Hello,
Sorry, I've just answered to Johannes without Cc to r-help.
Repeat:
See the difference.
# helper function
make.df <- function(){
x <- as.data.frame(matrix(1:24, ncol=6))
names(x) <- c(paste0("crast", 1:2), "A", "B", paste0("crast", 5:6))
x
}
DF <- make.df()
(ix <- grep("^crast", colnames(DF)))
DF[, ix] <- as.factor(DF[, ix])
# Error in sort.list(y) : 'x' must be atomic for 'sort.list'
# Have you called 'sort' on a list?
DF <- make.df()
DF[, ix] <- sapply(DF[, ix], as.factor)
str(DF) # crast1: chr "1" "2" "3" "4"
DF <- make.df()
for(i in ix)
DF[[i]] <- as.factor(DF[[i]])
str(DF) # crast1: Factor w/ 4 levels
DF <- make.df()
for(i in ix)
DF[, i] <- as.factor(DF[, i])
str(DF) # crast1: Factor w/ 4 levels
I had already noticed that the class of data.frames columns can only be
changed one by one.
I'm still not sure if this is true, if they can be changed collectively,
but since this works, I haven't looked further.
(My problem was to change from character to Date, when, for instance,
there are start dates and end dates.)
Maybe someone else has another way of doing it.
Hope this helps,
Rui Barradas
Em 01-06-2012 09:29, Johannes Radinger escreveu:
Hi,
I have a dataframe with around 100 columns. Now I want
to redefine some of the columns as factors (using as.factor).
Luckily all the names of the columns I want to redefine start with
"crast". Thus I thought I can use grep() for that purpose...
...I found an example for redefining a single column as factor
but that is not working with multiple columns I get from grep()...
what I tried so far:
df[, grep("^crast", colnames(df))]<- as.factor(df[, grep("^crast",
colnames(df))])
any suggestions?
cheers,
/Johannes
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