On Jan 29, 2010, at 12:43 PM, Muhammad Rahiz wrote:
Thanks David & Dennis,
I may have found something.
Given that the object xx is the product of unlist(x), to create a
2x2 matrix with subsets, I could do,
> y <- matrix(xx[c(1:4)], 2, 2).
This returns,
[,1] [,2]
[1,] -27.3 14.4
[2,] 29.0 -38.1
Much simpler to do:
> xx[ , , 1]
[,1] [,2]
[1,] -27.3 14.4
[2,] 29.0 -38.1
If I do,
> y2 <- matrix(xx[c(5:8)],2,2)
it returns,
[,1] [,2]
[1,] 14.4 29.0
[2,] -38.1 -3.4
The results are exactly what I want to achieve.
The question is, how can I incorporate the increment in a for loop
so that it becomes
c(1:4)
c(5:8)
c(9:12) and so on
How should I modify this code?
y <- # typeof ? for (i in 1:32){
y[[i]] <- matrix(xx[c(1:4)],2,2)
}
I don't get it. You had the data in a list. You wanted it out of that
list, and now you're going to put it back in another list??? ( y2
would not be the same as y[[2]] ) What's wrong with xx[ , , 2]?
Muhammad
David Winsemius wrote:
On Jan 29, 2010, at 9:45 AM, Dennis Murphy wrote:
Hi:
The problem, I'm guessing, is that you need to assign each of the
matrices
to an object.
There's undoubtedly a slick apply family solution for this (which
I want to
see, BTW!),
I don't have a method that would assign names but you could
populate an array of sufficient size and dimension. I populated a
three-element list with his data:
> dput(x)
list(structure(list(V1 = c(-27.3, 29), V2 = c(14.4, -38.1)), .Names
= c("V1",
"V2"), class = "data.frame", row.names = c("1", "2")),
structure(list(
V1 = c(14.4, -38.1), V2 = c(29, -3.4)), .Names = c("V1",
"V2"), class = "data.frame", row.names = c("1", "2")),
structure(list(
V1 = c(29, -3.4), V2 = c(-38.1, 55.1)), .Names = c("V1",
"V2"), class = "data.frame", row.names = c("1", "2")))
> xx <- array( , dim=c(2,2,3))
> xx[,,1:3] <- sapply(x, data.matrix)
> xx
, , 1
[,1] [,2]
[1,] -27.3 14.4
[2,] 29.0 -38.1
, , 2
[,1] [,2]
[1,] 14.4 29.0
[2,] -38.1 -3.4
, , 3
[,1] [,2]
[1,] 29.0 -38.1
[2,] -3.4 55.1
Without the more complex structure ready to accept the 2x2 arrays
I got this:
> sapply(x, data.matrix)
[,1] [,2] [,3]
[1,] -27.3 14.4 29.0
[2,] 29.0 -38.1 -3.4
[3,] 14.4 29.0 -38.1
[4,] -38.1 -3.4 55.1
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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