How to use cast or another function to create a binary table in R

Question

I am trying to create a list of factors that have a binary response and have been using cast.

DF2 <- cast(data.frame(DM), id ~ region)
names(DF2)[-1] <- paste("region", names(DF2)[-1], sep = "")

The problem I am getting is that the responses are the frequency of which the answer comes up, while I'm looking for just whether or not it matches.

For example I have:

What I'd like is:

id region1 region2 region3
1   0          1     1
2   0          1     0
3   1          0     0

joran · Accepted Answer · 2012-07-25 21:59:53Z

8

I kind of prefer dcast from reshape2:

library(reshape2)
dat <- read.table(text = "id region
 1   2
 1   3
 2   2
 3   1
 3   1",header = TRUE,sep = "")

dcast(dat,id~region,fun.aggregate = function(x){as.integer(length(x) > 0)})

  id 1 2 3
1  1 0 1 1
2  2 0 1 0
3  3 1 0 0

There may be a smoother way to do that, but I'll be honest I don't cast stuff all that often.

answered Jul 25, 2012 at 21:59

joran

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1 Comment

A5C1D2H2I1M1N2O1R2T1 Over a year ago

I love the reshape2 package, but sometimes things seem so slow. In comparison to your code, (table(df) > 0)+0 is almost 24 times faster on my system! Specifying value.var usually speeds things up a little bit, but I wonder if there are any other tips on how to speed up dcast.

thelatemail · Accepted Answer · 2014-02-18 02:09:42Z

5

Original data:

x <- data.frame(id=c(1,1,2,3,3), region=factor(c(2,3,2,1,1)))

> x
  id region
1  1      2
2  1      3
3  2      2
4  3      1
5  3      1

Group up the data:

aggregate(model.matrix(~ region - 1, data=x), x["id"], max)

Result:

  id region1 region2 region3
1  1       0       1       1
2  2       0       1       0
3  3       1       0       0

edited Feb 18, 2014 at 2:09

answered Jul 25, 2012 at 22:29

thelatemail

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2 Comments

A5C1D2H2I1M1N2O1R2T1 Over a year ago

I knew there had to be an aggregate solution, but it wasn't coming to me. +1.

thelatemail Over a year ago

@mrdwab Don't worry, it took me 12 hours for it to click!

A5C1D2H2I1M1N2O1R2T1 · Accepted Answer · 2012-07-26 03:23:24Z

4

Here's sort of a "tricky" way to do it in one line using table (the brackets are important). Assuming your data.frame is named df:

(table(df) > 0)+0
#    region
# id  1 2 3
#   1 0 1 1
#   2 0 1 0
#   3 1 0 0

table(df) > 0 gives us TRUE and FALSE; adding +0 converts the TRUE and FALSE to numbers.

answered Jul 26, 2012 at 3:23

A5C1D2H2I1M1N2O1R2T1

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Comments

Michael · Accepted Answer · 2012-07-25 22:27:21Z

1

No specialized functions are needed:

x <- data.frame(id=1:4, region=factor(c(3,2,1,2)))
x
   id region
1  1      3
2  2      2
3  3      1
4  4      2

x.bin <- data.frame(x$id, sapply(levels(x$region), `==`, x$region))
names(x.bin) <- c("id", paste("region", levels(x$region),sep=''))
x.bin

  id region1 region2 region3
1  1   FALSE   FALSE    TRUE
2  2   FALSE    TRUE   FALSE
3  3    TRUE   FALSE   FALSE
4  4   FALSE    TRUE   FALSE

Or for integer results:

x.bin2 <- data.frame(x$id,  
    apply(sapply(levels(x$region), `==`, x$region),2,as.integer)
) 
names(x.bin2) <- c("id", paste("region", levels(x$region),sep=''))
x.bin2


  id region1 region2 region3
1  1       0       0       1
2  2       0       1       0
3  3       1       0       0
4  4       0       1       0

edited Jul 25, 2012 at 22:27

answered Jul 25, 2012 at 22:16

Michael

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