Thanks for your kind replies
Jorge's answer helps.
warm wishes
Xueming
xmliu1...@gmail.com
From: Jorge I Velez
Date: 2014-04-26 23:41
To: xmliu1...@gmail.com
CC: r-help
Subject: Re: [R] Faster way to transform vector [3 8 4 6 1 5] to [2 6 3 5 1 4]
Hi Xueming,
Try
(1:length(bo))
Hi,
Perhaps,
rank(bo)
#[1] 2 6 3 5 1 4
A.K.
On Saturday, April 26, 2014 11:00 AM, "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 element
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 wrote:
> Hi,
>
> could anybody help me to find a fast way to fi
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
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 suppo
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