[R] filling list of data frames

[MacQueen, Don]

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 preventing them from being factors.

-Don

-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 6/27/13 1:19 AM, "Frederico Mestre" <mestre.frederico at gmail.com> wrote:

>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 slowly.
>
> 
>
>Is there any workaround  this? How can I avoid the use of for loops?
>
> 
>
>As an example:
>
> 
>
>output.list <- as.list(rep("", 100))#creation of a list
>
> 
>
>output.list[[1]] <- df1#first position
>
> 
>
> 
>
>for(I in 2:100){#following positions
>
> 
>
>df0 <- output.list[[i-1]]
>
> 
>
>df0_1 <- f1(df0)#function applied to the previous df
>
> 
>
>output.list[[i]] <- df0_1#new df
>
> 
>
>}
>
> 
>
>thanks,
>
> 
>
>Frederico 
>
> 
>
> 
>
>
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>
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