[R] function similar to ddply? + calculations based on previous row

[R. Michael Weylandt]

Hi,

I got a little lost in your verbal description, but if I understand
your code correctly (what is n?), I think I'd implement it with the
split function: step i) make a single (hopefully vectorized) pass to
get whatever split points you need (the offset is in this step); ii)
use split() to break your data.frame into a list efficiently; iii) use
lapply() to run your function on each element of the list.

Hope that helps,
Michael



On Wed, Feb 15, 2012 at 4:25 PM, Nerak <nerak.t at hotmail.com> wrote:
> I saw I made a little mistake in the loop,
> in the line
> test.number$numberb[y-Year[1]+1]<-length(which(test.starty==1 &
> test.f[(n+1)]== 1 ))
> it is n+1 instead of n-1 (like I wrote in the beginning)
>
> But the question I have about it is still the same.
> My excuzes
>
>
>
> Year<-data.frame(Date=c(1980,1980,1980,1980,1981,1981,1981,1981,1982,1982,1982,1982,1983,1983,1983,1983))
> test.b<-data.frame(C=c(0,0,0,0,5,2,0,0,0,15,12,10,6,0,0,0),B=c(0,0,0,0,9,6,2,0,0,24,20,16,2,0,0,0),F=c(0,0,0,0,6,5,1,0,0,18,16,12,10,5,1,0))
> test.start<-data.frame(C=c(0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0),B=c(0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0),F=c(0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0))
>
>
>
> test.2b<-test.b>1
> test.number<-data.frame(c(1980:1983))
> for (l in 1:nrow(test.b))
> {
>        for (y in 1980:1983)
>        {
>
>        test.f<-test.2b[which(Year == y),l]
>        test.starty<-test.start[which(Year ==y),l]
>        test.number$numberb[y-Year[1]+1]<-length(which(test.starty==1 &
> test.f[(n+1)]== 1 ))
>
>        }
>        test.number[,l+1]<-cbind(test.number$numberb)
>
> }
>
>
>
> --
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> Sent from the R help mailing list archive at Nabble.com.
>
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