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 of it, and finally permute back both foo
and bar
to the original order of foo
- that means inverse permutation:
foo <- c(1, 7, 3, 5, 2)
o <- order(foo)
foo <- foo[o] # Now foo is permuted, and sorted: foo == 1 2 3 5 7
bar = 2 * foo # bar == 2 4 6 10 14
And here should go your answer, such that we have the below desired final values:
foo == 1 7 3 5 2
bar == 2 14 6 10 4
How to do this?
Please simply do not reply: "you could do bar = 2 * foo
and not permute it". This is just a simple example. In some situations we have to sort foo
for efficiency (quick search over it) or something like that.
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
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