Dear 'R' friends
I have a sort of stupid question to ask.
I have a matrix say of the order 4 X 3 as
83 98 90
21 83 84
70 39 56
65 29 38
Is there any command in R which will reverse the order i.e. I need to have same
4 X 3 matrix but as given below
65 29 38
70
x[nrow(x):1,]
b
On Thu, Feb 18, 2010 at 11:45 AM, Amelia Livington
amelia_living...@yahoo.com wrote:
Dear 'R' friends
I have a sort of stupid question to ask.
I have a matrix say of the order 4 X 3 as
83 98 90
21 83 84
70 39 56
65 29 38
Is there any command in
Hello.
I am working with a set of variables which are in columns. Three of them
are of the same length while one has a different length. Typing
'data-matrix(c(ca$value, mo$value,b2$value, y1), ncol = 3)'
appears to read any three columns out of the four, though I can say exactly
which of these
On 02/01/2010 07:52 PM, ogbos okike wrote:
Hello.
I am working with a set of variables which are in columns. Three of them
are of the same length while one has a different length. Typing
'data-matrix(c(ca$value, mo$value,b2$value, y1), ncol = 3)'
appears to read any three columns out of the
Hi the list,
Is there a way to give a matrix to a C function, and then to use it as a
matrix ?
I write a function to print a matrix, but I use it as a vector :
1. void printMatrix(double *mTraj,int *nbCol, int *nbLigne){
2. int i=0,j=0;
3. for(i=0 ; i *nbLigne ; i++){
4. for(j=0
Christophe,
That's another question for the R-devel mailing list.
A few things however.
Short answer : no it is not possible. I don't think x[i,j] is even
syntactically valid in C or C++.
I'd suggest you to give a go at the .Call interface that lets you
manipulate R objects directly. So in
let consider following matrix :
mat - matrix(rnorm(45), 15)
Now I want to convert it to a list object mat_list, which will have 15
elements and each element will again be a matrix with 3 rows and 3 columns.
How can I do that?
Thanks,
--
View this message in context:
Try this:
mat_list - lapply(split(mat, seq(nrow(mat))), matrix, ncol = 3)
On Mon, Dec 28, 2009 at 9:20 AM, Ron_M ron_michae...@yahoo.com wrote:
let consider following matrix :
mat - matrix(rnorm(45), 15)
Now I want to convert it to a list object mat_list, which will have 15
elements and
re [R] matrix^(-1/2)
re the discussion in November on this thread. I don't know about expm but
the problem must be equivalent to solve(B^(1/2)) and a solution will exist
iff B is invertible and has a square root A with A%*%A = B. For 2x2
matrices necessary and sufficient conditions for B to have
Hello,
We are occasionally getting matrix results that appear to be corrupted... here
are the last several rows of an example, copy-pasted out of the R command
window. These are supposed to be floating point numbers.
[25015,] 1.820848e-01-3.2090e-06i
[25016,] 2.178046e-01-4.8140e-06i
On Dec 3, 2009, at 3:02 PM, Stephen Grubb wrote:
Hello,
We are occasionally getting matrix results that appear to be
corrupted... here are the last several rows of an example, copy-
pasted out of the R command window. These are supposed to be
floating point numbers.
[25015,]
On Oct 31, 2009, at 9:33 PM, David Winsemius wrote:
On Oct 31, 2009, at 4:39 PM, Kajan Saied wrote:
Dear R-Help Team,
as a R novice I have a (maybe for you very simple question), how do
I get
the following solved in R:
Let R be a n x n matrix:
\mid R\mid^{-\frac{1}{2}}
solve(A) gives
A question, a comment, and an alternative answer to matrix^(-1/2):
QUESTION:
What's the status of the expm package, mentioned in the email you
cited from Martin Maechler, dated Apr 5 19:52:09 CEST 2008? I tried both
install.packages('expm') and
On Sun, 1 Nov 2009, spencerg wrote:
A question, a comment, and an alternative answer to matrix^(-1/2):
QUESTION:
What's the status of the expm package, mentioned in the email you cited
from Martin Maechler, dated Apr 5 19:52:09 CEST 2008? I tried both
install.packages('expm') and
Hi, Chuck:
Thanks very much, but why do I get package 'expm' is not
available from
install.packages(expm,repos=http://R-Forge.R-project.org;)?
Best Wishes,
Spencer Graves
Charles C. Berry wrote:
On Sun, 1 Nov 2009, spencerg wrote:
A question, a comment, and an
On Nov 1, 2009, at 1:46 PM, spencerg wrote:
Hi, Chuck:
Thanks very much, but why do I get package 'expm' is not
available from install.packages(expm,repos=http://R-Forge.R-project.org
)?
In my case I think it was it is because there is no 2.10 branch to
either the:
On Sun, 1 Nov 2009, David Winsemius wrote:
On Nov 1, 2009, at 1:46 PM, spencerg wrote:
Hi, Chuck:
Thanks very much, but why do I get package 'expm' is not available
from install.packages(expm,repos=http://R-Forge.R-project.org;)?
In my case I think it was it is because there is no
On Nov 1, 2009, at 3:12 PM, Charles C. Berry wrote:
On Sun, 1 Nov 2009, David Winsemius wrote:
On Nov 1, 2009, at 1:46 PM, spencerg wrote:
Hi, Chuck:
Thanks very much, but why do I get package 'expm' is not
available
from install.packages(expm,repos=http://R-Forge.R-
Dear R-Help Team,
as a R novice I have a (maybe for you very simple question), how do I get
the following solved in R:
Let R be a n x n matrix:
\mid R\mid^{-\frac{1}{2}}
solve(A) gives me the inverse of the matrix R, however not the ^(-1/2) of
the matrix...
Thank you very much in advance!
On Oct 31, 2009, at 4:39 PM, Kajan Saied wrote:
Dear R-Help Team,
as a R novice I have a (maybe for you very simple question), how do
I get
the following solved in R:
Let R be a n x n matrix:
\mid R\mid^{-\frac{1}{2}}
solve(A) gives me the inverse of the matrix R, however not the
Hi Guys,
Im trying to find the exponential of a matrix.
Can someone please help me do this?
Thanks a lot
_
View photos of singles in your area Click Here
On Sep 30, 2009, at 9:18 PM, Kon Knafelman wrote:
Hi Guys,
Im trying to find the exponential of a matrix.
Can someone please help me do this?
Didn't we just do this one?
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
Kon Knafelman wrote:
Hi Guys,
Im trying to find the exponential of a matrix.
Can someone please help me do this?
Thanks a lot
There was actually a very recent discussion of this, and related operations.
See:
RSiteSearch('exponential of a matrix')
produced 982 matches.
RSiteSearch('{exponential of a matrix}')
produced 13.
David Winsemius wrote:
On Sep 30, 2009, at 9:18 PM, Kon Knafelman wrote:
Hi Guys,
Im trying to find the exponential of a matrix.
Can someone please help
On Sep 30, 2009, at 10:14 PM, spencerg wrote:
RSiteSearch('exponential of a matrix')
produced 982 matches.
RSiteSearch('{exponential of a matrix}')
produced 13.
Yes, I know.
David Winsemius wrote:
On Sep 30, 2009, at 9:18 PM, Kon Knafelman wrote:
Hi Guys,
Im trying to
Dear list,
As a minimal test of a more complex grid layout, I'm trying to find a
clean and efficient way to arrange text grobs in a rectangular layout.
The labels may be expressions, or text with a fontsize different of
the default, which means that the cell sizes should probably be
calculated
A few amendments might make this improved code more readable,
e = expression(alpha,testing very large width, hat(beta),
integral(f(x)*dx, a, b))
library(grid)
rowMax.units - function(u, nrow){ # rowMax with a fake matrix of units
matrix.indices - matrix(seq_along(u), nrow=nrow)
Hi
Sorry I don't seem to have explained what I'm trying to do very clearly.
The piece of code below multiplies the two matrices together a number of
times based on the value in the matmult(InitialPop,1) term in this case one
(year), this gives me the end population for the analysis.
If I understand you can use replicate:
replicate(10, matmult(InitialPop, 1))
On Fri, Sep 11, 2009 at 1:11 PM, RFish tomworthing...@talk21.com wrote:
Hi
Sorry I don't seem to have explained what I'm trying to do very clearly.
The piece of code below multiplies the two matrices together a
Dear All
I new to using R and am struggling with some matrix multiplication.
I have two matrices, one containing random numbers, these are multiplied
together to get another matrix which is different each time. When I put in
another for loop to repeat this process a multiple times the matrices
I am not sure what you mean by being the same each time. If I make
successive calls to the function, I get different results:
z
[,1]
[1,] 0.
[2,] 0.
[3,] 201.6382
[4,] 0.
[5,] 0.
[6,] 0.
[7,] 0.
matmult(InitialPop,1)
[,1]
[1,] 0.000
[2,]
RFish wrote:
I new to using R and am struggling with some matrix multiplication.
I'm not sure what you're trying to print, but you could place this vector in
an expression
Sorry I probably wasn't clear with my description. The reason i put for loop
in was that I want to do the matrix multiplication about 1000 times to get a
1000 different matrices. Therefore I was hoping the for loop would be able
to automate this then use write.table to write to an external
Dear friends,
I would like to solve the following regression problem:
y=c1 x1 + c2 x2 + + cn xn
where the y, xi are all matrices and the ci are constants that need to be
determined. The y, xi are distance matrices (symmetric). ci should be forced
to positive or null (i.e. non negative).
For the example df below this also works:
library(lattice); library(zoo)
xyplot(zoo(df[1:4], df$x), type = p)
On Sun, Sep 6, 2009 at 12:51 AM, David Winsemiusdwinsem...@comcast.net wrote:
I'm not exactly sure what structure df has. Here's my effort to duplicate
it:
df -
On Sat, Sep 5, 2009 at 10:43 PM, Bryan Hansonhan...@depauw.edu wrote:
Thanks David, your way of constructing df is much more compact than what I
was using, so I've incorporated it. I also had my rows and columns
transposed relative to how xyplot wanted them (though I had tested for that,
On Sep 6, 2009, at 12:51 AM, David Winsemius wrote:
I'm not exactly sure what structure df has. Here's my effort to
duplicate it:
df - data.frame(y=matrix(rnorm(24), nrow=6), x=1:6)
df
y.1y.2y.3y.4 x
1 0.1734636 0.2348417 -1.2375648 -1.3246439 1
2
Hello R Folks...
I have a list with the following structure:
str(df)
List of 3
$ y: num [1:4, 1:1242] -0.005379 0.029874 -0.023274 0.000655 -0.004537
..
$ x: num [1:1242] 501 503 505 507 509 ...
$ names: Factor w/ 4 levels PC Loading 1,..: 1 2 3 4
I want to plot each row of df$y
I'm not exactly sure what structure df has. Here's my effort to
duplicate it:
df - data.frame(y=matrix(rnorm(24), nrow=6), x=1:6)
df
y.1y.2y.3y.4 x
1 0.1734636 0.2348417 -1.2375648 -1.3246439 1
2 1.9551669 -1.1027262 -0.7307332 0.3953752 2
3 -0.7645778
Thanks David, your way of constructing df is much more compact than what I
was using, so I've incorporated it. I also had my rows and columns
transposed relative to how xyplot wanted them (though I had tested for that,
other problems interfered).
In my case, I may have varying numbers of y
Dear fellow R-users,
Say we have a matrix x, defined as follows
set.seed(50)
x - matrix(rbinom(100*5,1, p=0.75),nrow=100, ncol=5)
Now the interpretation of x is that each for of x is actually a sequence of
length 5, and i would like to transform x in such a way that I can describe the
On Tue, 1 Sep 2009, Gregory Gentlemen wrote:
Dear fellow R-users,
Say we have a matrix x, defined as follows
set.seed(50)
x - matrix(rbinom(100*5,1, p=0.75),nrow=100, ncol=5)
Now the interpretation of x is that each for of x is actually a sequence
of length 5, and i would like to transform
Hello,
Â
What function can I use for matrices addition? I couldnât find any
information about it in the manual or in the internet.
(A+B suits, when the number of matrixes is small, function sum() doesnât suit
for matrices addition, because it sums all variables in the matrices and
produces
On Thu, Aug 13, 2009 at 11:35 AM, Lina Rusyteliner...@yahoo.co.uk wrote:
Hi Lina,
What function can I use for matrices addition? I couldn’t find any
information about it in the manual or in the internet.
(A+B suits, when the number of matrixes is small, function sum() doesn’t suit
for
If A, B and C are matrices then:
L - list(A, B, C)
Reduce(+, L)
On Thu, Aug 13, 2009 at 5:35 AM, Lina Rusyteliner...@yahoo.co.uk wrote:
Hello,
What function can I use for matrices addition? I couldn’t find any
information about it in the manual or in the internet.
(A+B suits, when the
On Thu, 13 Aug 2009 15:11:03 +0200, Michael Knudsen micknud...@gmail.com
wrote:
On Thu, Aug 13, 2009 at 11:35 AM, Lina Rusyteliner...@yahoo.co.uk
wrote:
Hi Lina,
What function can I use for matrices addition? I couldn’t find any
information about it in the manual or in the internet.
Hi,
Is your matrix K symmetric? If yes, there is an analytical solution.
--- On Sat, 1/8/09, nhawrylyshyn nichlas.hawrylys...@gmail.com wrote:
From: nhawrylyshyn nichlas.hawrylys...@gmail.com
Subject: [R] Matrix Integral
To: r-help@r-project.org
Received: Saturday, 1 August, 2009, 12:15 AM
[mailto:cindy.g...@gmail.com]
Sent: Monday, August 10, 2009 7:05 PM
To: Nordlund, Dan (DSHS/RDA)
Cc: r-help@r-project.org
Subject: Re: [R] matrix power
Hi, Dan,
Yes, this is what I want. Is there better way to solve this?
Cindy
On Mon, Aug 10, 2009 at 2:52 PM, Nordlund, Dan (DSHS/RDA) nord
Hi, All,
If I have a symmetric matrix, how can I get the negative square root of the
matrx, ie. X^(-1/2) ?
Thanks,
Cindy
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
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-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of cindy Guo
Sent: Monday, August 10, 2009 2:32 PM
To: r-help@r-project.org
Subject: [R] matrix power
Hi, All,
If I have a symmetric matrix, how can I get the negative square
On 10-Aug-09 21:31:30, cindy Guo wrote:
Hi, All,
If I have a symmetric matrix, how can I get the negative square root
of the matrx, ie. X^(-1/2) ?
Thanks,
Cindy
X - matrix(c(2,1,1,2),nrow=2)
X
# [,1] [,2]
# [1,]21
# [2,]12
E - eigen(X)
V - E$values
Q -
Hi, Ted,
Thanks for the sample code. It is exactly what I want. But can I ask another
question? The matrix for which I want the negative square root is a
covariance matrix. I suppose it should be positive definite, so I can do
1/sqrt(V) as you wrote. But the covariance matrix I got in R using the
I posted this earlier, but am not certain that it was in fact posted, so I
will try again.
Hi, I have an equation Xcp and would like to look at errors affecting it. I
am applying errors of -.58 to .1 to R and -.45 to .47 to Xc. I would like
to look at all combinations. I set up a matrix
On 10-Aug-09 22:36:03, cindy Guo wrote:
Hi, Ted,
Thanks for the sample code. It is exactly what I want. But can
I ask another question? The matrix for which I want the negative
square root is a covariance matrix. I suppose it should be positive
definite, so I can do 1/sqrt(V) as you wrote.
Hi, Ted,
Now I understand the problem. Thank you for the explanation. It's very
helpful. I appreciate it.
Cindy
On Mon, Aug 10, 2009 at 3:58 PM, Ted Harding
ted.hard...@manchester.ac.ukwrote:
On 10-Aug-09 22:36:03, cindy Guo wrote:
Hi, Ted,
Thanks for the sample code. It is exactly what I
If its not important which of many solutions you use then
the generalized inverse can be used, say. Just use 0
for each small eigenvalue and 1/sqrt(x) for the others.
On Mon, Aug 10, 2009 at 6:36 PM, cindy Guocindy.g...@gmail.com wrote:
Hi, Ted,
Thanks for the sample code. It is exactly what
Sent: Monday, August 10, 2009 2:32 PM
To: r-help@r-project.org
Subject: [R] matrix power
Hi, All,
If I have a symmetric matrix, how can I get the negative square root of
the
matrx, ie. X^(-1/2) ?
Thanks,
Cindy
Cindy,
Just to be sure we are all on the same page
I think it may be important, but I am not sure. Actually I am trying to
program the adaptive nearest neighbor method proposed by Hastie and
Tibshirani. I am following the steps in the book 'The elements of
statistical learning' by Hastie, Tibshirani and Friedman, in which the local
metric is
Hi, I have an equation Xcp and would like to look at errors affecting it. I
am applying errors of -.58 to .1 to R and -.45 to .47 to Xc. I would like
to look at all combinations. I set up a matrix function as follows and it
does not work. On the last line, when it run, it does not seem like
Hi
I have dataset that consists of two columns
AB0.102
AC -0.002
BA -0.102
BC 0.270
CA 0.002
CB -0.270
I wish to create a matrix so that I can eventually plot the data.
A
B
C
A
1
: [R] matrix
Hi
I have dataset that consists of two columns
AB0.102
AC -0.002
BA -0.102
BC 0.270
CA 0.002
CB -0.270
I wish to create a matrix so that I can eventually plot the data.
A
B
C
A
1
0.102
You can try something about like this:
merge(transform(x, V1 = gsub(^[A-Z], , x$V1),
V3 = gsub([A-Z]$, , x$V1)),
data.frame(V1 = LETTERS[1:3],
V3 = LETTERS[1:3],
V2 = 1),
by = c(V1, V3), all = TRUE)
On Tue, Aug 4, 2009 at 8:35 AM,
My apologies, to elaborate
I carried out a correlation analysis and I want to plot the data (similar to
the graphs available in the corrplot package. However, the results are as
follows (I have many more combinations):
AB0.102
AC -0.002
BA -0.102
BC
Hi,
Any help on this would be appreciated:
I need to integrate where K is a 4x4 matrix, and SIGMA is a 4x4 matrix from
say a to b, i.e. 0 to 5:
integral MatrixExp(-K * s) %*% SIGMA %*% t(SIGMA) %*% MatrixExp(t(-K) s) ds
t is tranpose , %*% : matrix mult , MatrixExp : matrix exponential
I've
If m is a matrix and s is a logical matrix of the same dimensions, I'd
like to be able to update m with
m[s] - 0
If m2 is another matrix, I'd also like to be able to do
m[s] - m2
where elements of m for which s is TRUE get the corresponding element of
m2.
However, this doesn't work in R 2.7.1.
On Jul 16, 2009, at 6:41 PM, Ross Boylan wrote:
If m is a matrix and s is a logical matrix of the same dimensions, I'd
like to be able to update m with
m[s] - 0
If m2 is another matrix, I'd also like to be able to do
m[s] - m2
where elements of m for which s is TRUE get the corresponding
On Wed, Jul 15, 2009 at 3:42 AM, Nair, Murlidharan Tmn...@iusb.edu wrote:
Hi!!
I am trying to multiply 5 matrices and then using the inverse of that matrix
for further computation. I read about precision problems from the archives
and the suggestion was to use as.numeric while computing the
Hi Douglas,
And the big lesson, of course, is the first rule of numerical linear
algebra., Just because a formula is written in terms of the inverse
of a matrix doesn't mean that is a good way to calculate the result;
in fact, it is almost inevitably the worst way of calculating the
result.
A good discussion of this is provided by Gill, Murray and Wright
Num Lin Alg and Opt, section 4.7.2.
url:www.econ.uiuc.edu/~rogerRoger Koenker
emailrkoen...@uiuc.eduDepartment of Economics
vox: 217-333-4558University of Illinois
fax:
-Original Message-
From: dmba...@gmail.com [mailto:dmba...@gmail.com] On Behalf Of Douglas Bates
Sent: Wednesday, July 15, 2009 7:29 AM
To: Nair, Murlidharan T
Cc: r-help@r-project.org
Subject: Re: [R] Matrix multiplication precision
On Wed, Jul 15, 2009 at 3:42 AM, Nair, Murlidharan Tmn
Hi!!
I am trying to multiply 5 matrices and then using the inverse of that matrix
for further computation. I read about precision problems from the archives and
the suggestion was to use as.numeric while computing the products. I am still
having problems with the results. Here is how I am
Thanks everyone for your help!
Bill
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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,
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
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
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
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
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(
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
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),
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
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
,
r-help@r-project.org
Subject
RE: [R] Matrix inversion-different
answers from LAPACK and LINPACK
- Original Message -
From: avraham.ad...@guycarp.com
Date: Thursday, June 18, 2009 10:54 am
Subject: RE: [R] Matrix inversion-different answers from LAPACK and LINPACK
To: Ravi Varadhan rvarad...@jhmi.edu
Cc: r-help@r-project.org
Thank you. One question, though. In the case where I have closed
Hello.
I am trying to invert a matrix, and I am finding that I can get different
answers depending on whether I set LAPACK true or false using qr. I had
understood that LAPACK is, in general more robust and faster than LINPACK,
so I am confused as to why I am getting what seems to be invalid
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of avraham.ad...@guycarp.com
Sent: Wednesday, June 17, 2009 11:38 AM
To: r-help@r-project.org
Subject: [R] Matrix inversion
: Wednesday, June 17, 2009 11:38 AM
To: r-help@r-project.org
Subject: [R] Matrix inversion-different answers from LAPACK and LINPACK
Hello.
I am trying to invert a matrix, and I am finding that I can get different
answers depending on whether I set LAPACK true or false using qr. I had
understood
cc
Subject
RE: [R] Matrix inversion-different
-Original Message-
From: avraham.ad...@guycarp.com [mailto:avraham.ad...@guycarp.com]
Sent: Wednesday, June 17, 2009 1:10 PM
To: Ravi Varadhan
Cc: r-help@r-project.org
Subject: RE: [R] Matrix
Subject
RE: [R] Matrix inversion-different
answers from LAPACK and LINPACK
As you seem to be aware, the matrix is poorly conditioned:
kappa(PLLH,exact=TRUE)
[1] 115868900869
It might be worth your while to think about reparametrizing.
albyn
On Wed, Jun 17, 2009 at 11:37:48AM -0400, avraham.ad...@guycarp.com wrote:
Hello.
I am trying to invert a matrix, and I
On Wed, Jun 17, 2009 at 2:02 PM, Albyn Jonesjo...@reed.edu wrote:
As you seem to be aware, the matrix is poorly conditioned:
kappa(PLLH,exact=TRUE)
[1] 115868900869
It might be worth your while to think about reparametrizing.
Also, if it is to be a variance-covariance matrix then it must be
cc
avraham.ad...@guycarp.com,
r-help@r-project.org
06/17/2009 05:55 Subject
PMRe: [R] Matrix
-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of avraham.ad...@guycarp.com
Sent: Wednesday, June 17, 2009 6:11 PM
To: Douglas Bates
Cc: dmba...@gmail.com; r-help@r-project.org
Subject: Re: [R] Matrix inversion-different answers from LAPACK and LINPACK
I will be the first one to admit I
-project.org] On
Behalf Of avraham.ad...@guycarp.com
Sent: Wednesday, June 17, 2009 6:11 PM
To: Douglas Bates
Cc: dmba...@gmail.com; r-help@r-project.org
Subject: Re: [R] Matrix inversion-different answers from LAPACK and LINPACK
I will be the first one to admit I may be doing something stupid
Hello everyone,
I have a couple of fairly simple questions (I hope) the answers to which I
cannot find through the documentation at the moment.
1. I would like to delete the a row from a matrix if a certain elimination
criterion is met. I am familiar with x - x[-7,] (to remove row 7, for
Try this:
For the first and the second question:
transform(subset(d, row.names(d) != 2), row.names=NULL)
On Thu, Jun 11, 2009 at 3:53 PM, Payam Minoofar payam.minoo...@meissner.com
wrote:
Hello everyone,
I have a couple of fairly simple questions (I hope) the answers to which I
cannot
On Jun 11, 2009, at 2:53 PM, Payam Minoofar wrote:
Hello everyone,
I have a couple of fairly simple questions (I hope) the answers to
which I cannot find through the documentation at the moment.
1. I would like to delete the a row from a matrix if a certain
elimination criterion is
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of David Winsemius
Sent: Thursday, June 11, 2009 1:49 PM
To: Payam Minoofar
Cc: r-help@r-project.org
Subject: Re: [R] Matrix manipulation
On Jun 11, 2009, at 2:53 PM, Payam
Or perhaps:
M10[rowSums(M10 == 63) == 0, ]
On Thu, Jun 11, 2009 at 4:49 PM, David Winsemiusdwinsem...@comcast.net wrote:
On Jun 11, 2009, at 2:53 PM, Payam Minoofar wrote:
Hello everyone,
I have a couple of fairly simple questions (I hope) the answers to which I
cannot find through the
Hi,
Is there a way to convert a matrix into a vector representing all
permutations of values and column/row headings with native R functions?
I did this with 2 nested for loops and it took about 25 minutes to run
on a ~700x700 matrix. I'm assuming there must be a smarter way to do
this with
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