1
$\begingroup$

I have kind of 'distance matrix', every element of which is represented by number of interactions (contacts) between pair of positions on single path-like(curve) object.

It is stated, that number of interactions (contacts) is linearly proportional to the actual distance between two positions in space. # Closer they are, more interactions is observed.

So my 'distance matrix' does not actually contains Euclidean distances, but since these values are proportional to Euclidean distances, may I still use Classical (Torgerson's) metric MDS? If not, why?

Thing is, I do not completely understand why is it important to have Euclidean distances for Classical (Torgerson's) metric MDS and when exactly do I use non-metric ones.

What happens if this contacts are not linearly proportional to distances, which MDS I can use in this case?

$\endgroup$
  • $\begingroup$ Closer they are, more interactions is observed That looks like similarity, not distance. In what sence then values are proportional to Euclidean distances? $\endgroup$ – ttnphns Oct 6 '16 at 10:54
  • $\begingroup$ I do not completely understand why is it important to have Euclidean distances for Classical (Torgerson's) metric MDS That MDS method does not demand that dissimilarities be euclidean. However, any metric MDS transforms dissimilarities to disparaties just linearly. It follows then that, given that map is done in euclidean space, usually the fit will be better if these and those are kin - i.e. input dissimilarities are euclidean distances too. It is preference, not demand. See also similar question stats.stackexchange.com/q/208190/3277. $\endgroup$ – ttnphns Oct 6 '16 at 11:08
  • $\begingroup$ ttnphns, thank you for your answer. You are right, it is not distance. But if we multiply this 'similarities' by some constant (unknown) we would get real distances. That is what I mean by 'proportional'. $\endgroup$ – skkap Oct 7 '16 at 6:19
  • $\begingroup$ As I understand, as long as I only care about resulting distances being relatively close to the input distances and do not care about actual values of distances - I can use Classical (Torgerson's) metric MDS for that. Is it right? What I still cannot understand, in what case I would choose non-metric MDS, seems like metric one dose the job for any set of input data $\endgroup$ – skkap Oct 7 '16 at 6:22
  • $\begingroup$ Well, about so, roughly speaking. You see, TorgMDS or other metric MDS will try to reconstruct the dissimilarities as they are (but for linear transform). But what if these dissimilarities are noneuclidean (so that negative eigenvalues in matrix B are pronounced)? Reconstruction will then be not only imperfect - it may likely be nonuniformly imperfect: rank order of map distances may be far from that of dissimilarities. $\endgroup$ – ttnphns Oct 7 '16 at 8:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.