Re: [R] Matrix problem
On Apr 10, 2012, at 7:33 PM, Worik R wrote:
Friends
I am extracting sub-sets of the rows of a matrix. Generally the
result is
a matrix. But there is a special case. When the result returned is a
single row it is returned as a vector (in the example below an integer
vector). If there are 0, or more than 1 rows returned the result is a
matrix.
I am doing this in a function and I cannot be sure how many rows I am
removing. How can I do this in a general way that always returns a
matrix?
?[
M[1, , drop=FALSE]
a b c d
a1 0 3 2 1
class( M[1, , drop=FALSE] )
[1] matrix
M - matrix(0:3, nrow=3, ncol=4)
colnames(M) - c('a','b','c','d')
rownames(M) - c('a1','b2','c3')
N - M[M[,a]==0,]
O - M[M[,a]!=0,]
P - M[M[,a]==100,]
c(class(M), class(N), class(O), class(P))
[1] matrix integer matrix matrix
M
a b c d
a1 0 3 2 1
b2 1 0 3 2
c3 2 1 0 3
N
a b c d
0 3 2 1
O
a b c d
b2 1 0 3 2
c3 2 1 0 3
P
a b c d
cheers
Worik
[[alternative HTML version deleted]]
__
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.
David Winsemius, MD
West Hartford, CT
__
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 problem
Thank you.
That was exactly what I need.
Looking at '?[' I see...
drop: For matrices and arrays. If TRUE the result is coerced to
the lowest possible dimension (see the examples). This only
works for extracting elements, not for the replacement. See
drop for further details.
But that implies that in the case where 0 rows are returned it should be
coerced into zero dimensions. I am not quite sure what it would mean to be
coerced into zero dimensions, but if I read that without having seen the
actual behavior (impossible now) I would assume...
M[M[,a]==1000,] (from my example below)
would return NULL, which has class NULL rather than a matrix with zero
rows.
thanks
Worik
On Wed, Apr 11, 2012 at 11:54 AM, David Winsemius dwinsem...@comcast.netwrote:
On Apr 10, 2012, at 7:33 PM, Worik R wrote:
Friends
I am extracting sub-sets of the rows of a matrix. Generally the result is
a matrix. But there is a special case. When the result returned is a
single row it is returned as a vector (in the example below an integer
vector). If there are 0, or more than 1 rows returned the result is a
matrix.
I am doing this in a function and I cannot be sure how many rows I am
removing. How can I do this in a general way that always returns a
matrix?
?[
M[1, , drop=FALSE]
a b c d
a1 0 3 2 1
class( M[1, , drop=FALSE] )
[1] matrix
M - matrix(0:3, nrow=3, ncol=4)
colnames(M) - c('a','b','c','d')
rownames(M) - c('a1','b2','c3')
N - M[M[,a]==0,]
O - M[M[,a]!=0,]
P - M[M[,a]==100,]
c(class(M), class(N), class(O), class(P))
[1] matrix integer matrix matrix
M
a b c d
a1 0 3 2 1
b2 1 0 3 2
c3 2 1 0 3
N
a b c d
0 3 2 1
O
a b c d
b2 1 0 3 2
c3 2 1 0 3
P
a b c d
cheers
Worik
[[alternative HTML version deleted]]
__**
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/**
posting-guide.html http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
[[alternative HTML version deleted]]
__
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 problem
On Apr 10, 2012, at 8:01 PM, Worik R wrote:
Thank you.
That was exactly what I need.
Looking at '?[' I see...
drop: For matrices and arrays. If TRUE the result is coerced to
the lowest possible dimension (see the examples). This only
works for extracting elements, not for the replacement. See
drop for further details.
But that implies that in the case where 0 rows are returned it
should be coerced into zero dimensions. I am not quite sure what it
would mean to be coerced into zero dimensions, but if I read that
without having seen the actual behavior (impossible now) I would
assume...
M[M[,a]==1000,] (from my example below)
would return NULL, which has class NULL rather than a matrix with
zero rows.
Cue music:
http://www.youtube.com/watch?v=NzlG28B-R8Y
: That signpost up ahead ... You're entering into a land of shadow
and substance. You just crossedoer into the R Zen Zone.
... where the sounds of one hand clapping is the test question:
dim(M[0, ])
[1] 0 4
str(M[0, ])
int[0 , 1:4]
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : chr [1:4] a b c d
dim(M[1, ])
NULL
thanks
Worik
On Wed, Apr 11, 2012 at 11:54 AM, David Winsemius dwinsem...@comcast.net
wrote:
On Apr 10, 2012, at 7:33 PM, Worik R wrote:
Friends
I am extracting sub-sets of the rows of a matrix. Generally the
result is
a matrix. But there is a special case. When the result returned is a
single row it is returned as a vector (in the example below an integer
vector). If there are 0, or more than 1 rows returned the result is a
matrix.
I am doing this in a function and I cannot be sure how many rows I am
removing. How can I do this in a general way that always returns a
matrix?
?[
M[1, , drop=FALSE]
a b c d
a1 0 3 2 1
class( M[1, , drop=FALSE] )
[1] matrix
M - matrix(0:3, nrow=3, ncol=4)
colnames(M) - c('a','b','c','d')
rownames(M) - c('a1','b2','c3')
N - M[M[,a]==0,]
O - M[M[,a]!=0,]
P - M[M[,a]==100,]
c(class(M), class(N), class(O), class(P))
[1] matrix integer matrix matrix
M
a b c d
a1 0 3 2 1
b2 1 0 3 2
c3 2 1 0 3
N
a b c d
0 3 2 1
O
a b c d
b2 1 0 3 2
c3 2 1 0 3
P
a b c d
cheers
Worik
[[alternative HTML version deleted]]
__
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.
David Winsemius, MD
West Hartford, CT
David Winsemius, MD
West Hartford, CT
[[alternative HTML version deleted]]
__
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 problem to extract animal associations
set.seed(1) (DFid - data.frame( x = sample(1:20,10), y = sample(1:20,10), IDs = sapply(1:10,function(i) paste(ID,i,sep= require(spdep) coordinates(DFid) - ~x+y coords - coordinates(DFid) dnn4 - dnearneigh(DFid,0,4) summary(dnn4) plot(DFid) plot(dnn4,coords,add=T,col=2) nb2mat(dnn4, zero.policy=TRUE) This just one option from the multitude of spatial packages. HTH On Sun, Feb 26, 2012 at 4:55 PM, Ross Dwyer ross.dw...@uq.edu.au wrote: Dear List, I have been trying to extract associations from a matrix whereby individual locations are within a certain distance threshold from one another. I have been able to extract those individuals where there is 'no interaction' (i.e. where these individuals are not within a specified distance threshold from another individual) and give these individuals a unique Group ID containing that one individual. i.e. ID Group 1 ID1 1 2 ID3 2 3 ID4 3 4 ID5 4 5 ID7 5 6 ID8 6 7 ID9 7 What I need assistance with is allocating associations with a unique group id. i.e. If we have interactions between ID2_ID6, ID6_ID2, ID6_ID10, ID10_ID6 as in the example code... ID Group 1 ID1 1 2 ID3 2 3 ID4 3 4 ID5 4 5 ID7 5 6 ID8 6 7 ID9 7 ## 8 ID2 8 9 ID6 8 10 ID10 8 ## The code also needs to robust enough to recognize instances where we have an interaction in a separate group... i.e. ID11_ID12 should be in a separate group (Group 9) as they don't interact with IDs 2, 6, or 10 (not in below code!) 11 ID11 9 12 ID12 9 I've been trying to figure this out but have drawn a blank. My example code can be found below. Very best wishes, Ross Dr Ross Dwyer Postdoctoral Research Fellow University of Queensland ### require(stats) x - sample(1:20,10) y - sample(1:20,10) IDs - sapply(1:10,function(i) paste(ID,i,sep=)) (DFid - data.frame(x,y)) x y 1 7 20 2 5 3 3 12 5 4 3 12 5 18 19 6 2 1 7 19 15 8 20 11 9 13 14 10 1 2 (DMdist - dist(DFid, method = euclidean, + diag = FALSE, upper = TRUE)) 1 2 3 4 5 6 7 8 9 10 1 17.117243 15.811388 8.944272 11.045361 19.646883 13.00 15.811388 8.485281 18.973666 2 17.117243 7.280110 9.219544 20.615528 3.605551 18.439089 17.00 13.601471 4.123106 3 15.811388 7.280110 11.401754 15.231546 10.770330 12.206556 10.00 9.055385 11.401754 4 8.944272 9.219544 11.401754 16.552945 11.045361 16.278821 17.029386 10.198039 10.198039 5 11.045361 20.615528 15.231546 16.552945 24.083189 4.123106 8.246211 7.071068 24.041631 6 19.646883 3.605551 10.770330 11.045361 24.083189 22.022716 20.591260 17.029386 1.414214 7 13.00 18.439089 12.206556 16.278821 4.123106 22.022716 4.123106 6.082763 22.203603 8 15.811388 17.00 10.00 17.029386 8.246211 20.591260 4.123106 7.615773 21.023796 9 8.485281 13.601471 9.055385 10.198039 7.071068 17.029386 6.082763 7.615773 16.970563 10 18.973666 4.123106 11.401754 10.198039 24.041631 1.414214 22.203603 21.023796 16.970563 #Generate True/False matrix on those individuals 4 units apart DMTF - apply(as.matrix(DMdist), c(1,2), function(x) ifelse(x=4,T,F)) diag(DMTF)- NA #replace diagonal with NA dimnames(DMTF) - list(IDs, IDs) #add individual's name to matrix DMTF ID1 ID2 ID3 ID4 ID5 ID6 ID7 ID8 ID9 ID10 ID1 NA FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE ID2 FALSE NA FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE ID3 FALSE FALSE NA FALSE FALSE FALSE FALSE FALSE FALSE FALSE ID4 FALSE FALSE FALSE NA FALSE FALSE FALSE FALSE FALSE FALSE ID5 FALSE FALSE FALSE FALSE NA FALSE FALSE FALSE FALSE FALSE ID6 FALSE TRUE FALSE FALSE FALSE NA FALSE FALSE FALSE TRUE ID7 FALSE FALSE FALSE FALSE FALSE FALSE NA FALSE FALSE FALSE ID8 FALSE FALSE FALSE FALSE FALSE FALSE FALSE NA FALSE FALSE ID9 FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE NA FALSE ID10 FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE NA irow - as.character(gl(length(IDs),length(IDs),labels=IDs)) icol -rep(IDs, length(IDs)) AssocMatrix - matrix(data=paste(irow,_,icol,_,sep=), + nrow = length(IDs), ncol = length(IDs), + dimnames= list(IDs, IDs)) AssocMatrix ID1 ID2 ID3 ID4 ID5 ID6 ID7 ID8 ID9 ID10 ID1 ID1_ID1_ ID2_ID1_ ID3_ID1_ ID4_ID1_ ID5_ID1_ ID6_ID1_ ID7_ID1_ ID8_ID1_ ID9_ID1_ ID10_ID1_ ID2 ID1_ID2_ ID2_ID2_ ID3_ID2_ ID4_ID2_ ID5_ID2_ ID6_ID2_ ID7_ID2_ ID8_ID2_ ID9_ID2_ ID10_ID2_ ID3 ID1_ID3_ ID2_ID3_ ID3_ID3_ ID4_ID3_ ID5_ID3_ ID6_ID3_ ID7_ID3_ ID8_ID3_ ID9_ID3_
Re: [R] matrix problem
How about: y - c(1,1,1,3,2) m - matrix(0, nrow=length(y), ncol=4) m[y==1, ] - matrix(1:4, nrow=sum(y == 1), ncol=4, byrow=TRUE) or, depending on your actual problem y - c(1,1,1,3,2) m - matrix(0, nrow=length(y), ncol=4) m[y == 1,] - col(m[y == 1,]) Sarah On Mon, Jun 20, 2011 at 3:54 PM, Costis Ghionnis conigh...@gmail.com wrote: Hallo everyone! I have a problem about creating a matrix... Suppose we have a vector y-c(1,1,1,3,2) and a zero matrix, m ,with nrows=length(y) and ncol=4. The matrix would look like this: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 I want to change the first three rows with the vector c(1,2,3,4). I thought that with the command m[y==1,1:4]-c(1,2,3,4) i would get 1 2 3 4 1 2 3 4 1 2 3 4 0 0 0 0 0 0 0 0 but instead i am getting 1 4 3 2 2 1 4 3 3 2 1 4 0 0 0 0 0 0 0 0 It seems it is filling the data by col instead by row. I want to use this technique in more complicated problems. So i do not want to have to work with the transpose matrix. Do you know another way to make this work. Thank you... -- Sarah Goslee http://www.functionaldiversity.org __ 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 problem
On Jun 20, 2011, at 3:54 PM, Costis Ghionnis wrote: Hallo everyone! I have a problem about creating a matrix... Suppose we have a vector y-c(1,1,1,3,2) and a zero matrix, m ,with nrows=length(y) and ncol=4. The matrix would look like this: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 I want to change the first three rows with the vector c(1,2,3,4). I thought that with the command m[y==1,1:4]-c(1,2,3,4) i would get 1 2 3 4 1 2 3 4 1 2 3 4 0 0 0 0 0 0 0 0 Try: m[y==1,1:4]-rep( c(1,2,3,4), each= sum(y==1) ) m [,1] [,2] [,3] [,4] [1,]1234 [2,]1234 [3,]1234 [4,]0000 [5,]0000 -- David. but instead i am getting 1 4 3 2 2 1 4 3 3 2 1 4 0 0 0 0 0 0 0 0 It seems it is filling the data by col instead by row. I want to use this technique in more complicated problems. So i do not want to have to work with the transpose matrix. Do you know another way to make this work. 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. David Winsemius, MD West Hartford, CT __ 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 problem
Try this: Lines - '1 2 0.1 2 3 0.2 3 1 0.3' DF - read.table(textConnection(Lines)) m - matrix(0, ncol = nrow(DF), nrow = nrow(DF)) m[as.matrix(DF[1:2])] - DF[[3]] On Tue, Aug 10, 2010 at 3:03 PM, zhenjiang xu zhenjiang...@gmail.comwrote: Hi, I have a file like this: 1 2 0.1 2 3 0.2 3 1 0.3 And I want to read it to create a matrix like this: [,1] [,2][,3] [1,]0 0.1 0 [2,]0 00.2 [3,]0.300 How can I do it efficiently? Thanks. -- Best, Zhenjiang [[alternative HTML version deleted]] __ 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ 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 problem
Let me give you a not that efficient one... assume you have read the matrix (named as x) into R: n=dim(x)[1] y=matrix(0,n,n) for (i in 1:n) y[x[i,1],x[i,2]]=x[i,3] -- View this message in context: http://r.789695.n4.nabble.com/matrix-problem-tp2320193p2320219.html Sent from the R help mailing list archive at Nabble.com. __ 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 problem
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of zhenjiang xu Sent: Tuesday, August 10, 2010 11:03 AM To: R-help@r-project.org Subject: [R] matrix problem Hi, I have a file like this: 1 2 0.1 2 3 0.2 3 1 0.3 And I want to read it to create a matrix like this: [,1] [,2][,3] [1,]0 0.1 0 [2,]0 00.2 [3,]0.300 How can I do it efficiently? Thanks. Use a k-column matrix as a subscript into your k-dimensional output array. (k is 2 in your case.) E.g., 'input' is your matrix in a form that one can paste into an R session: input - cbind(c(1,2,3), c(2,3,1), c(.1,.2,.3)) size - max(input[,1:2]) # you may want something else here output - matrix(0.0, size, size) output[input[,1:2]] - input[,3] output [,1] [,2] [,3] [1,] 0.0 0.1 0.0 [2,] 0.0 0.0 0.2 [3,] 0.3 0.0 0.0 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -- Best, Zhenjiang [[alternative HTML version deleted]] __ 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 problem
Hi, I guess you just want to reshape your data to wide format. strs - Index Time Value 1 2 0.1 2 3 0.2 3 1 0.3 DF - read.table(textConnection(strs),header=T) rDF - reshape(DF, idvar=Index, timevar=Time, direction=wide) rDF[is.na(rDF)] - 0 - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/matrix-problem-tp2320193p2320287.html Sent from the R help mailing list archive at Nabble.com. __ 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 problem
Thanks everyone for your help! Bill __ 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 problem
Try this: m - matrix(0, nrow = 5, ncol = 4) diag(m) - x[1] diag(m[-1,]) - x[2] On Sat, Jul 4, 2009 at 12:17 PM, William Simpson william.a.simp...@gmail.com wrote: Can anybody please tell me a good way to do the following? Given a vector, number of rows and number of columns, return a matrix as follows. Easiest to give an example: x=c(1,2), nrow=5, ncol=4 return the matrix: 1 0 0 0 2 1 0 0 0 2 1 0 0 0 2 1 0 0 0 2 Thanks very much for any help! Bill __ 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ 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 problem
On Jul 4, 2009, at 11:17 AM, William Simpson wrote: Can anybody please tell me a good way to do the following? Given a vector, number of rows and number of columns, return a matrix as follows. Easiest to give an example: x=c(1,2), nrow=5, ncol=4 You ought to separate those assignments with semicolons: x=c(1,2); nrow=5; ncol=4 MM - matrix( c(x,rep(0, nrow-length(x)+1)), nrow=nrow, ncol=ncol) Warning message: In matrix(c(x, rep(0, nrow - length(x) + 1)), nrow = nrow, ncol = ncol) : data length [6] is not a sub-multiple or multiple of the number of rows [5] MM [,1] [,2] [,3] [,4] [1,]1000 [2,]2100 [3,]0210 [4,]0021 [5,]0002 OR-- M2 - diag( , nrow=5, ncol=4) M2[row(M2) == col(M2)+1] - 2 M2 [,1] [,2] [,3] [,4] [1,]1000 [2,]2100 [3,]0210 [4,]0021 [5,]0002 return the matrix: 1 0 0 0 2 1 0 0 0 2 1 0 0 0 2 1 0 0 0 2 Thanks very much for any help! Bill __ David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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 problem
Thanks everyone for the help.
I should have said that I want to do this generally, not as a one-off.
So I want a function to do it. Like this
tp-function(x, nr, nc)
{
matrix( c(x,rep(0, nr-length(x)+1)), nrow=nr, ncol=nc)
}
tp(x=c(1,2), nr=5, nc=4)
This one looks good -- the warning message is annoying though...
Bill
__
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 problem
On Sat, 4 Jul 2009, William Simpson wrote: Can anybody please tell me a good way to do the following? Not sure how you want cases like x = 1:3, nrow=2 , ncol=7 to be handled, but for the example you give, this works: mat - matrix(0,nr=nrow,nc=ncol) indx - outer( seq( from=0, length=length(x) ), seq( from=1, by=nrow+1, length=ncol ), + ) mat[ indx ] - x HTH, Chuck Given a vector, number of rows and number of columns, return a matrix as follows. Easiest to give an example: x=c(1,2), nrow=5, ncol=4 return the matrix: 1 0 0 0 2 1 0 0 0 2 1 0 0 0 2 1 0 0 0 2 Thanks very much for any help! Bill __ 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. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:cbe...@tajo.ucsd.edu UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ 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 problem
Dear William,
Here is one way using Henrique's solution:
Make - function(x, nR, nC){
m - matrix(0, nrow = nR, ncol = nC)
diag(m) - x[1]
diag(m[-1,]) - x[2]
m
}
Make(x = c(1,2), nR = 5, nC = 4)
HTH,
Jorge
On Sat, Jul 4, 2009 at 11:59 AM, William Simpson
william.a.simp...@gmail.com wrote:
Thanks everyone for the help.
I should have said that I want to do this generally, not as a one-off.
So I want a function to do it. Like this
tp-function(x, nr, nc)
{
matrix( c(x,rep(0, nr-length(x)+1)), nrow=nr, ncol=nc)
}
tp(x=c(1,2), nr=5, nc=4)
This one looks good -- the warning message is annoying though...
Bill
__
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.
[[alternative HTML version deleted]]
__
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 problem
On Jul 4, 2009, at 11:59 AM, William Simpson wrote:
Thanks everyone for the help.
I should have said that I want to do this generally, not as a one-off.
So I want a function to do it. Like this
tp-function(x, nr, nc)
{
matrix( c(x,rep(0, nr-length(x)+1)), nrow=nr, ncol=nc)
}
tp(x=c(1,2), nr=5, nc=4)
This one looks good -- the warning message is annoying though...
tp-function(x, nr, nc)
+ {suppressWarnings(
+ matrix( c(x,rep(0, nr-length(x)+1)), nrow=nr, ncol=nc) )
+ }
tp(x=c(1,2), nr=5, nc=4)
[,1] [,2] [,3] [,4]
[1,]1000
[2,]2100
[3,]0210
[4,]0021
[5,]0002
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
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 problem
Doesn't work:
Make(x=c(2,1,1,1),nR=5,nC=2)
[,1] [,2]
[1,]20
[2,]12
[3,]01
[4,]00
[5,]00
should be
[,1] [,2]
[1,]20
[2,]12
[3,]11
[4,]11
[5,]01
On Sat, Jul 4, 2009 at 5:04 PM, Jorge Ivan
Velezjorgeivanve...@gmail.com wrote:
Dear William,
Here is one way using Henrique's solution:
Make - function(x, nR, nC){
m - matrix(0, nrow = nR, ncol = nC)
diag(m) - x[1]
diag(m[-1,]) - x[2]
m
}
Make(x = c(1,2), nR = 5, nC = 4)
HTH,
Jorge
On Sat, Jul 4, 2009 at 11:59 AM, William Simpson
william.a.simp...@gmail.com wrote:
Thanks everyone for the help.
I should have said that I want to do this generally, not as a one-off.
So I want a function to do it. Like this
tp-function(x, nr, nc)
{
matrix( c(x,rep(0, nr-length(x)+1)), nrow=nr, ncol=nc)
}
tp(x=c(1,2), nr=5, nc=4)
This one looks good -- the warning message is annoying though...
Bill
__
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 problem
Try this:
foo - function(x, nrow, ncol){
m - matrix(0, nrow = nrow, ncol = ncol)
m[cbind(unlist(lapply(0:(ncol - 1), `+`, seq(x))),rep(1:ncol, each =
length(x)))] - x
m
}
foo(c(2, 1, 1, 1), nrow = 5, ncol = 2)
On Sat, Jul 4, 2009 at 1:26 PM, William Simpson william.a.simp...@gmail.com
wrote:
Doesn't work:
Make(x=c(2,1,1,1),nR=5,nC=2)
[,1] [,2]
[1,]20
[2,]12
[3,]01
[4,]00
[5,]00
should be
[,1] [,2]
[1,]20
[2,]12
[3,]11
[4,]11
[5,]01
On Sat, Jul 4, 2009 at 5:04 PM, Jorge Ivan
Velezjorgeivanve...@gmail.com wrote:
Dear William,
Here is one way using Henrique's solution:
Make - function(x, nR, nC){
m - matrix(0, nrow = nR, ncol = nC)
diag(m) - x[1]
diag(m[-1,]) - x[2]
m
}
Make(x = c(1,2), nR = 5, nC = 4)
HTH,
Jorge
On Sat, Jul 4, 2009 at 11:59 AM, William Simpson
william.a.simp...@gmail.com wrote:
Thanks everyone for the help.
I should have said that I want to do this generally, not as a one-off.
So I want a function to do it. Like this
tp-function(x, nr, nc)
{
matrix( c(x,rep(0, nr-length(x)+1)), nrow=nr, ncol=nc)
}
tp(x=c(1,2), nr=5, nc=4)
This one looks good -- the warning message is annoying though...
Bill
__
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.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
__
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: Problem with the code
Well, mat doesn't have any dimensions / isn't a matrix, and we don't
know what p is supposed to be. But leaving aside those little details,
do you perhaps want something like this:
x-c(23,67,2,87,9,63,8,2,35,6,91,41,22,3)
p - 5
mat- matrix(0, nrow=p, ncol=length(x))
for(j in 1:length(x))
{
for(i in 1:p)
mat[i,j]-x[j]^i
}
Two notes: I didn't try it out, and if that's what you want rather
than a toy example
of a larger problem, there are more elegant ways to do it in R.
Sarah
On Fri, Jan 9, 2009 at 6:42 PM, Bhargab Chattopadhyay
bharga...@yahoo.com wrote:
Hi,
Can any one please explain why the following code doesn't work? Or can anyone
suggest an alternative.
Suppose
x-c(23,67,2,87,9,63,8,2,35,6,91,41,22,3)
mat-0;
for(j in 1:length(x))
{
for(i in 1:p)
mat[i,j]-x[j]^i;
}
Actually I want to have a matrix with p columns such that each column will
have the elements of x^(column#).
Thanks in advance.
Bhargab
--
Sarah Goslee
http://www.functionaldiversity.org
__
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: Problem with the code
One of those more elegant ways:
outer(x, 1:p, ^)
Charlotte
On Fri, Jan 9, 2009 at 4:24 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
Well, mat doesn't have any dimensions / isn't a matrix, and we don't
know what p is supposed to be. But leaving aside those little details,
do you perhaps want something like this:
x-c(23,67,2,87,9,63,8,2,35,6,91,41,22,3)
p - 5
mat- matrix(0, nrow=p, ncol=length(x))
for(j in 1:length(x))
{
for(i in 1:p)
mat[i,j]-x[j]^i
}
Two notes: I didn't try it out, and if that's what you want rather
than a toy example
of a larger problem, there are more elegant ways to do it in R.
Sarah
On Fri, Jan 9, 2009 at 6:42 PM, Bhargab Chattopadhyay
bharga...@yahoo.com wrote:
Hi,
Can any one please explain why the following code doesn't work? Or can
anyone suggest an alternative.
Suppose
x-c(23,67,2,87,9,63,8,2,35,6,91,41,22,3)
mat-0;
for(j in 1:length(x))
{
for(i in 1:p)
mat[i,j]-x[j]^i;
}
Actually I want to have a matrix with p columns such that each column
will have the elements of x^(column#).
Thanks in advance.
Bhargab
--
Sarah Goslee
http://www.functionaldiversity.org
__
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: Problem with the code
Charlotte: I ran your code because I wasn't clear on it and your way
would cause more matrices than the person requested. So
I think the code below it, although not too short, does what the person
asked. Thanks though because I understand outer better now.
temp - matrix(c(1,2,3,4,5,6),ncol=2)
print(temp)
#One of those more elegant ways:
print(temp)
outer(temp,1:p,'^')One of those more elegant ways:
# THIS WAY I THINK GIVES WHAT THEY WANT
sapply(1:ncol(temp), function(.col) {
temp[,.col]^.col
})
On Fri, Jan 9, 2009 at 7:40 PM, Charlotte Wickham wrote:
One of those more elegant ways:
outer(x, 1:p, ^)
Charlotte
On Fri, Jan 9, 2009 at 4:24 PM, Sarah Goslee sarah.gos...@gmail.com
wrote:
Well, mat doesn't have any dimensions / isn't a matrix, and we don't
know what p is supposed to be. But leaving aside those little
details,
do you perhaps want something like this:
x-c(23,67,2,87,9,63,8,2,35,6,91,41,22,3)
p - 5
mat- matrix(0, nrow=p, ncol=length(x))
for(j in 1:length(x))
{
for(i in 1:p)
mat[i,j]-x[j]^i
}
Two notes: I didn't try it out, and if that's what you want rather
than a toy example
of a larger problem, there are more elegant ways to do it in R.
Sarah
On Fri, Jan 9, 2009 at 6:42 PM, Bhargab Chattopadhyay
bharga...@yahoo.com wrote:
Hi,
Can any one please explain why the following code doesn't work? Or
can anyone suggest an alternative.
Suppose
x-c(23,67,2,87,9,63,8,2,35,6,91,41,22,3)
mat-0;
for(j in 1:length(x))
{
for(i in 1:p)
mat[i,j]-x[j]^i;
}
Actually I want to have a matrix with p columns such that each
column will have the elements of x^(column#).
Thanks in advance.
Bhargab
--
Sarah Goslee
http://www.functionaldiversity.org
__
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.
__
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: Problem with the code
On Fri, Jan 9, 2009 at 6:36 PM, markle...@verizon.net wrote:
Charlotte: I ran your code because I wasn't clear on it and your way would
cause more matrices than the person requested.
Bhargab gave us
x-c(23,67,2,87,9,63,8,2,35,6,91,41,22,3)
and said: I want to have a matrix with p columns such that each
column will have the elements of x^(column#).
so, I think Charlotte's code was spot-on:
p - 3
outer(x, 1:p, '^')
[,1] [,2] [,3]
[1,] 23 529 12167
[2,] 67 4489 300763
[3,]24 8
[4,] 87 7569 658503
[5,]9 81729
[6,] 63 3969 250047
[7,]8 64512
[8,]24 8
[9,] 35 1225 42875
[10,]6 36216
[11,] 91 8281 753571
[12,] 41 1681 68921
[13,] 22 484 10648
[14,]39 27
Here's another way -- a bit less elegant, but a gentle
introduction to thinking in vectors rather than elements:
mat - matrix(0,nrow=length(x), ncol=p)
for(i in 1:p) mat[,i] - x^i
mat
[,1] [,2] [,3]
[1,] 23 529 12167
[2,] 67 4489 300763
[3,]24 8
[4,] 87 7569 658503
[5,]9 81729
[6,] 63 3969 250047
[7,]8 64512
[8,]24 8
[9,] 35 1225 42875
[10,]6 36216
[11,] 91 8281 753571
[12,] 41 1681 68921
[13,] 22 484 10648
[14,]39 27
best,
Kingsford Jones
So
I think the code below it, although not too short, does what the person
asked. Thanks though because I understand outer better now.
temp - matrix(c(1,2,3,4,5,6),ncol=2)
print(temp)
#One of those more elegant ways:
print(temp)
outer(temp,1:p,'^')One of those more elegant ways:
# THIS WAY I THINK GIVES WHAT THEY WANT
sapply(1:ncol(temp), function(.col) {
temp[,.col]^.col
})
On Fri, Jan 9, 2009 at 7:40 PM, Charlotte Wickham wrote:
One of those more elegant ways:
outer(x, 1:p, ^)
Charlotte
On Fri, Jan 9, 2009 at 4:24 PM, Sarah Goslee sarah.gos...@gmail.com
wrote:
Well, mat doesn't have any dimensions / isn't a matrix, and we don't
know what p is supposed to be. But leaving aside those little details,
do you perhaps want something like this:
x-c(23,67,2,87,9,63,8,2,35,6,91,41,22,3)
p - 5
mat- matrix(0, nrow=p, ncol=length(x))
for(j in 1:length(x))
{
for(i in 1:p)
mat[i,j]-x[j]^i
}
Two notes: I didn't try it out, and if that's what you want rather
than a toy example
of a larger problem, there are more elegant ways to do it in R.
Sarah
On Fri, Jan 9, 2009 at 6:42 PM, Bhargab Chattopadhyay
bharga...@yahoo.com wrote:
Hi,
Can any one please explain why the following code doesn't work? Or can
anyone suggest an alternative.
Suppose
x-c(23,67,2,87,9,63,8,2,35,6,91,41,22,3)
mat-0;
for(j in 1:length(x))
{
for(i in 1:p)
mat[i,j]-x[j]^i;
}
Actually I want to have a matrix with p columns such that each column
will have the elements of x^(column#).
Thanks in advance.
Bhargab
--
Sarah Goslee
http://www.functionaldiversity.org
__
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.
__
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: Problem with the code
Thanks Kingsford. I thought the column power was supposed to be just for
that column but you're probably correct. English has its oddities
because if one reads the actual sentence the person wrote it's still not
clear, atleast to me.
Actually I want to have a matrix with p columns such that each column
will have the elements of x^(column#)
Thanks and apologies to Charlotte for my incorrect correction.
On Fri, Jan 9, 2009 at 9:37 PM, Kingsford Jones wrote:
On Fri, Jan 9, 2009 at 6:36 PM, markle...@verizon.net wrote:
Charlotte: I ran your code because I wasn't clear on it and your way
would
cause more matrices than the person requested.
Bhargab gave us
x-c(23,67,2,87,9,63,8,2,35,6,91,41,22,3)
and said: I want to have a matrix with p columns such that each
column will have the elements of x^(column#).
so, I think Charlotte's code was spot-on:
p - 3
outer(x, 1:p, '^')
[,1] [,2] [,3]
[1,] 23 529 12167
[2,] 67 4489 300763
[3,]24 8
[4,] 87 7569 658503
[5,]9 81729
[6,] 63 3969 250047
[7,]8 64512
[8,]24 8
[9,] 35 1225 42875
[10,]6 36216
[11,] 91 8281 753571
[12,] 41 1681 68921
[13,] 22 484 10648
[14,]39 27
Here's another way -- a bit less elegant, but a gentle
introduction to thinking in vectors rather than elements:
mat - matrix(0,nrow=length(x), ncol=p)
for(i in 1:p) mat[,i] - x^i
mat
[,1] [,2] [,3]
[1,] 23 529 12167
[2,] 67 4489 300763
[3,]24 8
[4,] 87 7569 658503
[5,]9 81729
[6,] 63 3969 250047
[7,]8 64512
[8,]24 8
[9,] 35 1225 42875
[10,]6 36216
[11,] 91 8281 753571
[12,] 41 1681 68921
[13,] 22 484 10648
[14,]39 27
best,
Kingsford Jones
So
I think the code below it, although not too short, does what the
person
asked. Thanks though because I understand outer better now.
temp - matrix(c(1,2,3,4,5,6),ncol=2)
print(temp)
#One of those more elegant ways:
print(temp)
outer(temp,1:p,'^')One of those more elegant ways:
# THIS WAY I THINK GIVES WHAT THEY WANT
sapply(1:ncol(temp), function(.col) {
temp[,.col]^.col
})
On Fri, Jan 9, 2009 at 7:40 PM, Charlotte Wickham wrote:
One of those more elegant ways:
outer(x, 1:p, ^)
Charlotte
On Fri, Jan 9, 2009 at 4:24 PM, Sarah Goslee
sarah.gos...@gmail.com
wrote:
Well, mat doesn't have any dimensions / isn't a matrix, and we
don't
know what p is supposed to be. But leaving aside those little
details,
do you perhaps want something like this:
x-c(23,67,2,87,9,63,8,2,35,6,91,41,22,3)
p - 5
mat- matrix(0, nrow=p, ncol=length(x))
for(j in 1:length(x))
{
for(i in 1:p)
mat[i,j]-x[j]^i
}
Two notes: I didn't try it out, and if that's what you want rather
than a toy example
of a larger problem, there are more elegant ways to do it in R.
Sarah
On Fri, Jan 9, 2009 at 6:42 PM, Bhargab Chattopadhyay
bharga...@yahoo.com wrote:
Hi,
Can any one please explain why the following code doesn't work? Or
can
anyone suggest an alternative.
Suppose
x-c(23,67,2,87,9,63,8,2,35,6,91,41,22,3)
mat-0;
for(j in 1:length(x))
{
for(i in 1:p)
mat[i,j]-x[j]^i;
}
Actually I want to have a matrix with p columns such that each
column
will have the elements of x^(column#).
Thanks in advance.
Bhargab
--
Sarah Goslee
http://www.functionaldiversity.org
__
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.
__
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 problem
On 4/21/2008 5:54 AM, William Simpson wrote: Hi Everyone, I am running into a problem with matrices. I use R version 2.4.1 and an older version. The problem is this: m-matrix(ncol=3,nrow=4) m[,1:3]-runif(n=4) That does what I expect; it fills up the rows of the matrix with the data vector m [,1] [,2] [,3] [1,] 0.2083071 0.2083071 0.2083071 [2,] 0.5865763 0.5865763 0.5865763 [3,] 0.7901782 0.7901782 0.7901782 [4,] 0.8298317 0.8298317 0.8298317 But this doesn't work: m[1:4,]-runif(n=3) m [,1] [,2] [,3] [1,] 0.96864939 0.11656740 0.06182311 [2,] 0.11656740 0.06182311 0.96864939 [3,] 0.06182311 0.96864939 0.11656740 [4,] 0.96864939 0.11656740 0.06182311 I want it to fill up the columns of the matrix with the data vector. Does this help? matrix(runif(4), ncol=3, nrow=4) [,1] [,2] [,3] [1,] 0.60226296 0.60226296 0.60226296 [2,] 0.74104084 0.74104084 0.74104084 [3,] 0.70955138 0.70955138 0.70955138 [4,] 0.03136881 0.03136881 0.03136881 matrix(runif(3), ncol=3, nrow=4, byrow=TRUE) [,1] [,2] [,3] [1,] 0.7008625 0.8348078 0.1003123 [2,] 0.7008625 0.8348078 0.1003123 [3,] 0.7008625 0.8348078 0.1003123 [4,] 0.7008625 0.8348078 0.1003123 Maybe there is a better way to do what I want. I need to do both of the above. The matrices are large, so I need a fast method. Thanks very much for any help. Bill Simpson __ 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. -- Chuck Cleland, Ph.D. NDRI, Inc. (www.ndri.org) 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ 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 problem
On 21/04/2008, at 9:54 PM, William Simpson wrote:
Hi Everyone,
I am running into a problem with matrices. I use R version 2.4.1 and
an older version.
The problem is this:
m-matrix(ncol=3,nrow=4)
m[,1:3]-runif(n=4)
That does what I expect; it fills up the rows of the matrix with the
data vector
m
[,1] [,2] [,3]
[1,] 0.2083071 0.2083071 0.2083071
[2,] 0.5865763 0.5865763 0.5865763
[3,] 0.7901782 0.7901782 0.7901782
[4,] 0.8298317 0.8298317 0.8298317
But this doesn't work:
m[1:4,]-runif(n=3)
m
[,1] [,2] [,3]
[1,] 0.96864939 0.11656740 0.06182311
[2,] 0.11656740 0.06182311 0.96864939
[3,] 0.06182311 0.96864939 0.11656740
[4,] 0.96864939 0.11656740 0.06182311
I want it to fill up the columns of the matrix with the data vector.
Maybe there is a better way to do what I want. I need to do both of
the above. The matrices are large, so I need a fast method.
R fills arrays in ``reverse odometer order'' --- the first index (row
index in matrices) ticks over fastest; the second index second fastest.
Unless you tell it to do otherwise.
In other words matrices get filled up column by column. Unless other
instructions are given.
Thus m[] - runif(3) (note that the ``1:4'' is redundant) puts
the generated 3-vector into the first 3 entries of column 1, then
starts over again, putting the first entry of that vector into the
last entry of column 1, then the next two entries of the vector
into the first two entries of column 2, and so on.
What you want to do is
m - matrix(runif(3),nrow=4,ncol=3,byrow=TRUE)
i.e. it is unnecessary and counter-productive to create m beforehand
and then try to fill it up.
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
__
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 problem
Thanks very much Petr and Rold for your helpful replies. Cheers Bill __ 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.
