If you have a million levels is it really necessary to use a factor? I'm
not sure what advantages it will to have to a string in this circumstance
(especially since you don't seem to know the levels a priori but have to
learn them from the data).
Hadley
On Sunday, September 16, 2012, Sam Steingol
Hello,
The obvious simplification is to call union() only once. With 10M rows
it should save time.
Then I've asked myself whether unique() wouldn't be faster.
f1 <- function(x){
x[[1]] <- factor(x[[1]], levels = union(x[[1]], x[[2]]))
x[[2]] <- factor(x[[2]], levels = union(x[[1]], x
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