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))
df <- cbind(df, do.call(rbind, df_split)
seems to do the same (up to column names) but faster. However,
all the solutions rely on there being exactly two strings when
you split. The different solutions behave differently if this
assumption is violated and none of them really checks this. You
can, for instance, check this with all(sapply(df_split, length) == 2)
Best, Niels R. Hansen
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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