Hi Xueming,
Try
(1:length(bo))[rank(bo)]
In a function the above would be
f <- function(x){
N <- length(x)
(1:N)[rank(x)]
}
f(bo)
# [1] 2 6 3 5 1 4
HTH,
Jorge.-
On Sat, Apr 26, 2014 at 7:54 PM, xmliu1...@gmail.com <xmliu1...@gmail.com>wrote:
> Hi,
>
> could anybody help me to find a fast way to fix the following question?
>
> Given a verctor of length N, for example bo = [3 8 4 6 1 5],
> I want to drive a vector whose elements are 1, 2, ..., N and the order of
> elements is the same as that in verctor bo.
> In this example, the result is supposed to be bt = [2 6 3 5 1 4].
>
> I used the following code to solove this:
>
> bo <- c(3, 8, 4, 6, 1, 5)
> N <- length(bo)
> bt <- rep(0, N)
> M <- max(bo)
> temp <- bo
> for(i in 1 : N)
> {
> min <- M
> i_min <- 0
>
> for(j in 1 : N)
> {
> if(min >= temp[j])
> {
> min <- temp[j]
> i_min <-j
> }
> }
> bt[i_min] <- i
> temp[i_min] <- M+ 1
> }
> > bt
> [1] 2 6 3 5 1 4
>
> However, the time complexity is O(N2).
> When N is larger than 1000000, it takes too much time.
> Is there any faster way to fix it?
>
> best
> Xueming
>
>
>
> xmliu1...@gmail.com
> [[alternative HTML version deleted]]
>
> ______________________________________________
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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