R help - May 2010 - apply a function on elements of a list two by two

Dear all,

I want to apply a function to list elements, two by two. I hoped that combn
would help me out, but I can't get it to work. A nested for-loop works, but
seems highly inefficient when you have large lists. Is there a more
efficient way of approaching this?

# Make some toy data
data(iris)
test <- vector("list",3)
for (i in 1:3){
    x <- levels(iris$Species)[i]
    tmp <- dist(iris[iris$Species==x,-5])
    test[[i]] <- tmp
}
names(test) <- levels(iris$Species)

# nested for loop works
for(i in 1:2){
    for(j in (i+1):3){
        print(all.equal(test[[i]],test[[j]]))
    }
}

# combn doesn't work
combn(test,2,all.equal)

Cheers
Joris
-- 
Joris Meys
Statistical Consultant

Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control

Coupure Links 653
B-9000 Gent

tel : +32 9 264 59 87
Joris.Meys@Ugent.be
-------------------------------
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php

	[[alternative HTML version deleted]]

On May 8, 2010, at 9:43 AM, Joris Meys wrote:
> Dear all,
>
> I want to apply a function to list elements, two by two. I hoped  
> that combn
> would help me out, but I can't get it to work. A nested for-loop  
> works, but
> seems highly inefficient when you have large lists. Is there a more
> efficient way of approaching this?
>
> # Make some toy data
> data(iris)
> test <- vector("list",3)
> for (i in 1:3){
>    x <- levels(iris$Species)[i]
>    tmp <- dist(iris[iris$Species==x,-5])
>    test[[i]] <- tmp
> }
> names(test) <- levels(iris$Species)
>
> # nested for loop works
> for(i in 1:2){
>    for(j in (i+1):3){
>        print(all.equal(test[[i]],test[[j]]))
>    }
> }
>
> # combn doesn't work
> combn(test,2,all.equal)
all.equal takes two arguments:

 > all.equal(c(1,2))
Error in mode(current) :
   element 1 is empty;
    the part of the args list of 'is.expression' being evaluated was:
    (x)

So...:

 > combn(test,2, function(x) all.equal(x[[1]], x[[2]]))
      [,1]
[1,] "Attributes: < Component 4: 50 string mismatches >"
[2,] "Mean relative difference: 0.7888781"
      [,2]
[1,] "Attributes: < Component 4: 50 string mismatches >"
[2,] "Mean relative difference: 0.953595"
      [,3]
[1,] "Attributes: < Component 4: 50 string mismatches >"
[2,] "Mean relative difference: 0.6366219"

Thanks for posing this problem. It made me look at portions of combn's  
capacities about which I was completely unaware/
-- 
David.
>
> Cheers
> Joris
> -- 
> Joris Meys
> Statistical Consultant
>
> Ghent University
> Faculty of Bioscience Engineering
> Department of Applied mathematics, biometrics and process control
>
> Coupure Links 653
> B-9000 Gent
>
> tel : +32 9 264 59 87
> Joris.Meys at Ugent.be
> -------------------------------
> Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
>
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT

Am 08.05.2010 15:43, schrieb Joris Meys:
> Dear all,
>
> I want to apply a function to list elements, two by two. I hoped that combn
> would help me out, but I can't get it to work. A nested for-loop works,
but
> seems highly inefficient when you have large lists. Is there a more
> efficient way of approaching this?
>
> # Make some toy data
> data(iris)
> test<- vector("list",3)
> for (i in 1:3){
>      x<- levels(iris$Species)[i]
>      tmp<- dist(iris[iris$Species==x,-5])
>      test[[i]]<- tmp
> }
> names(test)<- levels(iris$Species)
# Using 'lapply' and 'split' is a little bit more flexible:
test <- lapply(split(iris[, -5], iris$Species), function(x) dist(x))
>
> # nested for loop works
> for(i in 1:2){
>      for(j in (i+1):3){
>          print(all.equal(test[[i]],test[[j]]))
>      }
> }
>
> # combn doesn't work
> combn(test,2,all.equal)
Sorry, no answer

HTH
Patrick
>
> Cheers
> Joris