0
$\begingroup$

I am looking for a metric for comparing gene count tables. These are long columns of data (a few millions genes by a few dozen samples), with all non-negative entries, about 90% of which are zeros. The goal is to compare the performance of several tools/algorithms that these tables originate from, by comparing the resulting tables among themselves or with the expected counts (in a case of sumulates data). In principle, one compares on a sample-by-sample basis, but comparing different samples might be also of interest, e.g., to filter out spurious correlations.

What I am using now is Spearman rank coefficient, taking account for the fact that some entries have identical ranks (certainly the zeros). I am looking for an approach more adapted to this setting (and preferably robust to outliers) and will appreciate suggestions.

$\endgroup$
1
$\begingroup$

The first idea that comes to mind is a similarity measure such as cosine. It's often used with sparse vectors (text represented as vectors over the vocabulary). There are many options for distance/similarity measures:

  • Basic set measures like overlap coefficient or Jaccard
  • Entropy-based measures such as KL divergence.
  • ...
$\endgroup$
2
  • $\begingroup$ Thank you for answering! I am not sure, if Jaccard has good generalizations to non-boolean vectors... As for Entropy-based measures (like mutual information), they are not very robust (although I forgot to mention this is my question). $\endgroup$ – Roger Vadim Jan 20 at 13:51
  • 1
    $\begingroup$ @Vadim there are variants of Jaccard for non-boolean vectors, but in general all these measures will give pretty similar results: cosine, euclidean distance, Jaccard, etc. The question of which measure makes sense in your task depends on the data, but for that you're the expert ;) For example Spearman correlation focuses on the rank between the values in the vector, so it doesn't take into account the values themselves (e.g. [2,100,67] would obtain perfect Spearman correlation with [20,22,21]) $\endgroup$ – Erwan Jan 20 at 15:12

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.