Find number of elements less than some number: Elegant/fastsolution needed
My colleague Sunduz Keles once mentioned a similar problem to me. She had a large sample from a reference distribution and a test sample (both real-valued in her case) and she wanted, for each element of the test sample, the proportion of the reference sample that was less than the element. It's a type of empirical p-value calculation. I forget the exact sizes of the samples (do you remember, Sunduz?) but they were in the tens of thousands or larger. Solutions in R tended to involve comparing every element in the test sample to every element in the reference sample but, of course, that is unnecessary. If you sort both samples then you can start the comparisons for a particular element of the test sample at the element of the reference sample where the last comparison failed. I was able to write a very short C++ function using the Rcpp package that provided about a 1000-fold increase in speed relative to the best I could do in R. I don't have the script on this computer so I will post it tomorrow when I am back on the computer at the office. Apologies for cross-posting to the Rcpp-devel list but I am doing so because this might make a good example of the usefulness of Rcpp and inline.
On Thu, Apr 14, 2011 at 4:45 PM, William Dunlap <wdunlap at tibco.com> wrote:
-----Original Message-----
From: r-help-bounces at r-project.org
[mailto:r-help-bounces at r-project.org] On Behalf Of Kevin Ummel
Sent: Thursday, April 14, 2011 12:35 PM
To: r-help at 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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