Re: [R] Find number of elements less than some number: Elegant/fast solution needed

From: Marc Schwartz <marc_schwartz_at_me.com>
Date: Thu, 14 Apr 2011 15:37:29 -0500

On Apr 14, 2011, at 2:34 PM, Kevin Ummel 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
>

I started working on a solution to your problem above and then noted the one below.

Here is one approach to the above:

> colSums(outer(y, x, "<"))

 [1] 0 0 0 0 1 2 2 3 3 4

>
> 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.

Something like the following might work, if I correctly understand the problem:

> match(x, x[-y])

 [1] 1 NA 2 3 NA NA 4 5 NA 6

HTH, Marc Schwartz



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