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
apply and table
3 messages · Jinsong Zhao, Peter Dalgaard, arun
On May 19, 2013, at 16:22 , Jinsong Zhao wrote:
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?
To clue the casual reader in, the former gives:
apply(z, 2, table, c("A", "B", "C"))
[[1]]
A B C
A 1 1 0
B 0 0 1
[[2]]
A B C
B 1 0 0
C 0 1 1
[[3]]
A B C
A 1 0 0
B 0 1 0
C 0 0 1
whereas the latter gives the first of the tables strung out as a 6x1 matrix.
This is a generic awkwardness of apply(). It tries to simplify the result (similar to sapply), so if the result for all columns have the same length (say, k), it converts them to a (k x C) matrix. If the results are incommensurable, it gives up and returns a list.
So if we modify the code to always give a 3x3 matrix, the following happens:
ABC <- LETTERS[1:3] apply(z, 2, function(x) table(factor(x, levels=ABC), ABC))
[,1] [,2] [,3] [1,] 1 0 1 [2,] 0 1 0 [3,] 0 0 0 [4,] 1 0 0 [5,] 0 0 1 [6,] 0 1 0 [7,] 0 0 0 [8,] 1 0 0 [9,] 0 1 1 (This, incidentally, also answers your question below.) You can't turn simplification off in apply(), but a passable workaround is
tapply(z, col(z), function(x) table(factor(x, levels=ABC), ABC))
$`1`
ABC
A B C
A 1 1 0
B 0 0 1
C 0 0 0
$`2`
ABC
A B C
A 0 0 0
B 1 0 0
C 0 1 1
$`3`
ABC
A B C
A 1 0 0
B 0 1 0
C 0 0 1
Another question: how to output the table in squared matrix (or data frame)? For example:
table(c("C", "B", "B"), c("A", "B", "C"))
A B C B 0 1 1 C 1 0 0 I hope to get the result something like: A B C A 0 0 0 B 0 1 1 C 1 0 0 Is there a way that can output that? Any suggestions will be really appreciated. Thanks in advance.
Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com
Hi,
May be this helps:
lev1<- unique(as.vector(z))
lapply(lapply(as.data.frame(z),factor,levels=lev1),table,lev1)
#$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 ______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.