Look into the rank function.
If there are duplicated values in the input vector its 'ties' argument
says how to deal with them. If there are ties I think your algorithm
puts the last one in the first position, e.g., it maps
c(101,101,101,102,102) to c(3,2,1,5,4). rank does not include this
option, but if that is really what you want to do you can use
myRank <- function (x) rev(rank(rev(x), ties = "first"))
On Sat, Apr 26, 2014 at 2:54 AM, 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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--
Bill Dunlap
TIBCO Software
wdunlap tibco.com
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