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  1. PO
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    <p>first of all, you can't have done it exactly this ways as ...</p> <pre><code>&gt; corperm = function(A,B,1000) { Error: unexpected numeric constant in "corperm = function(A,B,1000" </code></pre> <p>The third argument has no name but it should have one! Perhaps you meant</p> <pre><code>&gt; corperm &lt;- function(A, B, n=1000) { # etc </code></pre> <p>Then you need to think about what do you want to achieve. Initially you have two data sets with 3 variables and then you collapse them into two vectors and compute a correlation between the permuted vectors. Why does it make sense? The structure of permuted data set should be the same as the original data set. </p> <pre><code>obs = abs(cor(A,B)) tmp = sapply(1:n,function(z) {abs(cor(sample(A,replace=TRUE),B))}) return(1-sum(obs&gt;tmp)/n) </code></pre> <p>Why do you use replace=TRUE here? This makes sense if you would like to have bootstrap CI-s but (a) it'd be better to use a dedicated function then e.g boot from package boot, and (B) you'd need to do the same with B, i.e. sample(B, replace=TRUE).</p> <p>For permutation test you sample without replacement and it makes no difference whether you do it for both A and B or only A.</p> <p>And how to get the histogram? Well, hist(tmp) would draw you a histogram of the permuted values, and obs is absolute value of the observed correlation. </p> <p>HTHAB</p> <p>(edit) </p> <pre><code>corperm &lt;- function(x, y, N=1000, plot=FALSE){ reps &lt;- replicate(N, cor(sample(x), y)) obs &lt;- cor(x,y) p &lt;- mean(reps &gt; obs) # shortcut for sum(reps &gt; obs)/N if(plot){ hist(reps) abline(v=obs, col="red") } p } </code></pre> <p>Now you can use this on a single pair of variables:</p> <pre><code>corperm(A[,1], B[,1]) </code></pre> <p>To apply it to all pairs, use <code>for</code> or <code>mapply</code>. <code>for</code> is easier to understand so I wouldn't insist in using <code>mapply</code> to get all possible pairs.</p> <pre><code>res &lt;- matrix(NA, nrow=NCOL(A), ncol=NCOL(B)) for(iii in 1:3) for(jjj in 1:3) res[iii,jjj] &lt;- corperm(A[,iii], B[,jjj], plot=FALSE) rownames(res)&lt;-names(A) colnames(res) &lt;- names(B) print(res) </code></pre> <p>To make all histograms, use plot=TRUE above.</p>
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    1. POPermutations of correlation coefficients
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    1. USlebatsnok
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    1. POPermutations of correlation coefficients
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    1. CO@ lebatsnok : Could you please suggest how to go about permuting these correlations and obtain the histogram to show that the correlations still holds.I did the above following a tutorial.
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      1. USPaul
    2. COhi Paul. Could you state your exact research question? Are you interested in testing the significance of correlations between each pair of variables in data frames A and B? Or is it something else? Or what do you mean by "correlation still holds"?
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    3. CO@ lebastsnok : Yes, testing the significance of correlation between the pairs of variables in both dataframes.
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