Hi:
Here are a couple of ways. Since your data frame does not contain a
'c' in ID2, we redefine the factor to give it all five levels rather
than the observed four:
> df <- read.table(textConnection("
+ ID1 ID2 Value
+ a b 1
+ b d 1
+ c a 2
+ c e 1
+ d a 1
+ e d 2"), header = TRUE)
str(df)
> str(df)
'data.frame': 6 obs. of 3 variables:
$ ID1 : Factor w/ 5 levels "a","b","c","d",..: 1 2 3 3 4 5
$ ID2 : Factor w/ 4 levels "a","b","d","e": 2 3 1 4 1 3
$ Value: int 1 1 2 1 1 2
df$ID2 <- factor(df$ID2, levels = letters[1:5])
> str(df)
'data.frame': 6 obs. of 3 variables:
$ ID1 : Factor w/ 5 levels "a","b","c","d",..: 1 2 3 3 4 5
$ ID2 : Factor w/ 5 levels "a","b","c","d",..: 2 4 1 5 1 4
$ Value: int 1 1 2 1 1 2
Now we're good...
# (1) xtabs:
with(df, xtabs(Value ~ ID1 + ID2) + xtabs(Value ~ ID2 + ID1))
ID2
ID1 a b c d e
a 0 1 2 1 0
b 1 0 0 1 0
c 2 0 0 0 1
d 1 1 0 0 2
e 0 0 1 2 0
# (2) acast() in the reshape2 package:
library('reshape2')
v1 <- acast(df, ID1 ~ ID2, value_var = 'Value', drop = FALSE, fill = 0)
v2 <- acast(df, ID2 ~ ID1, value_var = 'Value', drop = FALSE, fill = 0)
v <- v1 + v2
v[v == 0L] <- NA
v
a b c d e
a NA 1 2 1 NA
b 1 NA NA 1 NA
c 2 NA NA NA 1
d 1 1 NA NA 2
e NA NA 1 2 NA
HTH,
Dennis
On Tue, Aug 2, 2011 at 10:00 AM, Jagz Bell <jagzb...@yahoo.com> wrote:
> Hi,
> I've tried to look through all the previous related Threads/posts but can't
> find a solution to what's probably a simple question.
>
> I have a data frame comprised of three columns e.g.:
>
> ID1 ID2 Value
> a b 1
> b d 1
> c a 2
> c e 1
> d a 1
> e d 2
>
> I'd like to convert the data to a matrix i.e.:
>
> a b c d e
> a n/a 1 2 1 n/a
> b 1 n/a n/a 1 n/a
> c 2 n/a n/a n/a 1
> d 1 1 n/a n/a 2
> e n/a n/a 1 2 n/a
>
> Any help is much appreciated,
>
> Jagz
>
> ______________________________________________
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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