[R] Choices from a matrix

davidr@rhotrading.com davidr at rhotrading.com
Fri May 6 22:17:24 CEST 2005


Bert, that was almost it, we just need one more as.matrix:

newX[,I]<-as.matrix(expand.grid(as.list(as.data.frame(X[,I]))))

This works as advertised.

Best regards,
David

p.s. I agree that the extra as.list is counterintuitive.
It might be nice if there were a matrix form of expand.grid.

-----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
>  
> 312-362-4963 (voice)
> 312-362-4941 (fax)
>  
> 
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