CODEWITHSUNDEEP
rpermutationcorrelation
Paul
Paul

Reputation: 1117

Permutations of correlation coefficients

My question is on the permutation of correlation coefficients.

        A<-data.frame(A1=c(1,2,3,4,5),B1=c(6,7,8,9,10),C1=c(11,12,13,14,15 ))

        B<-data.frame(A2=c(6,7,7,10,11),B2=c(2,1,3,8,11),C2=c(1,5,16,7,8))

          cor(A,B)

          #           A2        B2       C2
          # A1 0.9481224 0.9190183 0.459588
          # B1 0.9481224 0.9190183 0.459588
          # C1 0.9481224 0.9190183 0.459588

I obtained this correlation and then wanted to perform permutation tests to check if the correlation still holds.

I did the permutation as follows:

              A<-as.vector(t(A))
              B<-as.vector(t(B))

     corperm <- function(A,B,1000) {
     # n is the number of permutations
     # x and y are the vectors to correlate
    obs = abs(cor(A,B))
    tmp = sapply(1:n,function(z) {abs(cor(sample(A,replace=TRUE),B))})
   return(1-sum(obs>tmp)/n)
     }

The result was 

         [1] 0.645

and using "cor.test"

cor.test(A,B)

Pearson's product-moment correlation

data:  A and B
t = 0.4753, df = 13, p-value = 0.6425
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
-0.4089539  0.6026075
sample estimates:
cor 
0.1306868

How could I draw a plot or a histogram to show the actual correlation and the permuted correlation value from the permuted data ???

Upvotes: 0

Views: 4140

Answers (2)

lebatsnok
lebatsnok

Reputation: 6459

first of all, you can't have done it exactly this ways as ...

> corperm = function(A,B,1000) {
Error: unexpected numeric constant in "corperm = function(A,B,1000"

The third argument has no name but it should have one! Perhaps you meant

> corperm <- function(A, B, n=1000) {
# etc

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. 

obs = abs(cor(A,B))
tmp = sapply(1:n,function(z) {abs(cor(sample(A,replace=TRUE),B))})
return(1-sum(obs>tmp)/n)

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).

For permutation test you sample without replacement and it makes no difference whether you do it for both A and B or only A.

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.

HTHAB

(edit) 

corperm <- function(x, y, N=1000, plot=FALSE){
    reps <- replicate(N, cor(sample(x), y))
    obs <- cor(x,y)
    p <- mean(reps > obs) # shortcut for sum(reps > obs)/N
    if(plot){
        hist(reps)
        abline(v=obs, col="red")
        }
     p
     }

Now you can use this on a single pair of variables:

corperm(A[,1], B[,1])

To apply it to all pairs, use for or mapply. for is easier to understand so I wouldn't insist in using mapply to get all possible pairs.

res <- matrix(NA, nrow=NCOL(A), ncol=NCOL(B))
for(iii in 1:3) for(jjj in 1:3) res[iii,jjj] <- corperm(A[,iii], B[,jjj], plot=FALSE)
rownames(res)<-names(A)
colnames(res) <- names(B)
print(res)

To make all histograms, use plot=TRUE above.

Upvotes: 2

cying Jack
cying Jack

Reputation: 11

I think there is not much significance to do permutation test for correlation analysis of two variants, because the cor.test()function offers "p.value" which has the same effect as permutation test.

Upvotes: 0

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