Generating inverse permutation

Question

Suppose we are given a vector foo and we have to temporarily permute it (sort or re-order it), compute some vector bar on the base

Answer 1

To recover the original order, use order(o):

> (foo[o]*2)[order(o)]
[1]  2 14  6 10  4

Answer 2

This will work, because [order(o)] inverts the action of [o]:

foo <- c(1, 7, 3, 5, 2)
o <- order(foo)

(foo[o]*2)[order(o)]
# [1]  2 14  6 10  4

To show that it works more generally:

testAlgorithm <- function(foo) {
    o <- order(foo)
    identical(foo, foo[o][order(o)])
}

x <- c(1, 7, 3, 5, 2)
y <- c(1,2,5,7,4, 99, 88, 3, 0)
z <- sample(1000)                  ## (No ties)
zz <- sample(1000, replace=TRUE)   ## (Many ties)
all(sapply(list(x,y,z,zz), testAlgorithm))
[1] TRUE