Choices from a matrix

Thanks, Bert, but when I tried that, I didn't get a matrix:

newX

[[1]]
[1] 1 2 1 2

[[2]]
[1] 5 5 6 6

[[3]]
[1] 1 2 1 2

[[4]]
[1] 5 5 6 6

[[5]]
[1] 0

[[6]]
[1] 0

[[7]]
[1] 0

[[8]]
[1] 0

[[9]]
[1] 1 2 1 2

[[10]]
[1] 5 5 6 6

[[11]]
[1] 1 2 1 2

[[12]]
[1] 5 5 6 6

[[13]]
[1] 0

[[14]]
[1] 0

[[15]]
[1] 0

[[16]]
[1] 0

The matrix is in there, four times in fact. For larger problems, I believe it would be in 2^k times? It is certainly a one-liner, though, so if the matrix could be extracted simply, it could be of use.

Best regards,
David

-----Original Message-----
From: Berton Gunter [mailto:gunter.berton at gene.com] 
Sent: Friday, May 06, 2005 2:49 PM
To: David Reiner <davidr at rhotrading.com>; r-help at stat.math.ethz.ch
Subject: RE: [R] Choices from a matrix

If I understand you correctly, here's one way based on expand.grid().

I is just an index set, and so all you really need to do is generate your
2^k rows from the part of the matrix you're using in the right places via
replacement:  

e.g. newX<-matrix(0, ncol=ncol(X),nrow=2^length(I))
newX[,I]<-expand.grid(as.list(as.data.frame(X[,I]))) 


N.B. I tried to do the this without the explicit as.list() cast, but got an
error message. I would have thought that expand.grid should have recognized
that a data.frame IS a list without the cast.

-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
 
"The business of the statistician is to catalyze the scientific learning
process."  - George E. P. Box

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch 
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of 
davidr at rhotrading.com
Sent: Friday, May 06, 2005 12:06 PM
To: r-help at stat.math.ethz.ch
Subject: [R] Choices from a matrix

Could someone please suggest a more clever solution to the 
following problem than my loop below?

Given X a 2xN matrix X, and I a k-subset of N, 
Generate the (2^k)xN matrix Y with columns not in I all zero 
and the other columns with all choices of an entry from the 
first or second row of X.

For example, with
X <- matrix(1:8, nrow=2)
I <- c(1,3)

X is
1 3 5 7
2 4 6 8

and Y should be
1 0 5 0
2 0 5 0
1 0 6 0
2 0 6 0

The order of the rows is unimportant.
---
I solved this using a loop over the rows of Y after forming 
some preliminary matrices. I think it could be improved.

N <- NCOL(X)
k <- length(I)
G <- as.matrix(expand.grid(rep(list(c(1,2)),k)))
Y <- matrix(0,nc=N,nr=NROW(G))

for(i in 1:NROW(G)){
   ind <- rep(1,N)
   ind[I] <- G[i,]
   Y[i,] <- X[array(c(ind,1:N),dim=c(N,2))]
}
Y[,-I] <- 0




David L. Reiner
??
Rho Trading
440 S. LaSalle St -- Suite 620
Chicago?? IL?? 60605
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312-362-4963 (voice)
312-362-4941 (fax)
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