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
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