You might gain some speed by not creating df0 and df0_1:
for (i in 2:100) output.list[[i]] - f1(output.list[[i-1]])
You can also look at the structure of the data frames. For example, if
some of the elements in the data frames are factors but don't need to be
factors, you might gain by
Hello:
I have a list of data frames, built like this: the second df is a result of
a function applied to the first, and so on.
So the ith df is always dependent on the (i-1)th df. I've been doing this
using for loops. However I think I have too many for loops which is making
my code run
You might want to use Rprof to profile your code to understand where the time
is going; it might be in the function you are callingband therefore the for
loop might not be the issue.
Sent from my iPad
On Jun 27, 2013, at 4:19, Frederico Mestre mestre.freder...@gmail.com wrote:
Hello:
3 matches
Mail list logo
