cut() and table() will do it. Here's one less-than-elegant way to put
them together:
test <- structure(list(X1 = c(0, 1.3, 1.6, 2.1), X2 = c(1.3, 0, 1.9,
1.7), X3 = c(1.6, 1.9, 0, 2.2), X4 = c(2.1, 1.7, 2.2, 0)), .Names =
c("X1",
"X2", "X3", "X4"), class = "data.frame",
row.names = c("1", "2",
"3", "4"))
test1 <- apply(test, 2, cut, breaks=seq(0, 2.5, by=.5),
labels=c("0.5", "1", "1.5", "2",
"2.5"))
apply(test1, 2, function(x)table(factor(x, levels=c("0.5",
"1", "1.5",
"2", "2.5"))))
Sarah
On Tue, Jun 21, 2011 at 10:43 AM, Bleicher Niels (AFS)
<niels.bleicher at zuerich.ch> wrote:>
>
> Dear forumites
>
> As a newbie I try to figure out whether R can do a certain job quicker than
other programs and it seems so, but I don't find a solution to a seemingly
simple problem:
>
> I have built a matrix of distance with as.matrix(dist()) with several
hundred rows and columns
> ? ? ? ? ? ?1 ? ? ? ? ?2 ? ? ? ? ?3 ? ? ? ? ?4
> 1 ? ? ? ? ?0 ? ? ? ? ?1.3 ? ? ? 1.6 ? ? ? 2.1
> 2 ? ? ? ? ?1.3 ? ? ? 0 ? ? ? ? ?1.9 ? ? ? 1.7
> 3 ? ? ? ? ?1.6 ? ? ? 1.9 ? ? ? 0 ? ? ? ? ?2.2
> 4 ? ? ? ? ?2.1 ? ? ? 1.7 ? ? ? 2.2 ? ? ? 0
>
> Now I need for every column a frequency table of the type
>
> Class. ?1 ? ? ? ? ?2 ? ? ? ? ?3 ? ? ? ? ?4
> 0.5 ? ? ? 0 ? ? ? ? ?0 ? ? ? ? ?0 ? ? ? ? ?0
> 1 ? ? ? ? ?0 ? ? ? ? ?0 ? ? ? ? ?0 ? ? ? ? ?0
> 1.5 ? ? ? 1 ? ? ? ? ?1 ? ? ? ? ?0 ? ? ? ? ?0
> 2 ? ? ? ? ?1 ? ? ? ? ?2 ? ? ? ? ?2 ? ? ? ? ?1
> 2.5 ? ? ? 1 ? ? ? ? ?0 ? ? ? ? ?1 ? ? ? ? ?2
>
> Which function is appropriate here?
> Cheers
> Niels
>
--
Sarah Goslee
http://www.functionaldiversity.org