I did a thoroughly for the answer of this question but I found nothing.
Assume that I have two samples of paired observations with sample size 4. My goal is to perform an exact permutation set to compare the sample means, and to carry out that procedure I need to generate all the possible pair permutations, i.e. I have to generate 16 different permutations (2 possibilities for the first pair, two for the second... and so on).
That is, if my samples are:
#samples
x <- rnorm(4,0,1)
y <- rnorm(4,5,1)
#statistic over the original samples
dif0 <- mean(x) - mean(y)
I can generate all the possible permutations using 4 nested "for" loops, as follows:
d1 < c()
d2 <- c()
samp_ab <- data.frame("x"=x,"y"=y)
ind <- 1:2
for (i1 in 1:2)
{
d1[1] <- samp_ab[1,ind[i1]]
d2[1] <- samp_ab[1,ind[-i1]]
for (i2 in 1:2) {
d1[2] <- samp_ab[2,ind[i2]]
d2[2] <- samp_ab[2,ind[-i2]]
for (i3 in 1:2) {
d1[3] <- samp_ab[3,ind[i3]]
d2[3] <- samp_ab[3,ind[-i3]]
for (i4 in 1:2) { #loop4
d1[4] <- samp_ab[4,ind[i4]]
d2[4] <- samp_ab[4,ind[-i4]]
#compute the statistic
k = k + 1 #counter
S[k] <- mean(d1) - mean(d2)
}
}
}
}
There is any possibility for this code to be simplified? I thought about a function that "writes" the nested loops and then executes it... but I have no idea how to do it. I would like to obtain a general procedure for paired samples (or even for combining three elements) without writting many nested loops. For instance, I had to do it with n=10, which resulted in 10 loops... just to create 1024 (2^10) permutations.
Many thanks in advance.
