Timeline for Was it as valid to perform k-means on a distance matrix as on data matrix (text mining data)?
Current License: CC BY-SA 3.0
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| May 23, 2017 at 12:39 | history | edited | CommunityBot |
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
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| Jul 11, 2016 at 3:01 | vote | accept | Doug Fir | ||
| Jul 11, 2016 at 3:01 | vote | accept | Doug Fir | ||
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| Jul 10, 2016 at 14:26 | answer | added | gung - Reinstate Monica | timeline score: 7 | |
| Jul 10, 2016 at 14:03 | history | edited | gung - Reinstate Monica | CC BY-SA 3.0 |
light editing
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| Jul 10, 2016 at 1:47 | review | Close votes | |||
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| Jul 6, 2016 at 2:24 | history | edited | Doug Fir | CC BY-SA 3.0 |
I learnt something new since posting the question and added it. Prefer to leave up in case anyone else comes across in the future.
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| Jun 30, 2016 at 9:27 | comment | added | ttnphns |
I doubt I can help you, I'm not R user and don't know how its kmeans function work and what data it can and cannot take. Please contact with its documentation and perhaps with the articles' authors. If you find information that it is OK to input d matrix in it, please let us here know :-), 'couse it's intriguing.
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| Jun 30, 2016 at 9:22 | comment | added | Doug Fir | Also, I both of the articles I linked to use distance matrix as an input for kmeans which is really confusing since others are advising not to and to just use the dtm | |
| Jun 30, 2016 at 9:14 | comment | added | Doug Fir | @ttnphns so I should run kfit <- kmeans(dtm, 2). I don't even need the matrix m right? | |
| Jun 29, 2016 at 10:38 | history | tweeted | twitter.com/StackStats/status/748103365893173248 | ||
| Jun 29, 2016 at 9:29 | history | edited | ttnphns | CC BY-SA 3.0 |
edited title
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| Jun 29, 2016 at 8:47 | comment | added | ttnphns | (cont.) But as said before, my analysis (b) treated its input data as data matrix, not as distance matrix. Therefore, your analysis did it so too. I conclude that your K-means function is not designed to take in distance matrices (or you failed to play such an option if it does exist), it is standard K-means requiring data matrices. It is a mistake to try to feed it with a distance matrix. Your clustering results were therefore erroneous. So was my conclusion. | |
| Jun 29, 2016 at 8:47 | comment | added | ttnphns |
What I've done in SPSS with your data was this. I ran K-means with inputs (a) your document term matrix tdm; (b) with your euclidean distance matrix d. SPSS's K-means treats input always as cases X variables data and clusters the cases. As initial centres, I input in both analyses the output centres of your analysis - cluster means. Results: in analysis (b), but not in (a), I got final centres identical to the input centres. That means that K-means in (b) could not further improve the cluster centres, which implies that analysis (b) coincides with the k-means analysis done by you.
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| Jun 29, 2016 at 8:46 | comment | added | Doug Fir | Hi @ttnphns kfit is a variable of kfit <- kmeans(d, 2) in the example script I made. There's no actual kfit function | |
| Jun 29, 2016 at 8:12 | comment | added | ttnphns |
Where is kfit function documentation available? I've looked inside the tm library cran.r-project.org/web/packages/tm/tm.pdf and found no kfit there.
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| Jun 29, 2016 at 5:30 | history | edited | Doug Fir | CC BY-SA 3.0 |
Slightly modified wording
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| Jun 29, 2016 at 4:00 | history | edited | Doug Fir | CC BY-SA 3.0 |
spelling
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| Jun 29, 2016 at 2:16 | comment | added | Doug Fir | If downvoting please leave a comment letting me know why so I can try to amend | |
| Jun 29, 2016 at 1:14 | history | asked | Doug Fir | CC BY-SA 3.0 |