Re: [R] matrix values linked to vector index
Just a further suggestion: vec - c(3,2,5,0,1) mat - t(sapply(vec,=,1:max(vec))) ifelse(mat,1,0) Cheers, Christoph 2013/10/11 arun smartpink...@yahoo.com: Hi, In the example you showed: m1- matrix(0,length(vec),max(vec)) 1*!upper.tri(m1) #or m1[!upper.tri(m1)] - rep(rep(1,length(vec)),vec) #But, in a case like below, perhaps: vec1- c(3,4,5) m2- matrix(0,length(vec1),max(vec1)) indx - cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE)) m2[indx]- 1 m2 # [,1] [,2] [,3] [,4] [,5] #[1,]11100 #[2,]11110 #[3,]11111 A.K. Hi- I'd like to create a matrix of 0's and 1's where the number of 1's in each row defined by the value indexed in another vector, and where the (value-1) is back-filled by 0's. For example, given the following vector: vec= c(1,2,3) I'd like to produce a matrix with dimensions (length(vec), max(vec)): 1,0,0 1,1,0 1,1,1 Thank you! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] matrix values linked to vector index
Also:
mat+0
#or
mat*1
#However, sapply() based solutions would be slower in large matrices.
#Speed
makeMat3 - function(x,n){
if(is.numeric(x)){
x - as.integer(round(x))
x}
stopifnot(is.integer(x))
if(length(x)=n max(x)=n){
indx-rep(rep(c(1,0),max(length(x),n)),rbind(x,n-x))
m1- matrix(indx,nc=n,byr=TRUE)
}
else if(length(x) n) {
indx-rep(rep(c(1,0),n),c(as.vector(rbind(x,n-x)),rep(c(0,n),n-length(x
m1-matrix(indx[seq_len(length(indx)-(n*(n-length(x],nc=n,byr=TRUE)
}
else print(paste(Not possible: Number of columns less than the maximum value
of , max(x), or length of vector))
m1
}
set.seed(124)
xtest- sample(0:9,1e7,replace=TRUE)
system.time(res3- makeMat3(xtest,9))
# user system elapsed
# 2.000 0.528 2.533
system.time({res4- t(sapply(xtest,=,1:max(xtest)))
res4- res4*1})
# user system elapsed
# 42.648 0.728 43.461
identical(res3,res4)
#[1] TRUE
A.K.
On Sunday, October 13, 2013 3:55 AM, Christoph Häni ch.ha...@gmail.com wrote:
Just a further suggestion:
vec - c(3,2,5,0,1)
mat - t(sapply(vec,=,1:max(vec)))
ifelse(mat,1,0)
Cheers,
Christoph
2013/10/11 arun smartpink...@yahoo.com:
Hi,
In the example you showed:
m1- matrix(0,length(vec),max(vec))
1*!upper.tri(m1)
#or
m1[!upper.tri(m1)] - rep(rep(1,length(vec)),vec)
#But, in a case like below, perhaps:
vec1- c(3,4,5)
m2- matrix(0,length(vec1),max(vec1))
indx -
cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE))
m2[indx]- 1
m2
# [,1] [,2] [,3] [,4] [,5]
#[1,] 1 1 1 0 0
#[2,] 1 1 1 1 0
#[3,] 1 1 1 1 1
A.K.
Hi-
I'd like to create a matrix of 0's and 1's where the number of
1's in each row defined by the value indexed in another vector, and
where the (value-1) is back-filled by 0's.
For example, given the following vector:
vec= c(1,2,3)
I'd like to produce a matrix with dimensions (length(vec), max(vec)):
1,0,0
1,1,0
1,1,1
Thank you!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] matrix values linked to vector index
Modified Bert's solution for non-square matrices. For the tested vectors, it
worked. There, could still be some bugs.
x1- c(3,2,1,4)
x2- c(2,0,4,3,1)
x3 - c(2, 1, 2.2)
x4 - c(a,1,3)
makeMat3 - function(x,n){
if(is.numeric(x)){
x - as.integer(round(x))
x}
stopifnot(is.integer(x))
indx-rep(rep(c(1,0),n),c(as.vector(rbind(x,n-x)),rep(c(0,n),n-length(x
matrix(indx[seq_len(length(indx)-(n*(n-length(x],nc=n,byr=TRUE)
}
makeMat3(x1,4)
makeMat3(x1,5)
makeMat3(x1,6)
makeMat3(x1,7)
makeMat3(x2,7)
makeMat3(x2,6)
makeMat3(x2,4) # as length of vector n
#Error in rep(c(0, n), n - length(x)) : invalid 'times' argument
makeMat3(x3,4)
makeMat3(x3,5)
makeMat3(x4,4)
#Error: is.integer(x) is not TRUE
makeMat3(c(4,0,1,0,6),6)
A.K.
On Saturday, October 12, 2013 1:40 AM, Bert Gunter gunter.ber...@gene.com
wrote:
Your examples are the problem:
On Fri, Oct 11, 2013 at 2:43 PM, arun smartpink...@yahoo.com wrote:
Seems like a bug in the code:
x- c(3,4,1)
n- 3
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
## This can't work since x specifies 4 1's in the second row but you
have specified a 3 column matrix with n.
n- 4
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
Yes, this shows that my claim that non-square matrices also work is
false. I leave it as an exercise to fix it so that it works for
non-square matrices.
Cheers,
Bert
x2
[1] 2 0 4 3 1
matrix(rep(rep(c(1,0),n),rbind(x2,n-x2)),nc=n,byr=TRUE)
Error in rep(rep(c(1, 0), n), rbind(x2, n - x2)) :
invalid 'times' argument
A.K.
On Friday, October 11, 2013 5:17 PM, Bert Gunter gunter.ber...@gene.com
wrote:
simpler (and sloppier) but with **no looping or apply's **
**IFF* the matrix is structured as in the OP's example, then lower.tri
(or upper.tri) should be used:
n - 4 ## number of columns in matrix -- note that I changed it from
the example; does not have to be square
x - 1:3 ## the number of 1's per row
lower.tri(matrix(0,nr=length(x),nc=n),diagA=TRUE)+0
A general, fast, but **tricky** way to do it that depends on knowing
that a matrix is just a vector in column major order is to generate
the vector using rep and then structure it as a matrix. eg.
x - c(3,2,1,4) ## your vector of indices
n - 4 ## number of columns in matrix ## does not have to be square
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
[,1] [,2] [,3] [,4]
[1,] 1 1 1 0
[2,] 1 1 0 0
[3,] 1 0 0 0
[4,] 1 1 1 1
Cheers,
Bert
On Fri, Oct 11, 2013 at 1:41 PM, Dennis Murphy djmu...@gmail.com wrote:
Attempting to follow the OP's conditions and assuming I understood
them correctly, here is one way to wrap this up into a function:
makeMat - function(x)
{
stopifnot(is.integer(x))
nr - length(x)
nc - max(x)
# Initialize a matrix of zeros
m - matrix(0, nr, nc)
# Conditionally replace with ones
for(i in seq_len(nr)) if(x[i] != 0) m[i, 1:x[i]] - 1
m
}
## Examples:
x1 - 1:3
x2 - as.integer(c(2, 0, 4, 3, 1))
x3 - c(2, 1, 2.2)
makeMat(x1)
makeMat(x2)
makeMat(x3)
makeMat(4:6)
On Fri, Oct 11, 2013 at 9:49 AM, arun smartpink...@yahoo.com wrote:
Hi,
In the example you showed:
m1- matrix(0,length(vec),max(vec))
1*!upper.tri(m1)
#or
m1[!upper.tri(m1)] - rep(rep(1,length(vec)),vec)
#But, in a case like below, perhaps:
vec1- c(3,4,5)
m2- matrix(0,length(vec1),max(vec1))
indx -
cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE))
m2[indx]- 1
m2
# [,1] [,2] [,3] [,4] [,5]
#[1,] 1 1 1 0 0
#[2,] 1 1 1 1 0
#[3,] 1 1 1 1 1
A.K.
Hi-
I'd like to create a matrix of 0's and 1's where the number of
1's in each row defined by the value indexed in another vector, and
where the (value-1) is back-filled by 0's.
For example, given the following vector:
vec= c(1,2,3)
I'd like to produce a matrix with dimensions (length(vec), max(vec)):
1,0,0
1,1,0
1,1,1
Thank you!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
--
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
__
R-help@r-project.org mailing list
Re: [R] matrix values linked to vector index
This seems to do it, but as I mentioned in my original post, my
solution was tricky. Thinking about it some more, I now realize that
it was too tricky and has the additional flaw of using underlying
representations of objects rather than their exposed interfaces -- i.e
it treats a matrix as a vector.
Here is, I think, a much better solution that treats a matrix as a
matrix by making use of a not-often-enough-used technique (mea culpa!)
of matrix indexing. It obviously needs to be cleaned up to check
inputs, etc. , but I think it should work. Feel free to publish and
clean up bugs.
makemx - function(x,n)
{
out - matrix(0, nr=length(x), nc=n)
ix - cbind(rep(seq_along(x),x),unlist(sapply(x,seq_len)))
out[ix]- 1
out
}
makemx(c(3,2,1,4),4)
[,1] [,2] [,3] [,4]
[1,]1110
[2,]1100
[3,]1000
[4,]1111
makemx(c(3,2,1,4),5)
[,1] [,2] [,3] [,4] [,5]
[1,]11100
[2,]11000
[3,]10000
[4,]11110
Cheers,
Bert
On Sat, Oct 12, 2013 at 1:02 AM, arun smartpink...@yahoo.com wrote:
Modified Bert's solution for non-square matrices. For the tested vectors, it
worked. There, could still be some bugs.
x1- c(3,2,1,4)
x2- c(2,0,4,3,1)
x3 - c(2, 1, 2.2)
x4 - c(a,1,3)
makeMat3 - function(x,n){
if(is.numeric(x)){
x - as.integer(round(x))
x}
stopifnot(is.integer(x))
indx-rep(rep(c(1,0),n),c(as.vector(rbind(x,n-x)),rep(c(0,n),n-length(x
matrix(indx[seq_len(length(indx)-(n*(n-length(x],nc=n,byr=TRUE)
}
makeMat3(x1,4)
makeMat3(x1,5)
makeMat3(x1,6)
makeMat3(x1,7)
makeMat3(x2,7)
makeMat3(x2,6)
makeMat3(x2,4) # as length of vector n
#Error in rep(c(0, n), n - length(x)) : invalid 'times' argument
makeMat3(x3,4)
makeMat3(x3,5)
makeMat3(x4,4)
#Error: is.integer(x) is not TRUE
makeMat3(c(4,0,1,0,6),6)
A.K.
On Saturday, October 12, 2013 1:40 AM, Bert Gunter gunter.ber...@gene.com
wrote:
Your examples are the problem:
On Fri, Oct 11, 2013 at 2:43 PM, arun smartpink...@yahoo.com wrote:
Seems like a bug in the code:
x- c(3,4,1)
n- 3
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
## This can't work since x specifies 4 1's in the second row but you
have specified a 3 column matrix with n.
n- 4
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
Yes, this shows that my claim that non-square matrices also work is
false. I leave it as an exercise to fix it so that it works for
non-square matrices.
Cheers,
Bert
x2
[1] 2 0 4 3 1
matrix(rep(rep(c(1,0),n),rbind(x2,n-x2)),nc=n,byr=TRUE)
Error in rep(rep(c(1, 0), n), rbind(x2, n - x2)) :
invalid 'times' argument
A.K.
On Friday, October 11, 2013 5:17 PM, Bert Gunter gunter.ber...@gene.com
wrote:
simpler (and sloppier) but with **no looping or apply's **
**IFF* the matrix is structured as in the OP's example, then lower.tri
(or upper.tri) should be used:
n - 4 ## number of columns in matrix -- note that I changed it from
the example; does not have to be square
x - 1:3 ## the number of 1's per row
lower.tri(matrix(0,nr=length(x),nc=n),diagA=TRUE)+0
A general, fast, but **tricky** way to do it that depends on knowing
that a matrix is just a vector in column major order is to generate
the vector using rep and then structure it as a matrix. eg.
x - c(3,2,1,4) ## your vector of indices
n - 4 ## number of columns in matrix ## does not have to be square
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
[,1] [,2] [,3] [,4]
[1,]1110
[2,]1100
[3,]1000
[4,]1111
Cheers,
Bert
On Fri, Oct 11, 2013 at 1:41 PM, Dennis Murphy djmu...@gmail.com wrote:
Attempting to follow the OP's conditions and assuming I understood
them correctly, here is one way to wrap this up into a function:
makeMat - function(x)
{
stopifnot(is.integer(x))
nr - length(x)
nc - max(x)
# Initialize a matrix of zeros
m - matrix(0, nr, nc)
# Conditionally replace with ones
for(i in seq_len(nr)) if(x[i] != 0) m[i, 1:x[i]] - 1
m
}
## Examples:
x1 - 1:3
x2 - as.integer(c(2, 0, 4, 3, 1))
x3 - c(2, 1, 2.2)
makeMat(x1)
makeMat(x2)
makeMat(x3)
makeMat(4:6)
On Fri, Oct 11, 2013 at 9:49 AM, arun smartpink...@yahoo.com wrote:
Hi,
In the example you showed:
m1- matrix(0,length(vec),max(vec))
1*!upper.tri(m1)
#or
m1[!upper.tri(m1)] - rep(rep(1,length(vec)),vec)
#But, in a case like below, perhaps:
vec1- c(3,4,5)
m2- matrix(0,length(vec1),max(vec1))
indx -
cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE))
m2[indx]- 1
m2
# [,1] [,2] [,3] [,4] [,5]
#[1,]11100
#[2,]11110
#[3,]
Re: [R] matrix values linked to vector index
This looks better. My previous solution (makeMatrix2) also did the matrix
indexing without using sapply() route. Replacing the max(x) by n for
non-symmetric matrix:
makeMatrix2- function(x,n){ #including n
if(is.numeric(x)){
x - as.integer(round(x))
x}
stopifnot(is.integer(x))
m1- matrix(0,length(x),n) #change max(x) to n
indx -
cbind(rep(seq_along(x),x),seq_len(sum(x))-rep(cumsum(c(0L,x[-length(x)])),x))
m1[indx]- 1
m1}
identical(makeMatrix2(x3,4),makemx(x3,4))
#[1] TRUE
identical(makeMatrix2(x1,5),makemx(x1,5))
#[1] TRUE
identical(makeMatrix2(x2,7),makemx(x2,7))
#[1] TRUE
A.K.
On Saturday, October 12, 2013 11:37 AM, Bert Gunter gunter.ber...@gene.com
wrote:
This seems to do it, but as I mentioned in my original post, my
solution was tricky. Thinking about it some more, I now realize that
it was too tricky and has the additional flaw of using underlying
representations of objects rather than their exposed interfaces -- i.e
it treats a matrix as a vector.
Here is, I think, a much better solution that treats a matrix as a
matrix by making use of a not-often-enough-used technique (mea culpa!)
of matrix indexing. It obviously needs to be cleaned up to check
inputs, etc. , but I think it should work. Feel free to publish and
clean up bugs.
makemx - function(x,n)
{
out - matrix(0, nr=length(x), nc=n)
ix - cbind(rep(seq_along(x),x),unlist(sapply(x,seq_len)))
out[ix]- 1
out
}
makemx(c(3,2,1,4),4)
[,1] [,2] [,3] [,4]
[1,] 1 1 1 0
[2,] 1 1 0 0
[3,] 1 0 0 0
[4,] 1 1 1 1
makemx(c(3,2,1,4),5)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 1 0 0
[2,] 1 1 0 0 0
[3,] 1 0 0 0 0
[4,] 1 1 1 1 0
Cheers,
Bert
On Sat, Oct 12, 2013 at 1:02 AM, arun smartpink...@yahoo.com wrote:
Modified Bert's solution for non-square matrices. For the tested vectors, it
worked. There, could still be some bugs.
x1- c(3,2,1,4)
x2- c(2,0,4,3,1)
x3 - c(2, 1, 2.2)
x4 - c(a,1,3)
makeMat3 - function(x,n){
if(is.numeric(x)){
x - as.integer(round(x))
x}
stopifnot(is.integer(x))
indx-rep(rep(c(1,0),n),c(as.vector(rbind(x,n-x)),rep(c(0,n),n-length(x
matrix(indx[seq_len(length(indx)-(n*(n-length(x],nc=n,byr=TRUE)
}
makeMat3(x1,4)
makeMat3(x1,5)
makeMat3(x1,6)
makeMat3(x1,7)
makeMat3(x2,7)
makeMat3(x2,6)
makeMat3(x2,4) # as length of vector n
#Error in rep(c(0, n), n - length(x)) : invalid 'times' argument
makeMat3(x3,4)
makeMat3(x3,5)
makeMat3(x4,4)
#Error: is.integer(x) is not TRUE
makeMat3(c(4,0,1,0,6),6)
A.K.
On Saturday, October 12, 2013 1:40 AM, Bert Gunter gunter.ber...@gene.com
wrote:
Your examples are the problem:
On Fri, Oct 11, 2013 at 2:43 PM, arun smartpink...@yahoo.com wrote:
Seems like a bug in the code:
x- c(3,4,1)
n- 3
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
## This can't work since x specifies 4 1's in the second row but you
have specified a 3 column matrix with n.
n- 4
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
Yes, this shows that my claim that non-square matrices also work is
false. I leave it as an exercise to fix it so that it works for
non-square matrices.
Cheers,
Bert
x2
[1] 2 0 4 3 1
matrix(rep(rep(c(1,0),n),rbind(x2,n-x2)),nc=n,byr=TRUE)
Error in rep(rep(c(1, 0), n), rbind(x2, n - x2)) :
invalid 'times' argument
A.K.
On Friday, October 11, 2013 5:17 PM, Bert Gunter gunter.ber...@gene.com
wrote:
simpler (and sloppier) but with **no looping or apply's **
**IFF* the matrix is structured as in the OP's example, then lower.tri
(or upper.tri) should be used:
n - 4 ## number of columns in matrix -- note that I changed it from
the example; does not have to be square
x - 1:3 ## the number of 1's per row
lower.tri(matrix(0,nr=length(x),nc=n),diagA=TRUE)+0
A general, fast, but **tricky** way to do it that depends on knowing
that a matrix is just a vector in column major order is to generate
the vector using rep and then structure it as a matrix. eg.
x - c(3,2,1,4) ## your vector of indices
n - 4 ## number of columns in matrix ## does not have to be square
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
[,1] [,2] [,3] [,4]
[1,] 1 1 1 0
[2,] 1 1 0 0
[3,] 1 0 0 0
[4,] 1 1 1 1
Cheers,
Bert
On Fri, Oct 11, 2013 at 1:41 PM, Dennis Murphy djmu...@gmail.com wrote:
Attempting to follow the OP's conditions and assuming I understood
them correctly, here is one way to wrap this up into a function:
makeMat - function(x)
{
stopifnot(is.integer(x))
nr - length(x)
nc - max(x)
# Initialize a matrix of zeros
m - matrix(0, nr, nc)
# Conditionally replace with ones
for(i in seq_len(nr)) if(x[i] != 0) m[i,
Re: [R] matrix values linked to vector index
Some speed comparison:
set.seed(124)
xtest- sample(0:9,1e7,replace=TRUE)
system.time({res1 - makemx(xtest,9)})
# user system elapsed
# 51.124 0.812 52.039
system.time({res2 - makeMatrix2(xtest,9)})
# user system elapsed
# 3.460 0.168 3.631
identical(res1,res2)
#[1] TRUE
Also, it looks like there is some bugs still in the makeMat3
system.time({res3 - makeMat3(xtest,9)})
#Error in rep(c(0, n), n - length(x)) : invalid 'times' argument
#Timing stopped at: 0.616 0 0.616
A.K.
On Saturday, October 12, 2013 12:06 PM, arun smartpink...@yahoo.com wrote:
This looks better. My previous solution (makeMatrix2) also did the matrix
indexing without using sapply() route. Replacing the max(x) by n for
non-symmetric matrix:
makeMatrix2- function(x,n){ #including n
if(is.numeric(x)){
x - as.integer(round(x))
x}
stopifnot(is.integer(x))
m1- matrix(0,length(x),n) #change max(x) to n
indx -
cbind(rep(seq_along(x),x),seq_len(sum(x))-rep(cumsum(c(0L,x[-length(x)])),x))
m1[indx]- 1
m1}
identical(makeMatrix2(x3,4),makemx(x3,4))
#[1] TRUE
identical(makeMatrix2(x1,5),makemx(x1,5))
#[1] TRUE
identical(makeMatrix2(x2,7),makemx(x2,7))
#[1] TRUE
A.K.
On Saturday, October 12, 2013 11:37 AM, Bert Gunter gunter.ber...@gene.com
wrote:
This seems to do it, but as I mentioned in my original post, my
solution was tricky. Thinking about it some more, I now realize that
it was too tricky and has the additional flaw of using underlying
representations of objects rather than their exposed interfaces -- i.e
it treats a matrix as a vector.
Here is, I think, a much better solution that treats a matrix as a
matrix by making use of a not-often-enough-used technique (mea culpa!)
of matrix indexing. It obviously needs to be cleaned up to check
inputs, etc. , but I think it should work. Feel free to publish and
clean up bugs.
makemx - function(x,n)
{
out - matrix(0, nr=length(x), nc=n)
ix - cbind(rep(seq_along(x),x),unlist(sapply(x,seq_len)))
out[ix]- 1
out
}
makemx(c(3,2,1,4),4)
[,1] [,2] [,3] [,4]
[1,] 1 1 1 0
[2,] 1 1 0 0
[3,] 1 0 0 0
[4,] 1 1 1 1
makemx(c(3,2,1,4),5)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 1 0 0
[2,] 1 1 0 0 0
[3,] 1 0 0 0 0
[4,] 1 1 1 1 0
Cheers,
Bert
On Sat, Oct 12, 2013 at 1:02 AM, arun smartpink...@yahoo.com wrote:
Modified Bert's solution for non-square matrices. For the tested vectors, it
worked. There, could still be some bugs.
x1- c(3,2,1,4)
x2- c(2,0,4,3,1)
x3 - c(2, 1, 2.2)
x4 - c(a,1,3)
makeMat3 - function(x,n){
if(is.numeric(x)){
x - as.integer(round(x))
x}
stopifnot(is.integer(x))
indx-rep(rep(c(1,0),n),c(as.vector(rbind(x,n-x)),rep(c(0,n),n-length(x
matrix(indx[seq_len(length(indx)-(n*(n-length(x],nc=n,byr=TRUE)
}
makeMat3(x1,4)
makeMat3(x1,5)
makeMat3(x1,6)
makeMat3(x1,7)
makeMat3(x2,7)
makeMat3(x2,6)
makeMat3(x2,4) # as length of vector n
#Error in rep(c(0, n), n - length(x)) : invalid 'times' argument
makeMat3(x3,4)
makeMat3(x3,5)
makeMat3(x4,4)
#Error: is.integer(x) is not TRUE
makeMat3(c(4,0,1,0,6),6)
A.K.
On Saturday, October 12, 2013 1:40 AM, Bert Gunter gunter.ber...@gene.com
wrote:
Your examples are the problem:
On Fri, Oct 11, 2013 at 2:43 PM, arun smartpink...@yahoo.com wrote:
Seems like a bug in the code:
x- c(3,4,1)
n- 3
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
## This can't work since x specifies 4 1's in the second row but you
have specified a 3 column matrix with n.
n- 4
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
Yes, this shows that my claim that non-square matrices also work is
false. I leave it as an exercise to fix it so that it works for
non-square matrices.
Cheers,
Bert
x2
[1] 2 0 4 3 1
matrix(rep(rep(c(1,0),n),rbind(x2,n-x2)),nc=n,byr=TRUE)
Error in rep(rep(c(1, 0), n), rbind(x2, n - x2)) :
invalid 'times' argument
A.K.
On Friday, October 11, 2013 5:17 PM, Bert Gunter gunter.ber...@gene.com
wrote:
simpler (and sloppier) but with **no looping or apply's **
**IFF* the matrix is structured as in the OP's example, then lower.tri
(or upper.tri) should be used:
n - 4 ## number of columns in matrix -- note that I changed it from
the example; does not have to be square
x - 1:3 ## the number of 1's per row
lower.tri(matrix(0,nr=length(x),nc=n),diagA=TRUE)+0
A general, fast, but **tricky** way to do it that depends on knowing
that a matrix is just a vector in column major order is to generate
the vector using rep and then structure it as a matrix. eg.
x - c(3,2,1,4) ## your vector of indices
n - 4 ## number of columns in matrix ## does not have to be square
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
[,1] [,2]
Re: [R] matrix values linked to vector index
Apologies, in case it is double posting (I think my previous message didn't
make it to the list)
Hope the new version of `makeMat3` is bug free:
makeMat3 - function(x,n){
if(is.numeric(x)){
x - as.integer(round(x))
x}
stopifnot(is.integer(x))
if(length(x)=n max(x)=n){
indx-rep(rep(c(1,0),max(length(x),n)),rbind(x,n-x))
m1- matrix(indx,nc=n,byr=TRUE)
}
else if(length(x) n) {
indx-rep(rep(c(1,0),n),c(as.vector(rbind(x,n-x)),rep(c(0,n),n-length(x
m1-matrix(indx[seq_len(length(indx)-(n*(n-length(x],nc=n,byr=TRUE)
}
else print(paste(Not possible: Number of columns less than the maximum value
of , max(x), or length of vector))
m1
}
system.time(res3- makeMat3(xtest,9))
# user system elapsed
# 1.908 0.328 2.237
identical(res3,res2)
#[1] TRUE
identical(makemx(x3,4),makeMat3(x3,4))
#[1] TRUE
identical(makemx(x2,5),makeMat3(x2,5))
#[1] TRUE
identical(makemx(x2,14),makeMat3(x2,14))
#[1] TRUE
makemx(x2,3)
#Error in out[ix] - 1 : subscript out of bounds
makeMat3(x2,3)
#[1] Not possible: Number of columns less than the maximum value of 4 or
length of vector
#Error in makeMat3(x2, 3) : object 'm1' not found
A.K.
On Saturday, October 12, 2013 12:28 PM, arun smartpink...@yahoo.com wrote:
Some speed comparison:
set.seed(124)
xtest- sample(0:9,1e7,replace=TRUE)
system.time({res1 - makemx(xtest,9)})
# user system elapsed
# 51.124 0.812 52.039
system.time({res2 - makeMatrix2(xtest,9)})
# user system elapsed
# 3.460 0.168 3.631
identical(res1,res2)
#[1] TRUE
Also, it looks like there is some bugs still in the makeMat3
system.time({res3 - makeMat3(xtest,9)})
#Error in rep(c(0, n), n - length(x)) : invalid 'times' argument
#Timing stopped at: 0.616 0 0.616
A.K.
On Saturday, October 12, 2013 12:06 PM, arun smartpink...@yahoo.com wrote:
This looks better. My previous solution (makeMatrix2) also did the matrix
indexing without using sapply() route. Replacing the max(x) by n for
non-symmetric matrix:
makeMatrix2- function(x,n){ #including n
if(is.numeric(x)){
x - as.integer(round(x))
x}
stopifnot(is.integer(x))
m1- matrix(0,length(x),n) #change max(x) to n
indx -
cbind(rep(seq_along(x),x),seq_len(sum(x))-rep(cumsum(c(0L,x[-length(x)])),x))
m1[indx]- 1
m1}
identical(makeMatrix2(x3,4),makemx(x3,4))
#[1] TRUE
identical(makeMatrix2(x1,5),makemx(x1,5))
#[1] TRUE
identical(makeMatrix2(x2,7),makemx(x2,7))
#[1] TRUE
A.K.
On Saturday, October 12, 2013 11:37 AM, Bert Gunter gunter.ber...@gene.com
wrote:
This seems to do it, but as I mentioned in my original post, my
solution was tricky. Thinking about it some more, I now realize that
it was too tricky and has the additional flaw of using underlying
representations of objects rather than their exposed interfaces -- i.e
it treats a matrix as a vector.
Here is, I think, a much better solution that treats a matrix as a
matrix by making use of a not-often-enough-used technique (mea culpa!)
of matrix indexing. It obviously needs to be cleaned up to check
inputs, etc. , but I think it should work. Feel free to publish and
clean up bugs.
makemx - function(x,n)
{
out - matrix(0, nr=length(x), nc=n)
ix - cbind(rep(seq_along(x),x),unlist(sapply(x,seq_len)))
out[ix]- 1
out
}
makemx(c(3,2,1,4),4)
[,1] [,2] [,3] [,4]
[1,] 1 1 1 0
[2,] 1 1 0 0
[3,] 1 0 0 0
[4,] 1 1 1 1
makemx(c(3,2,1,4),5)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 1 0 0
[2,] 1 1 0 0 0
[3,] 1 0 0 0 0
[4,] 1 1 1 1 0
Cheers,
Bert
On Sat, Oct 12, 2013 at 1:02 AM, arun smartpink...@yahoo.com wrote:
Modified Bert's solution for non-square matrices. For the tested vectors, it
worked. There, could still be some bugs.
x1- c(3,2,1,4)
x2- c(2,0,4,3,1)
x3 - c(2, 1, 2.2)
x4 - c(a,1,3)
makeMat3 - function(x,n){
if(is.numeric(x)){
x - as.integer(round(x))
x}
stopifnot(is.integer(x))
indx-rep(rep(c(1,0),n),c(as.vector(rbind(x,n-x)),rep(c(0,n),n-length(x
matrix(indx[seq_len(length(indx)-(n*(n-length(x],nc=n,byr=TRUE)
}
makeMat3(x1,4)
makeMat3(x1,5)
makeMat3(x1,6)
makeMat3(x1,7)
makeMat3(x2,7)
makeMat3(x2,6)
makeMat3(x2,4) # as length of vector n
#Error in rep(c(0, n), n - length(x)) : invalid 'times' argument
makeMat3(x3,4)
makeMat3(x3,5)
makeMat3(x4,4)
#Error: is.integer(x) is not TRUE
makeMat3(c(4,0,1,0,6),6)
A.K.
On Saturday, October 12, 2013 1:40 AM, Bert Gunter gunter.ber...@gene.com
wrote:
Your examples are the problem:
On Fri, Oct 11, 2013 at 2:43 PM, arun smartpink...@yahoo.com wrote:
Seems like a bug in the code:
x- c(3,4,1)
n- 3
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
## This can't work since x specifies 4 1's in the second row but you
have specified a 3 column matrix with n.
n- 4
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
Re: [R] matrix values linked to vector index
Hi, In the example you showed: m1- matrix(0,length(vec),max(vec)) 1*!upper.tri(m1) #or m1[!upper.tri(m1)] - rep(rep(1,length(vec)),vec) #But, in a case like below, perhaps: vec1- c(3,4,5) m2- matrix(0,length(vec1),max(vec1)) indx - cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE)) m2[indx]- 1 m2 # [,1] [,2] [,3] [,4] [,5] #[1,] 1 1 1 0 0 #[2,] 1 1 1 1 0 #[3,] 1 1 1 1 1 A.K. Hi- I'd like to create a matrix of 0's and 1's where the number of 1's in each row defined by the value indexed in another vector, and where the (value-1) is back-filled by 0's. For example, given the following vector: vec= c(1,2,3) I'd like to produce a matrix with dimensions (length(vec), max(vec)): 1,0,0 1,1,0 1,1,1 Thank you! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] matrix values linked to vector index
Attempting to follow the OP's conditions and assuming I understood
them correctly, here is one way to wrap this up into a function:
makeMat - function(x)
{
stopifnot(is.integer(x))
nr - length(x)
nc - max(x)
# Initialize a matrix of zeros
m - matrix(0, nr, nc)
# Conditionally replace with ones
for(i in seq_len(nr)) if(x[i] != 0) m[i, 1:x[i]] - 1
m
}
## Examples:
x1 - 1:3
x2 - as.integer(c(2, 0, 4, 3, 1))
x3 - c(2, 1, 2.2)
makeMat(x1)
makeMat(x2)
makeMat(x3)
makeMat(4:6)
On Fri, Oct 11, 2013 at 9:49 AM, arun smartpink...@yahoo.com wrote:
Hi,
In the example you showed:
m1- matrix(0,length(vec),max(vec))
1*!upper.tri(m1)
#or
m1[!upper.tri(m1)] - rep(rep(1,length(vec)),vec)
#But, in a case like below, perhaps:
vec1- c(3,4,5)
m2- matrix(0,length(vec1),max(vec1))
indx -
cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE))
m2[indx]- 1
m2
# [,1] [,2] [,3] [,4] [,5]
#[1,]11100
#[2,]11110
#[3,]11111
A.K.
Hi-
I'd like to create a matrix of 0's and 1's where the number of
1's in each row defined by the value indexed in another vector, and
where the (value-1) is back-filled by 0's.
For example, given the following vector:
vec= c(1,2,3)
I'd like to produce a matrix with dimensions (length(vec), max(vec)):
1,0,0
1,1,0
1,1,1
Thank you!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] matrix values linked to vector index
simpler (and sloppier) but with **no looping or apply's **
**IFF* the matrix is structured as in the OP's example, then lower.tri
(or upper.tri) should be used:
n - 4 ## number of columns in matrix -- note that I changed it from
the example; does not have to be square
x - 1:3 ## the number of 1's per row
lower.tri(matrix(0,nr=length(x),nc=n),diagA=TRUE)+0
A general, fast, but **tricky** way to do it that depends on knowing
that a matrix is just a vector in column major order is to generate
the vector using rep and then structure it as a matrix. eg.
x - c(3,2,1,4) ## your vector of indices
n - 4 ## number of columns in matrix ## does not have to be square
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
[,1] [,2] [,3] [,4]
[1,]1110
[2,]1100
[3,]1000
[4,]1111
Cheers,
Bert
On Fri, Oct 11, 2013 at 1:41 PM, Dennis Murphy djmu...@gmail.com wrote:
Attempting to follow the OP's conditions and assuming I understood
them correctly, here is one way to wrap this up into a function:
makeMat - function(x)
{
stopifnot(is.integer(x))
nr - length(x)
nc - max(x)
# Initialize a matrix of zeros
m - matrix(0, nr, nc)
# Conditionally replace with ones
for(i in seq_len(nr)) if(x[i] != 0) m[i, 1:x[i]] - 1
m
}
## Examples:
x1 - 1:3
x2 - as.integer(c(2, 0, 4, 3, 1))
x3 - c(2, 1, 2.2)
makeMat(x1)
makeMat(x2)
makeMat(x3)
makeMat(4:6)
On Fri, Oct 11, 2013 at 9:49 AM, arun smartpink...@yahoo.com wrote:
Hi,
In the example you showed:
m1- matrix(0,length(vec),max(vec))
1*!upper.tri(m1)
#or
m1[!upper.tri(m1)] - rep(rep(1,length(vec)),vec)
#But, in a case like below, perhaps:
vec1- c(3,4,5)
m2- matrix(0,length(vec1),max(vec1))
indx -
cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE))
m2[indx]- 1
m2
# [,1] [,2] [,3] [,4] [,5]
#[1,]11100
#[2,]11110
#[3,]11111
A.K.
Hi-
I'd like to create a matrix of 0's and 1's where the number of
1's in each row defined by the value indexed in another vector, and
where the (value-1) is back-filled by 0's.
For example, given the following vector:
vec= c(1,2,3)
I'd like to produce a matrix with dimensions (length(vec), max(vec)):
1,0,0
1,1,0
1,1,1
Thank you!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] matrix values linked to vector index
Thanks Dennis. I noticed I didn't take the 0 value into consideration and
also didn't check the unsorted vector.vec1- c(2,4,1)
is.numeric(vec1)
#[1] TRUE
makeMat(as.integer(vec1))
makeMatrix2- function(x){
if(is.numeric(x)){
x - as.integer(round(x))
x}
stopifnot(is.integer(x))
m1- matrix(0,length(x),max(x))
indx -
cbind(rep(seq_along(x),x),seq_len(sum(x))-rep(cumsum(c(0L,x[-length(x)])),x))
m1[indx]- 1
m1}
identical(makeMat(x1),makeMatrix2(x1))
#[1] TRUE
identical(makeMat(x2),makeMatrix2(x2))
#[1] TRUE
identical(makeMat(x3),makeMatrix2(x3))
#Error: is.integer(x) is not TRUE
makeMatrix2(x3)
# [,1] [,2]
#[1,] 1 1
#[2,] 1 0
#[3,] 1 1
identical(makeMat(4:6),makeMatrix2(4:6))
#[1] TRUE
x4 - c(a,1,3)
identical(makeMat(x4),makeMatrix2(x4))
#Error: is.integer(x) is not TRUE
A.K.
On Friday, October 11, 2013 4:41 PM, Dennis Murphy djmu...@gmail.com wrote:
Attempting to follow the OP's conditions and assuming I understood
them correctly, here is one way to wrap this up into a function:
makeMat - function(x)
{
stopifnot(is.integer(x))
nr - length(x)
nc - max(x)
# Initialize a matrix of zeros
m - matrix(0, nr, nc)
# Conditionally replace with ones
for(i in seq_len(nr)) if(x[i] != 0) m[i, 1:x[i]] - 1
m
}
## Examples:
x1 - 1:3
x2 - as.integer(c(2, 0, 4, 3, 1))
x3 - c(2, 1, 2.2)
makeMat(x1)
makeMat(x2)
makeMat(x3)
makeMat(4:6)
On Fri, Oct 11, 2013 at 9:49 AM, arun smartpink...@yahoo.com wrote:
Hi,
In the example you showed:
m1- matrix(0,length(vec),max(vec))
1*!upper.tri(m1)
#or
m1[!upper.tri(m1)] - rep(rep(1,length(vec)),vec)
#But, in a case like below, perhaps:
vec1- c(3,4,5)
m2- matrix(0,length(vec1),max(vec1))
indx -
cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE))
m2[indx]- 1
m2
# [,1] [,2] [,3] [,4] [,5]
#[1,] 1 1 1 0 0
#[2,] 1 1 1 1 0
#[3,] 1 1 1 1 1
A.K.
Hi-
I'd like to create a matrix of 0's and 1's where the number of
1's in each row defined by the value indexed in another vector, and
where the (value-1) is back-filled by 0's.
For example, given the following vector:
vec= c(1,2,3)
I'd like to produce a matrix with dimensions (length(vec), max(vec)):
1,0,0
1,1,0
1,1,1
Thank you!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] matrix values linked to vector index
Seems like a bug in the code:
x- c(3,4,1)
n- 3
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
n- 4
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
x2
[1] 2 0 4 3 1
matrix(rep(rep(c(1,0),n),rbind(x2,n-x2)),nc=n,byr=TRUE)
Error in rep(rep(c(1, 0), n), rbind(x2, n - x2)) :
invalid 'times' argument
A.K.
On Friday, October 11, 2013 5:17 PM, Bert Gunter gunter.ber...@gene.com wrote:
simpler (and sloppier) but with **no looping or apply's **
**IFF* the matrix is structured as in the OP's example, then lower.tri
(or upper.tri) should be used:
n - 4 ## number of columns in matrix -- note that I changed it from
the example; does not have to be square
x - 1:3 ## the number of 1's per row
lower.tri(matrix(0,nr=length(x),nc=n),diagA=TRUE)+0
A general, fast, but **tricky** way to do it that depends on knowing
that a matrix is just a vector in column major order is to generate
the vector using rep and then structure it as a matrix. eg.
x - c(3,2,1,4) ## your vector of indices
n - 4 ## number of columns in matrix ## does not have to be square
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
[,1] [,2] [,3] [,4]
[1,] 1 1 1 0
[2,] 1 1 0 0
[3,] 1 0 0 0
[4,] 1 1 1 1
Cheers,
Bert
On Fri, Oct 11, 2013 at 1:41 PM, Dennis Murphy djmu...@gmail.com wrote:
Attempting to follow the OP's conditions and assuming I understood
them correctly, here is one way to wrap this up into a function:
makeMat - function(x)
{
stopifnot(is.integer(x))
nr - length(x)
nc - max(x)
# Initialize a matrix of zeros
m - matrix(0, nr, nc)
# Conditionally replace with ones
for(i in seq_len(nr)) if(x[i] != 0) m[i, 1:x[i]] - 1
m
}
## Examples:
x1 - 1:3
x2 - as.integer(c(2, 0, 4, 3, 1))
x3 - c(2, 1, 2.2)
makeMat(x1)
makeMat(x2)
makeMat(x3)
makeMat(4:6)
On Fri, Oct 11, 2013 at 9:49 AM, arun smartpink...@yahoo.com wrote:
Hi,
In the example you showed:
m1- matrix(0,length(vec),max(vec))
1*!upper.tri(m1)
#or
m1[!upper.tri(m1)] - rep(rep(1,length(vec)),vec)
#But, in a case like below, perhaps:
vec1- c(3,4,5)
m2- matrix(0,length(vec1),max(vec1))
indx -
cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE))
m2[indx]- 1
m2
# [,1] [,2] [,3] [,4] [,5]
#[1,] 1 1 1 0 0
#[2,] 1 1 1 1 0
#[3,] 1 1 1 1 1
A.K.
Hi-
I'd like to create a matrix of 0's and 1's where the number of
1's in each row defined by the value indexed in another vector, and
where the (value-1) is back-filled by 0's.
For example, given the following vector:
vec= c(1,2,3)
I'd like to produce a matrix with dimensions (length(vec), max(vec)):
1,0,0
1,1,0
1,1,1
Thank you!
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] matrix values linked to vector index
Your examples are the problem:
On Fri, Oct 11, 2013 at 2:43 PM, arun smartpink...@yahoo.com wrote:
Seems like a bug in the code:
x- c(3,4,1)
n- 3
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
## This can't work since x specifies 4 1's in the second row but you
have specified a 3 column matrix with n.
n- 4
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
#Error in rep(rep(c(1, 0), n), rbind(x, n - x)) : invalid 'times' argument
Yes, this shows that my claim that non-square matrices also work is
false. I leave it as an exercise to fix it so that it works for
non-square matrices.
Cheers,
Bert
x2
[1] 2 0 4 3 1
matrix(rep(rep(c(1,0),n),rbind(x2,n-x2)),nc=n,byr=TRUE)
Error in rep(rep(c(1, 0), n), rbind(x2, n - x2)) :
invalid 'times' argument
A.K.
On Friday, October 11, 2013 5:17 PM, Bert Gunter gunter.ber...@gene.com
wrote:
simpler (and sloppier) but with **no looping or apply's **
**IFF* the matrix is structured as in the OP's example, then lower.tri
(or upper.tri) should be used:
n - 4 ## number of columns in matrix -- note that I changed it from
the example; does not have to be square
x - 1:3 ## the number of 1's per row
lower.tri(matrix(0,nr=length(x),nc=n),diagA=TRUE)+0
A general, fast, but **tricky** way to do it that depends on knowing
that a matrix is just a vector in column major order is to generate
the vector using rep and then structure it as a matrix. eg.
x - c(3,2,1,4) ## your vector of indices
n - 4 ## number of columns in matrix ## does not have to be square
matrix(rep(rep(c(1,0),n),rbind(x,n-x)),nc=n,byr=TRUE)
[,1] [,2] [,3] [,4]
[1,]1110
[2,]1100
[3,]1000
[4,]1111
Cheers,
Bert
On Fri, Oct 11, 2013 at 1:41 PM, Dennis Murphy djmu...@gmail.com wrote:
Attempting to follow the OP's conditions and assuming I understood
them correctly, here is one way to wrap this up into a function:
makeMat - function(x)
{
stopifnot(is.integer(x))
nr - length(x)
nc - max(x)
# Initialize a matrix of zeros
m - matrix(0, nr, nc)
# Conditionally replace with ones
for(i in seq_len(nr)) if(x[i] != 0) m[i, 1:x[i]] - 1
m
}
## Examples:
x1 - 1:3
x2 - as.integer(c(2, 0, 4, 3, 1))
x3 - c(2, 1, 2.2)
makeMat(x1)
makeMat(x2)
makeMat(x3)
makeMat(4:6)
On Fri, Oct 11, 2013 at 9:49 AM, arun smartpink...@yahoo.com wrote:
Hi,
In the example you showed:
m1- matrix(0,length(vec),max(vec))
1*!upper.tri(m1)
#or
m1[!upper.tri(m1)] - rep(rep(1,length(vec)),vec)
#But, in a case like below, perhaps:
vec1- c(3,4,5)
m2- matrix(0,length(vec1),max(vec1))
indx -
cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE))
m2[indx]- 1
m2
# [,1] [,2] [,3] [,4] [,5]
#[1,]11100
#[2,]11110
#[3,]11111
A.K.
Hi-
I'd like to create a matrix of 0's and 1's where the number of
1's in each row defined by the value indexed in another vector, and
where the (value-1) is back-filled by 0's.
For example, given the following vector:
vec= c(1,2,3)
I'd like to produce a matrix with dimensions (length(vec), max(vec)):
1,0,0
1,1,0
1,1,1
Thank you!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
--
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
