How can I make the following line of code run faster? Is there a way to do this calculation as a matrix vs vector than vector vs vector?

In the code below, f is a matrix with one more row than vector. f has many columns and vector has only one column.

pearson = Map[Correlation[f[[2 ;;, #]], vector] &, Range[5, Dimensions[f][[2]]]];

1 Answer 1


Let's separate sampling from the computation.

 vector = RandomReal[1, 100000];
 f = RandomReal[1, {100001, 520}];

Here is doing it your way:

In[18]:= AbsoluteTiming[
 pearson2 = 
   Map[Correlation[f[[2 ;;, #]], vector] &, 
    Range[5, Dimensions[f][[2]]]];]

Out[18]= {5.865713, Null}

Of course you lose of standartization of vector. The solution is to use Correlation of two matrices:

In[19]:= AbsoluteTiming[
 pearson1 = Correlation[f[[2 ;;, 5 ;;]], Transpose[{vector}]];

Out[19]= {3.229262, Null}

Results are equal, of course:

In[20]:= Norm[pearson1 - pearson2]

Out[20]= 7.37439*10^-16
  • $\begingroup$ excellent.. that is exactly what I was looking for. thanks a lot! $\endgroup$
    – preeti
    Nov 22, 2012 at 16:22

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