Timeline for Within the context of a document term matrix, what exactly are x and y axis in kmeans clustering?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jan 20, 2017 at 15:08 | comment | added | Bryan Goggin | strictly positive values | |
| Jan 20, 2017 at 1:27 | comment | added | Han | Are both the x axis and y axis euclidian distances between the words in the document. How can there be negative values? | |
| Jul 2, 2016 at 20:23 | history | migrated | from stackoverflow.com (revisions) | ||
| Jun 26, 2016 at 20:54 | comment | added | Has QUIT--Anony-Mousse | It will mistakenly lead people to believe that k-means works on distance matrixes. It should be made clear that k-means expects to see the raw data, not a distance matrix. Otherwise you don't get e.g. a centroid. | |
| Jun 26, 2016 at 14:38 | comment | added | Bryan Goggin |
Thanks for down-voting my answer for a completely off topic reason. I was answering Doug's question, as asked, about the article's code and the use of dist function. If you have a problem with how the article uses k-means then take that up with them.
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| Jun 26, 2016 at 8:23 | comment | added | Has QUIT--Anony-Mousse | kneans should be run on the data matrix. Running it on the distance matrix has a different semantic, does not yield meaningful centers, and is O(n^2), i.e. much more expensive and will not scale. | |
| Jun 24, 2016 at 3:30 | vote | accept | Doug Fir | ||
| Jun 24, 2016 at 3:29 | comment | added | Doug Fir | Can't underline enough how helpful this has been. I get it now - thank you so much. You're last comment above really nails where I was getting confused | |
| Jun 24, 2016 at 2:26 | comment | added | Bryan Goggin |
In the first example (the quick brown fox, etc) there are 6 possible dimensions. However, for the dimension "the" both vectors have the value of 1. So the formula would be the square root of (1-1)^2 ("the") + (1-0)^2 ("quick")+ ... +(0-1)^2 ("dog") which is sqrt(0+1+1+1+1+1) = sqrt(5) = 2.236068.
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| Jun 24, 2016 at 2:17 | comment | added | Doug Fir | Thanks, this is very helpful. A small follow up, if you will. In the first matrix you showed distance between V1 and V2 dist(words) as 2.23. Looking at the distance formula at the top of the answer, how did that come to be? I tried √4 ( which is 2) minus √3 (which is 1.73) = 0.27. √4 and 3 because the formula √x1 +xn. How would one calculate the 2.23 distance? Again, really grateful for your time in explaining this so far | |
| Jun 24, 2016 at 1:49 | comment | added | Bryan Goggin | I edited my answer. Hopefully that helps. | |
| Jun 24, 2016 at 0:36 | comment | added | Doug Fir | Thanks for taking the time to answer. "The function dist computed the euclidean distance between vectors". At the point d is created d <- dist(t(dtmss), method="euclidian"), what is the vector(s) at this point in time? I looked at head(d) str(d) and it's a class object. The first value in there on my actual data is 68.19091 and the label is "someword". What does that mean? What is 68.19091 in this context? My cognitive abilities are just not sharp enough | |
| Jun 24, 2016 at 0:19 | history | answered | Bryan Goggin | CC BY-SA 3.0 |