[R] apply and table
arun
smartpink111 at yahoo.com
Sun May 19 20:04:04 CEST 2013
Hi,
May be this helps:
lev1<- unique(as.vector(z))
lapply(lapply(as.data.frame(z),factor,levels=lev1),table,lev1)
#$V1
#
# A B C
#A 1 1 0
#B 0 0 1
#C 0 0 0
#
#$V2
#
# A B C
#A 0 0 0
#B 1 0 0
#C 0 1 1
#
#$V3
#
# A B C
#A 1 0 0
#B 0 1 0
#C 0 0 1
#or
library(plyr)
llply(alply(z,2,factor,levels=lev1),table,lev1)
#$`1`
# lev1
# A B C
#A 1 1 0
#B 0 0 1
#C 0 0 0
#
#$`2`
# lev1
# A B C
#A 0 0 0
#B 1 0 0
#C 0 1 1
#
#$`3`
# lev1
# A B C
#A 1 0 0
#B 0 1 0
#C 0 0 1
A.K.
----- Original Message -----
From: Jinsong Zhao <jszhao at yeah.net>
To: R help <r-help at r-project.org>
Cc:
Sent: Sunday, May 19, 2013 10:22 AM
Subject: [R] apply and table
Hi there,
I have the following code:
z <- matrix(c("A", "A", "B", "B", "C", "C", "A", "B", "C"), ncol = 3)
apply(z, 2, table, c("A", "B", "C"))
which give correct results.
However, the following code:
apply(z[,1,drop=FALSE], 2, table, c("A", "B", "C"))
which does not give what I expect. I have been thought it should give
the same result as:
apply(z, 2, table, c("A", "B", "C"))[[1]]
What's the difference? Does apply not apply to column vector?
Another question: how to output the table in squared matrix (or data
frame)? For example:
> table(c("C", "B", "B"), c("A", "B", "C"))
A B C
B 0 1 1
C 1 0 0
I hope to get the result something like:
A B C
A 0 0 0
B 0 1 1
C 1 0 0
Is there a way that can output that?
Any suggestions will be really appreciated. Thanks in advance.
Regards,
Jinsong
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