(This post is a repost of a question I posted yesterday (now deleted), but I've tried to scale back volume of words and simplify what I'm asking)

I'm hoping to get some help interpreting a kmeans script and output I have created. This is in the context of text analysis. I created this script after reading several articles online on text analysis. I have linked to some of them them below.

Sample r script and corpus of text data I will refer to throughout this post:

library(tm) # for text mining

## make a example corpus
# make a df of documents a to i
a <- "dog dog cat carrot"
b <- "phone cat dog"
c <- "phone book dog"
d <- "cat book trees"
e <- "phone orange"
f <- "phone circles dog"
g <- "dog cat square"
h <- "dog trees cat"
i <- "phone carrot cat"
j <- c(a,b,c,d,e,f,g,h,i)
x <- data.frame(j)    

# turn x into a document term matrix (dtm)
docs <- Corpus(DataframeSource(x))
dtm <- DocumentTermMatrix(docs)

# create distance matrix for clustering
m <- as.matrix(dtm)
d <- dist(m, method = "euclidean")

# kmeans clustering
kfit <- kmeans(d, 2)
#plot – need library cluster
clusplot(m, kfit$cluster)

That's it for the script. Below are the output of some of the variables in the script:

Here's x, the data frame x that was transformed into a corpus:

    1 dog dog cat carrot
    2      phone cat dog
    3     phone book dog
    4     cat book trees
    5       phone orange
    6  phone circles dog
    7     dog cat square
    8      dog trees cat
    9   phone carrot cat

An here's the resulting document term matrix dtm:

    > inspect(dtm)
<<DocumentTermMatrix (documents: 9, terms: 9)>>
Non-/sparse entries: 26/55
Sparsity           : 68%
Maximal term length: 7
Weighting          : term frequency (tf)

Docs book carrot cat circles dog orange phone square trees
   1    0      1   1       0   2      0     0      0     0
   2    0      0   1       0   1      0     1      0     0
   3    1      0   0       0   1      0     1      0     0
   4    1      0   1       0   0      0     0      0     1
   5    0      0   0       0   0      1     1      0     0
   6    0      0   0       1   1      0     1      0     0
   7    0      0   1       0   1      0     0      1     0
   8    0      0   1       0   1      0     0      0     1
   9    0      1   1       0   0      0     1      0     0

And here is the distance matrix d

> d
         1        2        3        4        5        6        7        8
2 1.732051                                                               
3 2.236068 1.414214                                                      
4 2.645751 2.000000 2.000000                                             
5 2.828427 1.732051 1.732051 2.236068                                    
6 2.236068 1.414214 1.414214 2.449490 1.732051                           
7 1.732051 1.414214 2.000000 2.000000 2.236068 2.000000                  
8 1.732051 1.414214 2.000000 1.414214 2.236068 2.000000 1.414214         
9 2.236068 1.414214 2.000000 2.000000 1.732051 2.000000 2.000000 2.000000

Here is the result, kfit:

> kfit
K-means clustering with 2 clusters of sizes 5, 4

Cluster means:
         1        2        3        4        5        6        7        8        9
1 2.253736 1.194938 1.312096 2.137112 1.385641 1.312096 1.930056 1.930056 1.429253
2 1.527463 1.640119 2.059017 1.514991 2.384158 2.171389 1.286566 1.140119 2.059017

Clustering vector:
1 2 3 4 5 6 7 8 9 
2 1 1 2 1 1 2 2 1 

Within cluster sum of squares by cluster:
[1] 13.3468 12.3932
 (between_SS / total_SS =  29.5 %)

Available components:

[1] "cluster"      "centers"      "totss"        "withinss"     "tot.withinss" "betweenss"    "size"         "iter"        
[9] "ifault"      

Here is the resulting plot: enter image description here

I have several questions about this:

  1. In calculating my distance matrix d (a parameter used in kfit calculation) I did this: d <- dist(m, method = "euclidean"). Another article I encountered did this: d <- dist(t(m), method = "euclidean"). Then, separately on a SO question I posted recently someone commented "kmeans should be run on the data matrix, not on the distance matrix!". Presumably they mean kmeans() should take m instead of d as input. Of these 3 variations which/who is "right". Or, assuming all are valid in one way or another, which would be the conventional way to go in setting up an initial baseline model?
  2. As I understand it, when kmeans function is called on d, what happens is that 2 random centroids are chosen (in this case k=2). Then r will look at each row in d and determine which documents are closest to which centroid. Based on the matrix d above, what would that actually look like? For example if the first random centroid was 1.5 and the second was 2, then how would document 4 be assigned? In the matrix d doc4 is 2.645751 2.000000 2.000000 so (in r) mean(c(2.645751,2.000000,2.000000)) = 2.2 so in the first iteration of kmeans in this example doc4 is assigned to the cluster with value 2 since it's closer to that than to 1.5. After this the mean of the cluster is reclauculated as a new centroid and the docs reassigned where appropriate. Is this right or have I completely missed the point?
  3. In the kfit output above what is "cluster means"? E.g., Doc3 cluster 1 has a value of 1.312096. What is this number in this context? [edit, since looking at this again a few days after posting I can see that it's the distance of each document to the final cluster centers. So the lowest number (closest) is what determines which cluster each doc is assigned].
  4. In the kfit output above, "clustering vector" looks like it's just what cluster each doc was assigned to. OK.
  5. In the kfit output above, "Within cluster sum of squares by cluster". What is that? 13.3468 12.3932 (between_SS / total_SS = 29.5 %). A measure of the variance within each cluster, presumably meaning a lower number implies a stronger grouping as opposed to a more sparse one. Is that a fair statement? What about the percentage given 29.5%. What's that? Is 29.5% "good". Would a lower or higher number be preferred in any instance of kmeans? If I experimented with different numbers of k, what would I be looking for to determine if the increasing/decreasing number of clusters has helped or hindered the analysis?
  6. The screenshot of the plot goes from -1 to 3. What is being measured here? As opposed to education and earnings, height and weight, what is the number 3 at the top of the scale in this context?
  7. In the plot the message "These two components explain 50.96% of the point variability" I already found some detailed info here (in case anyone else comes across this post - just for completeness of understanding kmeans output wanted to add here.).

Here's some of the articles I read that helped me to create this script:

  • 3
    $\begingroup$ If downvoting please leave a comment letting me know why so I can try to amend $\endgroup$
    – Doug Fir
    Commented Jun 29, 2016 at 2:16
  • $\begingroup$ 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. $\endgroup$
    – ttnphns
    Commented Jun 29, 2016 at 8:12
  • $\begingroup$ Hi @ttnphns kfit is a variable of kfit <- kmeans(d, 2) in the example script I made. There's no actual kfit function $\endgroup$
    – Doug Fir
    Commented Jun 29, 2016 at 8:46
  • $\begingroup$ 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. $\endgroup$
    – ttnphns
    Commented Jun 29, 2016 at 8:47
  • $\begingroup$ (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. $\endgroup$
    – ttnphns
    Commented Jun 29, 2016 at 8:47

1 Answer 1


To understand how the kmeans() function works, you need to read the documentation and/or inspect the underlying code. That said, I am sure it does not take a distance matrix without even bothering. You could write your own function to do k-means clustering from a distance matrix, but it would be an awful hassle.

The k-means algorithm is meant to operate over a data matrix, not a distance matrix. It only minimizes squared Euclidean distances (cf. Why does k-means clustering algorithm use only Euclidean distance metric?). It is only sensible when you could have Euclidean distances as a meaningful distance metric. This has always been the case since the algorithm was invented, but few people seem to be aware of this, with the result that k-means is probably the most mis-used algorithm in machine learning.

Euclidean distance doesn't make any sense for sparse categorical data (text mining), so I wouldn't even try anything like this. You first need to figure out what distance metric is appropriate for your data (@ttnphns explains some possible measures here: What is the optimal distance function for individuals when attributes are nominal?). Then you can compute the distance matrix and use a clustering algorithm that can operate over one (e.g., k-medians / PAM, various hierarchical algorithms, etc.).

  • $\begingroup$ Thank you @gung for the pretty assertive answer, this is what I was seeking since I was a bit uncertain and went down a rabbit hole of internet research which seemed to confuse me further. I will look into k-medians and the other algorithms you mentioned. $\endgroup$
    – Doug Fir
    Commented Jul 11, 2016 at 3:03
  • 2
    $\begingroup$ Also, it seems there's a gap in the blogoshpere around this since more than one article use the distance matrix as an input to kmeans, perhaps one blogger copied from another and that had a cascading effect to several misinformed articles. $\endgroup$
    – Doug Fir
    Commented Jul 11, 2016 at 3:05

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