on 08/09/2008 06:52 AM Dan Davison wrote:
On Sat, Aug 09, 2008 at 06:29:59AM -0500, Marc Schwartz wrote:
on 08/09/2008 06:01 AM [EMAIL PROTECTED] wrote:
Hi;
If we have a matrix A, and a vector X, where length(X)=nrow(A), and X
contains a wanted column for each row in A, in row ascending order. How
would be the most effective way to extract the desired vector V (with
length(V)=nrow(A))?
A <- matrix(1:20, 4, 5)
A
[,1] [,2] [,3] [,4] [,5]
[1,] 1 5 9 13 17
[2,] 2 6 10 14 18
[3,] 3 7 11 15 19
[4,] 4 8 12 16 20
# Create an arbitrary set of indices, one for each row in A
X <- c(2, 5, 1, 4)
X
[1] 2 5 1 4
Presumably you want:
V <- c(A[1, 2], A[2, 5], A[3, 1], A[4, 4])
V
[1] 5 18 3 16
If so, then:
sapply(seq(nrow(A)), function(i) A[i, X[i]])
[1] 5 18 3 16
Or
A[cbind(seq(nrow(A)), X)]
[1] 5 18 3 16
Dan
Better (and faster) solution Dan.
I can't blame the lack of coffee on missing that one this morning. I
have had a full pot already over the past 6 hours, working on shifting
my internal clock and getting ready to begin my journey to Dortmund
later tonight...
Safe travels to all who are going.
Marc
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