[R] apply and table

[Jinsong Zhao]

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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[peter dalgaard]

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....@cbs.dk Priv: PDa...@gmail.com

[arun]

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.