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WG: R-Problem
3 messages · Bleicher Niels (AFS), Sarah Goslee, Alexander Engelhardt
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
Am 21.06.2011 16:43, schrieb Bleicher Niels (AFS):
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
Hi, I used apply() with hist(). For some reason the breakpoints had to start at -0.5. Try this: x <- matrix(c(0,1.3,1.6,2.1,1.3,0,1.9,1.7,1.6,1.9,0,2.2,2.1,1.7,2.2,0), nrow=4) cutpoints <- seq(-0.5, 2.5, by=0.5) mx <- apply(x,2,function(x,...) hist(x,plot=FALSE, breaks=cutpoints)$counts) rownames(x) <- cutpoints[-1] colnames(x) <- 1:ncol(x) x Best, Alex