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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.

vector=RandomReal[1,100000];
f=RandomReal[1,{100001,520}];
pearson = Map[Correlation[f[[2 ;;, #]], vector] &, Range[5, Dimensions[f][[2]]]];
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Let's separate sampling from the computation.

BlockRandom[SeedRandom[123];
 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
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  • $\begingroup$ excellent.. that is exactly what I was looking for. thanks a lot! $\endgroup$ – preeti Nov 22 '12 at 16:22

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