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.m...@ugent.be
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