as question, as I can see the speed is higher for the later method, why use the first one? Thanks.
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If you are trying to drop unused levels, all you need to do is:
x <- factor(x)
IRTFM
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so factor(x) is equivalent to as.factor(as.character(x))? – lokheart Jan 17 '11 at 07:20
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Yes, factor(x) will have the same result. – IRTFM Jan 17 '11 at 07:32
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@lokheart: I think they will not have the same result if you have to preserve the order of your factors. – daroczig Jan 17 '11 at 10:43
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New in R (from version 2.12.0) is the function droplevels() to do the same thing. It is implemented as:
> base:::droplevels.factor
function (x, ...)
factor(x)
<environment: namespace:base>
So I would use that function out of preference. It is a generic function in R with methods for objects of class "factor" and "data.frame", the latter is useful when there are many factors in the data frame that need levels dropping.
Gavin Simpson
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The two command does quite the same but not exactly, especially when you have the preserve the original order of the factors. In some cases you cannot use: as.factor(as.character(f)). See:
par(mfrow=c(2,3))
f <- factor(c("D", "B", "C", "K", "A"), levels=c("K", "B", "C", "D"))[2:4]
plot(f, main="Original factor")
f.fc <- as.factor(as.character(f))
plot(f.fc, main="as.factor(as.character(f))")
f.d <- drop.levels(f)
plot(f.d, main="drop.levels(f)")
f.d <- drop.levels(f, reorder=FALSE)
plot(f.d, main="drop.levels(f, reorder=FALSE))")
f.f <- factor(f)
plot(f.f, main="factor(f)")
as.factor(as.character(f)) and drop.levels(f) does the same and they do not preserve the original order of factors, they both re-level the text in ABC order. I you want to preserve the order you can use the reorder=FALSE option in drop.levels().
This is the default behavior of factor().
daroczig
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