Hi,
sapply(colnames(m)[-ncol(m)],function(i) {x1<-cbind(m[,i],m[,ncol(m)]);
length(which(x1[,1]!=0 & x1[,2]!=0))})
#A B C
#2 1 2
A.K.
----- Original Message -----
From: Hermann Norpois <hnorpois at gmail.com>
To: r-help <r-help at r-project.org>
Cc:
Sent: Friday, June 14, 2013 3:51 PM
Subject: [R] combination of columns in a matrix
Hello,
I have a matrix m and I want to know how often does 1 (or !0) simultanously
appear in A and REF, B and REF, C and REF.
So actually I wish to automate following expression:> length (which (m[,1]!=0&m[,4]!=0))
[1] 2> length (which (m[,2]!=0&m[,4]!=0))
[1] 1
Thanks
Hermann
> m
? A B C REF
r1 1 0 0? 1
r2 1 1 0? 0
r3 0 0 1? 1
r4 0 0 1? 1
r5 1 1 0? 1> dput (m)
structure(c(1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0,
1, 1, 1), .Dim = c(5L, 4L), .Dimnames = list(c("r1", "r2",
"r3",
"r4", "r5"), c("A", "B", "C",
"REF")))
??? [[alternative HTML version deleted]]
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