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

From: Gustavo Carvalho <gustavo.bio+R_at_gmail.com>
Date: Thu, 14 Apr 2011 18:23:13 -0300

This might be a bit quicker with larger vectors:

f <- function(x, y) sum(x > y)
vf <- Vectorize(f, "x")
vf(x, y)

On Thu, Apr 14, 2011 at 5:37 PM, Marc Schwartz <marc_schwartz_at_me.com> wrote:
> 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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