Hello,
>
> I use a for loop to take into account the years.
> Then, I want to do this as well for the other objects and I use again
> another loop around it.
> The script works well, but it takes a lot of time.
>
Look at the first line in the inner loop:
> for (l in 1:3)
> {
> for (y in 1980:1983)
> {
> test.2<-ifelse(test[,l]>1,1,0)
> [... etc ...]
It doesn't depend on 'y', could be put outside that loop. This alone would
save time.
What also saves time is to see that you're choosing values 1 and 0 depending
on the 'ifelse' condition.
To simply do
test.2 <- test[, l] > 1 # Use TRUE = 1 and FALSE = 0
makes it 20 times faster (!) (Tested with a much larger 'test' data.frame.)
>
> Does somebody knows a way to do this? I was thinking about some kind of
> form of apply, ...
With no loops, only *apply:
# After creating the data frames save a copy of the original 'test.1'
# (This line should be before the loops)
test.1b <- test.1
# Now, after the loops and after attributing names to the result, try the
following.
# Could this become 'unique(Year$Date)' ?
y <- 1980:1983
# This is a matrix, not vector by vector like above. It's the matrix 'xx' in
the nested 'apply'
test.2b <- test > 1
# This is the index 'jj' into 'xx'
test.3b <- sapply(y, function(yy) which(Year$Date == yy))
Length <- t(apply(test.3b, 2, function(jj)
apply(test.2b, 2, function(xx) sum(xx[jj]))))
# Maybe we don't need to save 'test.1b', if it's possible to cbind(y,
Length)
test.1b <- cbind(test.1b, Length)
names(test.1b) <- c("year", "C", "B", "F")
all.equal(test.1, test.1b)
I bet this is faster.
A final note, it's not a very good idea to use R's objects as variables
names .
For instance, 'length'. Prefer 'Length', it's not a object/function.
Hope this helps,
Rui Barradas
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