large numbers in K means

I am trying to find the kmeans of 1 column with 1048575 numbers rows

and I am getting the error below

Error using kmeans (line 277)

X must have more rows than the number of clusters.

Error in svia (line 18)

[idx, C, miro , D] = kmeans(bibi, K);

Can anyone assist me in finding a solution for this.

Thanks in advance

Tino

Answer 1

[Guillaume]

What's unclear about the error? You must specify less cluster than the number of rows in your matrix/vector. So make sure that K is less than size(bibi, 1) (or numel(bibi) if a vector).

[Walter Roberson]

It would be a problem for kmeans if bibi is a row vector instead of a column vector.

[Guillaume]

Not according to the doc: "If X is a numeric vector, then kmeans treats it as an n-by-1 data matrix, regardless of its orientation."

Answer 2

[John D'Errico]

x = rand(1048575,1);
size(x)
ans =
     1048575           1
     
[IDX, C] = kmeans(x, 10);
C
Warning: Failed to converge in 100 iterations. 
> In kmeans/loopBody (line 479)
  In internal.stats.parallel.smartForReduce (line 136)
  In kmeans (line 343) 
C =
      0.74809
     0.049586
      0.94968
       0.2486
      0.54773
      0.64779
      0.34794
      0.14901
      0.44755
      0.84915

It works fine, athough It seems starved for iterations. But that is trivially solved.

However, if I make a mistake, and pass in a ROW vector instead, then I get the same error that you did.

[IDX, C] = kmeans(x', 10);
Error using kmeans (line 277)
X must have more rows than the number of clusters.

[Guillaume]

That's countrary to the documentation which says: "If X is a numeric vector, then kmeans treats it as an n-by-1 data matrix, regardless of its orientation".

I don't have the required toolbox but if that's not what happens then it should be reported as a bug.

[John D'Errico]

That is indeed the statement, in R2019a. So it is indeed a bug, since the behavior runs contrary to the documentation.