Another way:
x <- list(c(9, 5, 7, 2, 14, 4, 4, 3), c(3, 6, 25, 2, 14, 3, 3, 4),
c(28, 4, 14, 3, 14, 2, 4, 5), 28)
apply(array(unlist(lapply(x, '[', 1:max(sapply(x, length, dim = c(1, 8,
4)), 1:2, mean, na.rm = TRUE)
or
rowMeans(as.data.frame(lapply(x, '[', 1:max(sapply(x, length, na.r
Hi Duncan,
If vectors of unequal length is the problem, one way to go, using your
example, would be:
# your example
x <- list(c(9, 5, 7, 2, 14, 4, 4, 3), c(3, 6, 25, 2, 14, 3, 3, 4),
c(28, 4, 14, 3, 14, 2, 4, 5), 28)
x
# maximum number of components
k <- max(sapply(x, length))
k
# expanding
Hi,
Ah! Thanks for your help! I have even seen Reduce before but for some reason
just didn't make the connection. This helps a ton! Thanks again.
Kind regards,
Greg
On Oct 29, 2010, at 6:53 PM, Jorge Ivan Velez wrote:
> Hi Greg,
>
> Here are two ways of doing it:
>
> > mylist <- list(x = rpoi
Aplogies to the original poster - I got names wrong in a cut and paste
Hi Jorge
I tried your methods for all (which work for complete rows) and then I
remove the first value of $y and repeated; both fail because of
the unequal numbers
The problem is when there are unequal numbers in the rows
Hi Jorge
I tried your methods for all (which work for complete rows) and then I
remove the first value of $y and repeated; both fail because of
the unequal numbers
The problem is when there are unequal numbers in the rows and trying to
make a matrix of them.
I was trying some things with Gr
Hi Greg,
Here are two ways of doing it:
> mylist <- list(x = rpois(10, 10), y = rpois(10, 20), z = rpois(10, 5))
> mylist
$x
[1] 3 13 14 16 10 7 3 5 12 14
$y
[1] 17 16 26 13 23 24 16 28 23 12
$z
[1] 2 6 5 5 5 1 9 11 6 4
>
> colMeans(do.call(rbind, mylist), na.rm = TRUE)
[1] 7
Hi Everyone,
I have a list of vectors like this (in this case it's 3 vectors but assume the
vector count and the length of each vector is not known):
[[1]]
[1] 9 5 7 2 14 4 4 3
[[2]]
[1] 3 6 25 2 14 3 3 4
[[3]]
[1] 28 4 14 3 14 2 4 5
What I want to do is take the average ver
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