Peter,
Thank you. I'll be sure to include a as.character.
Greg
On 4/7/10 11:40 AM, Peter Ehlers wrote:
On 2010-04-07 11:40, Greg Hirson wrote:
Dear list,
I have come across this issue:
combn(letters[1:5], 3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] "a" "a" "a" "a" "a" "a" "b" "b" "b" "c"
[2,] "b" "b" "b" "c" "c" "d" "c" "c" "d" "d"
[3,] "c" "d" "e" "d" "e" "e" "d" "e" "e" "e"
combn(factor(letters[1:5]), 3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] "a" "1" "1" "1" "1" "1" "2" "2" "2" "3"
[2,] "b" "2" "2" "3" "3" "4" "3" "3" "4" "4"
[3,] "c" "4" "5" "4" "5" "5" "4" "5" "5" "5"
I'm not sure why the first combination uses the factor labels and the
remaining combinations use the factor value. In the second case, I
expected that the labels would be used. In any event, I think it could
be consistent - either labels or values.
Is there reason to expect the second case to return the value it did?
Yes, that's how the code is written. Note the reference
to '1st combination' on the help page.
combn() is not intended for factors; the help page says:
x vector source for combinations, or integer n for x <- seq(n).
and factors are not vectors.
Two things will work with factors:
1. use combn(as.character(yourfactor), m)
2. use combn(yourfactor, m, simplify = FALSE) which will return a list.
-Peter
This occurs in R 2.10.1
Thanks,
Greg
--
Greg Hirson
ghir...@ucdavis.edu
Graduate Student
Agricultural and Environmental Chemistry
1106 Robert Mondavi Institute North
One Shields Avenue
Davis, CA 95616
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