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I am calculating Spearman correlations multiple times between individual data vectors that are not necessarily of the same length. So for example, X1=[1,2,3], Y1=[2,3,1] and X2=[10,2,3,4,5], Y2=[2,3,1,1,1], correlating X1 with Y1 and X2 with Y2. I now want to retrieve an average correlation coefficient.

I am aware that a common approach is a Fisher transformation on the individual coefficients; averaging the transformed values; and a back transformation.

However, I have two issues:

  1. Some of my correlation coefficients are 1 where the Fisher transformation is not defined. Is there some common way of handling that case?

  2. Do I need to cope somehow with the different vector lengths? I found some short information about that in the following paper, but wonder whether this is necessary or there is another approach.

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My answer is mostly more questions:

Correlations based on very small samples are naturally not very trustworthy. In particular, a Spearman correlation of 1 could just mean that your sample is not large enough to catch the uncertainty in the data.

If it makes sense to average the correlations, why not pool the data and calculate an overall correlation?

Why do you think you need an average correlation any way? Why not just report a distribution of correlations?

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  • $\begingroup$ I suppose if all else fails the OP could use the median. $\endgroup$ – mdewey Jun 20 '16 at 11:36
  • $\begingroup$ Pooling the data does not really make sense as the samples are heterogenous and this would result in non-independent samples. One idea could be to use a mixed-effects model though with a random intercept. $\endgroup$ – fsociety Jun 20 '16 at 12:33
  • $\begingroup$ I'd reverse your point: if heterogeneity is par for the course, what use or meaning would there be in an average correlation? $\endgroup$ – Nick Cox Jun 20 '16 at 14:01
  • $\begingroup$ With heterogeneity I do not mean that there is heterogeneity in correlation, but rather a general clustering effect that warrants to not treat the whole data as equal. Having something like X1=[100,100] and X2=[1,2] and correlate both with a vector like Y1=[1,1] should be treated independently, not pooled. After some thinking, I think a mixed effect model is suitable. $\endgroup$ – fsociety Jun 21 '16 at 6:52

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