add all that have ABCD etc. using the dictionary.
answered Mar 28 at 19:30
Anony-MousseAnony-Mousse
5,165625
answered Mar 28 at 19:30
Anony-MousseAnony-Mousse
5,165625
answered Mar 28 at 19:30
Anony-MousseAnony-Mousse
5,165625
answered Mar 28 at 19:30
Anony-MousseAnony-Mousse
5,165625
5,165625
$begingroup$
@wackojacko1997 I think this is the solution. Noting that if each key is a string, strengths need to be sorted alphabetically to place ABCD and CABD in the same group.
$endgroup$
– Esmailian
Mar 30 at 22:25
$begingroup$
While I need to think about the coding with the Dictionary a little bit, this answer does makes sense to me. When I look at @QuantifiedMe's answer (which I perceive as essentially the same thing, but using prime numbers), I think I can use that even without coding (directly in Excel). I'm inclined to mark this the answer, though, as the more general approach.
$endgroup$
– wackojacko1997
Apr 3 at 20:20
$begingroup$
Yes, he is suggesting the same thing, using a prime factor coding instead of a string coding.
$endgroup$
– Anony-Mousse
Apr 3 at 21:16
$begingroup$
Okay, thank you. I am accepting this answer as the general case then.
$endgroup$
– wackojacko1997
Apr 4 at 1:18
add a comment |
$begingroup$
@wackojacko1997 I think this is the solution. Noting that if each key is a string, strengths need to be sorted alphabetically to place ABCD and CABD in the same group.
$endgroup$
– Esmailian
Mar 30 at 22:25
$begingroup$
While I need to think about the coding with the Dictionary a little bit, this answer does makes sense to me. When I look at @QuantifiedMe's answer (which I perceive as essentially the same thing, but using prime numbers), I think I can use that even without coding (directly in Excel). I'm inclined to mark this the answer, though, as the more general approach.
$endgroup$
– wackojacko1997
Apr 3 at 20:20
$begingroup$
Yes, he is suggesting the same thing, using a prime factor coding instead of a string coding.
$endgroup$
– Anony-Mousse
Apr 3 at 21:16
$begingroup$
Okay, thank you. I am accepting this answer as the general case then.
$endgroup$
– wackojacko1997
Apr 4 at 1:18
$begingroup$
@wackojacko1997 I think this is the solution. Noting that if each key is a string, strengths need to be sorted alphabetically to place ABCD and CABD in the same group.
$endgroup$
– Esmailian
Mar 30 at 22:25
$begingroup$
@wackojacko1997 I think this is the solution. Noting that if each key is a string, strengths need to be sorted alphabetically to place ABCD and CABD in the same group.
$endgroup$
– Esmailian
Mar 30 at 22:25
$begingroup$
While I need to think about the coding with the Dictionary a little bit, this answer does makes sense to me. When I look at @QuantifiedMe's answer (which I perceive as essentially the same thing, but using prime numbers), I think I can use that even without coding (directly in Excel). I'm inclined to mark this the answer, though, as the more general approach.
$endgroup$
– wackojacko1997
Apr 3 at 20:20
$begingroup$
While I need to think about the coding with the Dictionary a little bit, this answer does makes sense to me. When I look at @QuantifiedMe's answer (which I perceive as essentially the same thing, but using prime numbers), I think I can use that even without coding (directly in Excel). I'm inclined to mark this the answer, though, as the more general approach.
$endgroup$
– wackojacko1997
Apr 3 at 20:20
$begingroup$
Yes, he is suggesting the same thing, using a prime factor coding instead of a string coding.
$endgroup$
– Anony-Mousse
Apr 3 at 21:16
$begingroup$
Yes, he is suggesting the same thing, using a prime factor coding instead of a string coding.
$endgroup$
– Anony-Mousse
Apr 3 at 21:16
$begingroup$
Okay, thank you. I am accepting this answer as the general case then.
$endgroup$
– wackojacko1997
Apr 4 at 1:18
$begingroup$
Okay, thank you. I am accepting this answer as the general case then.
$endgroup$
– wackojacko1997
Apr 4 at 1:18
add a comment |
$begingroup$
Assign each of the 34 traits a unique prime number.
Compute the product of the 5 prime numbers of each person.
Compare every person's value to find a match.
To find 4 matching traits out of 5, make the product from 4 of the 5 traits. You'll find 5 unique combinations. 1*2*3*4 , 1*2*3*5, 1*2*4*5, 2*3*4*5, and 1*3*4*5. Compare the values again to find the 4th degree matches.
edited Apr 2 at 1:18
Stephen Rauch♦
1,52551330
answered Apr 1 at 18:55
QuantifiedMeQuantifiedMe
111
$endgroup$
$begingroup$
I like this approach for the simplicity and the ease of employing it.
$endgroup$
– wackojacko1997
Apr 3 at 20:21
add a comment |
$begingroup$
Assign each of the 34 traits a unique prime number.
Compute the product of the 5 prime numbers of each person.
Compare every person's value to find a match.
To find 4 matching traits out of 5, make the product from 4 of the 5 traits. You'll find 5 unique combinations. 1*2*3*4 , 1*2*3*5, 1*2*4*5, 2*3*4*5, and 1*3*4*5. Compare the values again to find the 4th degree matches.
edited Apr 2 at 1:18
Stephen Rauch♦
1,52551330
answered Apr 1 at 18:55
QuantifiedMeQuantifiedMe
111
$endgroup$
$begingroup$
I like this approach for the simplicity and the ease of employing it.
$endgroup$
– wackojacko1997
Apr 3 at 20:21
add a comment |
$begingroup$
Assign each of the 34 traits a unique prime number.
Compute the product of the 5 prime numbers of each person.
Compare every person's value to find a match.
To find 4 matching traits out of 5, make the product from 4 of the 5 traits. You'll find 5 unique combinations. 1*2*3*4 , 1*2*3*5, 1*2*4*5, 2*3*4*5, and 1*3*4*5. Compare the values again to find the 4th degree matches.
edited Apr 2 at 1:18
Stephen Rauch♦
1,52551330
answered Apr 1 at 18:55
QuantifiedMeQuantifiedMe
111
$endgroup$
Assign each of the 34 traits a unique prime number.
Compute the product of the 5 prime numbers of each person.
Compare every person's value to find a match.
To find 4 matching traits out of 5, make the product from 4 of the 5 traits. You'll find 5 unique combinations. 1*2*3*4 , 1*2*3*5, 1*2*4*5, 2*3*4*5, and 1*3*4*5. Compare the values again to find the 4th degree matches.
edited Apr 2 at 1:18
Stephen Rauch♦
1,52551330
answered Apr 1 at 18:55
QuantifiedMeQuantifiedMe
111
edited Apr 2 at 1:18
Stephen Rauch♦
1,52551330
edited Apr 2 at 1:18
Stephen Rauch♦
1,52551330
edited Apr 2 at 1:18
Stephen Rauch♦
1,52551330
1,52551330
answered Apr 1 at 18:55
QuantifiedMeQuantifiedMe
111
answered Apr 1 at 18:55
QuantifiedMeQuantifiedMe
111
answered Apr 1 at 18:55
QuantifiedMeQuantifiedMe
111
111
$begingroup$
I like this approach for the simplicity and the ease of employing it.
$endgroup$
– wackojacko1997
Apr 3 at 20:21
add a comment |
$begingroup$
I like this approach for the simplicity and the ease of employing it.
$endgroup$
– wackojacko1997
Apr 3 at 20:21
$begingroup$
I like this approach for the simplicity and the ease of employing it.
$endgroup$
– wackojacko1997
Apr 3 at 20:21
$begingroup$
I like this approach for the simplicity and the ease of employing it.
$endgroup$
– wackojacko1997
Apr 3 at 20:21
add a comment |
$begingroup$
You can try k-modes or ROCK which are specifically made to work with categorical values. I don't have experience with them myself but you can look at:
Implementations:
- K-Modes
- ROCK
answered Mar 28 at 17:24
Simon LarssonSimon Larsson
734114
$endgroup$
add a comment |
$begingroup$
You can try k-modes or ROCK which are specifically made to work with categorical values. I don't have experience with them myself but you can look at:
Implementations:
- K-Modes
- ROCK
answered Mar 28 at 17:24
Simon LarssonSimon Larsson
734114
$endgroup$
add a comment |
$begingroup$
You can try k-modes or ROCK which are specifically made to work with categorical values. I don't have experience with them myself but you can look at:
Implementations:
- K-Modes
- ROCK
answered Mar 28 at 17:24
Simon LarssonSimon Larsson
734114
$endgroup$
You can try k-modes or ROCK which are specifically made to work with categorical values. I don't have experience with them myself but you can look at:
Implementations:
- K-Modes
- ROCK
answered Mar 28 at 17:24
Simon LarssonSimon Larsson
734114
answered Mar 28 at 17:24
Simon LarssonSimon Larsson
734114
answered Mar 28 at 17:24
Simon LarssonSimon Larsson
734114
answered Mar 28 at 17:24
Simon LarssonSimon Larsson
734114
734114
add a comment |
add a comment |
$begingroup$
If I were you, I would approach this as an Association Mining problem. You most likely will have to pre-process your data for this type of analysis, but it shouldn't be too difficult.
Here is an example in R
answered Mar 28 at 20:45
Rajat S. SubediRajat S. Subedi
1
$endgroup$
add a comment |
$begingroup$
If I were you, I would approach this as an Association Mining problem. You most likely will have to pre-process your data for this type of analysis, but it shouldn't be too difficult.
Here is an example in R
answered Mar 28 at 20:45
Rajat S. SubediRajat S. Subedi
1
$endgroup$
add a comment |
$begingroup$
If I were you, I would approach this as an Association Mining problem. You most likely will have to pre-process your data for this type of analysis, but it shouldn't be too difficult.
Here is an example in R
answered Mar 28 at 20:45
Rajat S. SubediRajat S. Subedi
1
$endgroup$
If I were you, I would approach this as an Association Mining problem. You most likely will have to pre-process your data for this type of analysis, but it shouldn't be too difficult.
Here is an example in R
answered Mar 28 at 20:45
Rajat S. SubediRajat S. Subedi
1
answered Mar 28 at 20:45
Rajat S. SubediRajat S. Subedi
1
answered Mar 28 at 20:45
Rajat S. SubediRajat S. Subedi
1
answered Mar 28 at 20:45
Rajat S. SubediRajat S. Subedi
1
1
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