Assuming every row is split into exactly two values by whatever string
you choose as split, one fancy exercise in R data structures is
dfsplit = function(df, split)
as.data.frame(
t(
structure(dim=c(2, nrow(df)),
unlist(
strsplit(split=split,
as.matrix(df))))))
so that if your data frame is
df = data.frame(c('1 2', '3 4', '5 6'))
then
dfsplit(df, ' ')
# V1 V2
# 1 1 2
# 2 3 4
# 3 5 6
renaming the columns left as an exercise.
vQ
On 01/18/2011 05:22 PM, Peter Ehlers wrote:
On 2011-01-18 08:14, Ivan Calandra wrote:
Hi,
I guess it's not the nicest way to do it, but it should work for you:
#create some sample data
df<- data.frame(a=c("A B", "C D", "A C", "A D", "B D"),
stringsAsFactors=FALSE)
#split the column by space
df_split<- strsplit(df$a, split=" ")
#place the first element into column a1 and the second into a2
for (i in 1:length(df_split[[1]])){
df[i+1]<- unlist(lapply(df_split, FUN=function(x) x[i]))
names(df)[i+1]<- paste("a",i,sep="")
}
I hope people will give you more compact solutions.
HTH,
Ivan
You can replace the loop with
df <- transform(df, a1 = sapply(df_split, "[[", 1),
a2 = sapply(df_split, "[[", 2))
Peter Ehlers
Le 1/18/2011 16:30, boris pezzatti a écrit :
Dear all,
how can I perform a string operation like strsplit(x," ") on a column
of a dataframe, and put the first or the second item of the split into
a new dataframe column?
(so that on each row it is consistent)
Thanks
Boris
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