His is better, but you can also use a for loop...
out<-data.frame(rows=1:3)
for(i in 1:3){
if(l[[i]][3]=='Message 1') {
out$V1[i]<-l[[i]][1]
} else {
out$V1[i]<-NA
}
}
but shouldn't if your list is very long
On Tue, Aug 23, 2011 at 9:35 AM, Henrique Dallazuanna wrote:
> Try this:
Try this:
subset(as.data.frame(do.call(rbind, lapply(l, "[", , 1))), row3 == "Message 1")
On Tue, Aug 23, 2011 at 1:28 PM, Lara Poplarski wrote:
> Hi all,
>
> I have an object that looks (roughly) like the following:
>
> l <- list(a = matrix(rnorm(9), 3), b = matrix(rnorm(9), 3), c =
> matrix(rn
Hi all,
I have an object that looks (roughly) like the following:
l <- list(a = matrix(rnorm(9), 3), b = matrix(rnorm(9), 3), c =
matrix(rnorm(9), 3))
l$a[3,] <- sample(c("Message 1", "Message 2", "Message 3"))
l$b[3,] <- sample(c("Message 1", "Message 2", "Message 3"))
l$c[3,] <- sample(c("Mess
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