> -----Original Message-----
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Kevin Ummel
> Sent: Thursday, April 14, 2011 12:35 PM
> To: r-help@r-project.org
> Subject: [R] Find number of elements less than some number:
> Elegant/fastsolution needed
>
> 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.
If x is sorted findInterval(x, y) would do it for <= but you
want strict <. Try
f2 <- function(x, y) length(y) - findInterval(-x, rev(-sort(y)))
where your version is
f0 <- function(x, y) sapply(x,FUN=function(i) {length(which(y<i))})
E.g.,
> x <- sort(sample(1e6, size=1e5))
> y <- sample(1e6, size=1e4, replace=TRUE)
> system.time(r0 <- f0(x,y))
user system elapsed
7.900 0.310 8.211
> system.time(r2 <- f2(x,y))
user system elapsed
0.000 0.000 0.004
> identical(r0, r2)
[1] TRUE
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
>
> 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]]
>
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