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davidr@rhotrading.com

2005-May-06 19:06 UTC

[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)
??

Berton Gunter

2005-May-06 19:48 UTC

head link

[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)
> ??
> 
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide! 
> http://www.R-project.org/posting-guide.html
>

davidr@rhotrading.com

2005-May-06 20:11 UTC

head link

[R] 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
> ??
> 312-362-4963 (voice)
> 312-362-4941 (fax)
> ??
> 
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide! 
> http://www.R-project.org/posting-guide.html
>

davidr@rhotrading.com

2005-May-06 20:17 UTC

head link

[R] Choices from a matrix

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)
> ??
> 
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide! 
> http://www.R-project.org/posting-guide.html
>

Berton Gunter

2005-May-06 20:27 UTC

head link

[R] Choices from a matrix

Quite right (ALWAYS TEST, Bert!), because expand.grid() returns a
data.frame. But all you have to do is create newX as a data.frame:

newX<-data.frame(matrix(0, ncol=ncol(X),nrow=2^length(I)))

and **now** it will work:
>X<-matrix(1:6,nr=2)
>I<- c(1,3)
>newX<-data.frame(matrix(0, ncol=ncol(X),nrow=2^length(I)))
>newX[,I]<-expand.grid(as.list(as.data.frame(X[,I])))
>newX
  X1 X2 X3
1  1  0  5
2  2  0  5
3  1  0  6
4  2  0  6

Note that newX is a data.frame -- you may wish to recast it as a matrix. Of
course, I'm still not sure that this is what you wanted.

-- 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: bgunter 
> Sent: Friday, May 06, 2005 12:49 PM
> To: 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)
> > ??
> > 
> > ______________________________________________
> > R-help at stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide! 
> > http://www.R-project.org/posting-guide.html
> > 
>

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