This is faster for small vectors
> system.time(for (i in 1:1000)
+ colSums(outer(y, x, `<`))
+ )
user system elapsed
0.11 0.00 0.11
> system.time(for (i in 1:1000)
+ sapply(x,FUN=function(i) {length(which(y<i))})
+ )
user system elapsed
0.14 0.00 0.16
>
> colSums(outer(y, x, `<`))
[1] 0 0 0 0 1 2 2 3 3 4
> sapply(x,FUN=function(i) {length(which(y<i))})
[1] 0 0 0 0 1 2 2 3 3 4
>
Rich
On Thu, Apr 14, 2011 at 3:34 PM, Kevin Ummel <kevinum...@gmail.com> wrote:
> Take vector x and a subset y:
>
> x=1:10
>
> y=c(4,5,7,9)
>
> For each value in 'x', I want to know how many elements in 'y' are less
> than 'x'.
>
> An example would be:
>
> sapply(x,FUN=function(i) {length(which(y<i))})
> [1] 0 0 0 0 1 2 2 3 3 4
>
> But this solution is far too slow when x and y have lengths in the
> millions.
>
> I'm certain an elegant (and computationally efficient) solution exists, but
> I'm in the weeds at this point.
>
> Any help is much appreciated.
>
> Kevin
>
> University of Manchester
>
>
>
>
>
>
>
>
> Take two vectors x and y, where y is a subset of x:
>
> x=1:10
>
> y=c(2,5,6,9)
>
> If y is removed from x, the original x values now have a new placement
> (index) in the resulting vector (new):
>
> new=x[-y]
>
> index=1:length(new)
>
> The challenge is: How can I *quickly* and *efficiently* deduce the new
> 'index' value directly from the original 'x' value -- using only 'y' as an
> input?
>
> In practice, I have very large matrices containing the 'x' values, and I
> need to convert them to the corresponding 'index' if the 'y' values are
> removed.
>
>
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
> and provide commented, minimal, self-contained, reproducible code.
>
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