[R] how to do away for loop using functionals?

Michael Hannon jmhannon.ucdavis at gmail.com
Wed Oct 14 11:02:25 CEST 2015


I've done a simple-minded transliteration of your code into code using nested
lapply's.  I doubt that it buys you much in terms of performance (or even
clarity, which is really one of the main advantages of the `apply` family).


> A
        [,1]      [,2]     [,3]     [,4]     [,5]
[1,] 3.06097  6.507521 10.99610 12.05556 15.10388
[2,]     Inf 11.818495 15.85044 16.69465 19.70425
[3,]     Inf       Inf      Inf 19.14779 22.30343
[4,]     Inf       Inf      Inf      Inf 26.11170
[5,]     Inf       Inf      Inf      Inf 28.29882

> B
        [,1]      [,2]     [,3]     [,4]     [,5]
[1,] 3.06097  6.507521 10.99610 12.05556 15.10388
[2,]     Inf 11.818495 15.85044 16.69465 19.70425
[3,]     Inf       Inf      Inf 19.14779 22.30343
[4,]     Inf       Inf      Inf      Inf 26.11170
[5,]     Inf       Inf      Inf      Inf 28.29882
> all.equal(A, B)
[1] TRUE

If I happen to think of a more-elegant approach, I'll let you know.

-- Mike

Appendix: code
==============

###### Anne's code

getResult <- function(d) {

      #examplefunction

     weighted.mean(x=d[,1], w=d[,2])

}

#example data setup

n=20;

set.seed(1)

g=rep(1:5,each=4)

df=as.data.frame(cbind( sort(rnorm(mean=15,sd=10, n)),runif(n), rbinom(n, 1,
0.4) , g )); df

getResult(df)

i0=c(1,2,4,5,5)

ng= length(unique(g))



#initiation of result matrix

A=matrix(Inf, ng, ng); A

for(i in 1:ng)

{              cat("i:",i,"")

                for(j in i0[i]:ng) {

                                ok= !is.na(match(g,i:j)); cat("j:",j,"\n");

                                A[i,j]=getResult(d=df[ok,])

                } #endfor (j)

} #end for (i)
A

###### Mike's code

n <- 20;
set.seed(1)
g <- rep(1:5,each=4)
df <- as.data.frame(cbind(sort(rnorm(mean=15,sd=10, n)),
                          runif(n),
                          rbinom(n, 1, 0.4),
                          g )); df
getResult(df)
i0 <- c(1,2,4,5,5)
ng <- length(unique(g))

B <- matrix(Inf, ng, ng);

invisible(lapply(1:ng, function(i) {
                     lapply(i0[i]:ng, function(j) {
                                ok <- !is.na(match(g, i:j))
                                B[i, j] <<- getResult(df[ok, ])
                            })
                 }))

B
all.equal(A, B)


On Mon, Oct 12, 2015 at 5:55 PM, Annie Hawk via R-help
<r-help at r-project.org> wrote:
> HI R-experts,
>
>
> I am trying to speed up my calculation of the A results below and replace the for loop withsome functionals like lapply.  After manyreadings, trial and error, I still have no success.  Would anyone please give me some hints onthat?
>
> Thank you in advance.
>
> Anne
>
>
> The program is this, I have a complicated function and itneeds to operate on some subsets of a dataset many times, depending on thevalues of group.  I simplify the functionand dataset for this example run.
>
> getResult <- function(d) {
>
>       #examplefunction
>
>      weighted.mean(x=d[,1], w=d[,2])
>
> }
>
>
>
> #example data setup
>
> n=20;
>
> set.seed(1)
>
> g=rep(1:5,each=4)
>
> df=as.data.frame(cbind( sort(rnorm(mean=15,sd=10, n)),runif(n), rbinom(n, 1, 0.4) , g )); df
>
> getResult(df)
>
> i0=c(1,2,4,5,5)
>
> ng= length(unique(g))
>
>
>
> #initiation of result matrix
>
> A=matrix(Inf, ng, ng); A
>
> for(i in 1:ng)
>
> {              cat("i:",i,"")
>
>                 for(jin i0[i]:ng) {
>
>                                 ok= !is.na(match(g,i:j)); cat("j:",j,"\n");
>
>                                 A[i,j]=getResult(d=df[ok,])
>
>                 } #endfor (j)
>
> } #end for (i)
>
> Is there an elegant way to remove the for loop here?  I try to make it flat for faster run but Icannot figure out how to subset the observations faster without error to apply the functiongetResult.  Any hint is appreciated.
>
>
>
>
>
> on another note, is there a more elegant way to initiate the list as follows?
>
> mylist=list(); w=rep(4,5)
>
> for (i in 1:5) mylist[[i]]=w[i:5]
>
>
>
>
>         [[alternative HTML version deleted]]
>
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