[R] Faster way to transform vector [3 8 4 6 1 5] to [2 6 3 5 1 4]

[arun]

Hi,
Perhaps,
rank(bo)
#[1] 2 6 3 5 1 4
A.K.




On Saturday, April 26, 2014 11:00 AM, "xmliu1988 at gmail.com" <xmliu1988 at 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



xmliu1988 at gmail.com
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