[R] Faster Solution for a simple code?

[jim holtman]

try this:

> x
       V1      V2 V3
1 5000000 3200000  0
2 5100000 3100000  0
3 5200000 3100000  0
4 5200000 3200000  0
> y
       V1      V2 V3
1 5000000 3200000  1
2 5000000 3200000  1
3 5200000 3100000  1
4 5200000 3000000  1
> z <- merge(x, y, by=c("V1", "V2"), all.x=TRUE)
> t(sapply(split(z, z[,1:2], drop=TRUE), function(.grp){
+     if (any(is.na(.grp))) return(c(.grp[1,1], .grp[1,2], 0))
+     c(.grp[1,1], .grp[1,2], nrow(.grp))
+ }))
                   [,1]    [,2] [,3]
5100000.3100000 5100000 3100000    0
5200000.3100000 5200000 3100000    1
5000000.3200000 5000000 3200000    2
5200000.3200000 5200000 3200000    0
>


On Mon, Apr 13, 2009 at 1:06 PM, Chris82 <rubenbauar at gmx.de> wrote:
>
> Hi R-users,
>
> I create a simple code to check out how often the same numbers in y occur in
> x. For example 5000000 320000 occurs two times.
> But the code with the loop is extremly slow. x have 6100 lines and y
> sometimes more than 50000 lines.
>
> Is there any alternative code to create with R?
>
> thanks.
>
>
> x
>
> 5000000 3200000 0
> 5100000 3100000 0
> 5200000 3100000 0
> 5200000 3200000 0
>
>
> lengthx <- length(x[,1])
>
> y
>
> 5000000 3200000 1
> 5000000 3200000 1
> 5200000 3100000 1
> 5200000 3000000 1
>
>
> langthy <- length(y[,1])
>
> for (i in 1:lengthx){
> for (j in 1:lengthy){
> if (x[i,1] == y[j,1]){
> if (x[i,2] == y[j,2]){
> x[i,3] <- x[i,3] + 1
> }
> }
> }
> }
> x
>
> 1  5000000    3200000      2
> 2  5100000    3100000      0
> 3  5200000    3100000      1
> 4  5200000    3200000      0
> --
> View this message in context: http://www.nabble.com/Faster-Solution-for-a-simple-code--tp23024985p23024985.html
> Sent from the R help mailing list archive at Nabble.com.
>
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?