R data frame format group - r

Question

I have data frame in this format-
ABC 2
ABC 4
ABC 6
DEF 10
DEF 20
How can I get this to-
ABC 2 4 6
DEF 10 20
I tried the aggregate function, but it needs functions like mean/sum as params. How can I just display the values directly in the row.

Answer 1

df <- read.table(sep=" ", header=F, text="
ABC 2
ABC 4
ABC 6
DEF 10
DEF 20")
unstack(df, form=V2~V1)
# $ABC
# [1] 2 4 6
#
# $DEF
# [1] 10 20
unstack produces a list in this case as the columns don't have the same length. In case of the same length:
df <- read.table(sep=" ", header=F, text="
ABC 2
ABC 4
ABC 6
DEF 10
DEF 20
DEF 20")
t(unstack(df, form=V2~V1))
# [,1] [,2] [,3]
# ABC 2 4 6
# DEF 10 20 20

Answer 2

Well, what are the observations? Are they suppose to measure the same thing for each category?
You can't actually get a data frame exactly as you have posted, because the number of observations for each category is different. But you could do that if you add an "NA" to the "DEF".
Like this:
ABC 2 4 6
DEF 10 20 NA
If that is what you want, you could just use reshape2's dcast.
But you have to name the observations:
library(reshape2)
df <- data.frame(obs =c(1:3, 1:2),
categories = c(rep("ABC", 3), rep("DEF",2)),
values=c(2,4,6,10,20), stringsAsFactors=FALSE)
df2 <- dcast(df, categories~obs)
df2
# categories 1 2 3
# 1 ABC 2 4 6
# 2 DEF 10 20 NA

Answer 3

To add to your alternatives:
This seems to be a basic "long to wide" reshape problem, but it is missing a "time" variable. It's easy to recreate one by using ave:
ave(as.character(df$V1), df$V1, FUN = seq_along)
# [1] "1" "2" "3" "1" "2"
df$time <- ave(as.character(df$V1), df$V1, FUN = seq_along)
Once you have a "time" variable, using reshape is pretty straightforward:
reshape(df, idvar="V1", timevar="time", direction = "wide")
# V1 V2.1 V2.2 V2.3
# 1 ABC 2 4 6
# 4 DEF 10 20 NA
If, instead, you wanted a list, there is no need for the time variable. Just use split:
split(df$V2, df$V1)
# $ABC
# [1] 2 4 6
#
# $DEF
# [1] 10 20
#
Similarly, if your data were balanced, split plus rbind could get you what you need. Using the sample data from #lukeA:
df <- read.table(sep=" ", header=F, text="
ABC 2
ABC 4
ABC 6
DEF 10
DEF 20
DEF 20")
do.call(rbind, split(df$V2, df$V1))
# [,1] [,2] [,3]
# ABC 2 4 6
# DEF 10 20 20

Answer 4

You want to obtain a sparse matrix? The two rows in your example have different lengths. Try a function producing a list:
mat<-cbind(
c("ABC","ABC","ABC","DEF","DEF"),
c(2,4,6,10,20)
)
count<-function(mat){
values<-unique(mat[,1])
outlist<-list()
for(v in values){
outlist[[v]]<-mat[mat[,1]==v,2]
}
return(outlist)
}
count(mat)
Which will give you this result:
$ABC
[1] "2" "4" "6"
$DEF
[1] "10" "20"