Thanks a lot, Gabor - it's perfect!
Dimitri
On Fri, Dec 19, 2008 at 6:24 PM, Gabor Grothendieck
wrote:
> Try this:
>
>> L <- list(data.frame(A=2, B=3, C=4),
> + data.frame(A=2, B=1, C=3, D=2, E=4, F=5),
> + data.frame(A=1, B=2, C=4, D=3, E=2, F=4, G=5, H=4, I=2))
>
>> library(plyr)
>> do.call(rbi
Try this:
> L <- list(data.frame(A=2, B=3, C=4),
+ data.frame(A=2, B=1, C=3, D=2, E=4, F=5),
+ data.frame(A=1, B=2, C=4, D=3, E=2, F=4, G=5, H=4, I=2))
> library(plyr)
> do.call(rbind.fill, L)
A B C D E F G H I
1 2 3 4 NA NA NA NA NA NA
2 2 1 3 2 4 5 NA NA NA
3 1 2 4 3 2 4 5 4 2
Hi Dimitri,
Not exceptionally elegant, but this should do it:
alldata <- as.data.frame(matrix(as.numeric(NA),nrow=99,ncol=9))
colnames(alldata) <- colnames(L[[?]])
( where ? is a 9-column data frame with the correct names )
foreach (i in 1:99) {
alldata[i,colnames(L[[i]])] <- L[[i]]
}
Hello, everyone!
I have list L that contains 99 data frames. All data frames have only
one row, but a different number of columns. Some data frames have 3
columns, some - 6 columns, and some - 9 columns. The names of the
first 3 columns are identical in all 99 data frames (e.g., A, B, and
C). The
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