Re: [R] Exponentiate a matrix
Out of curiosity and possibly for later use: is there an R-function that
does matrix logarithms?
Best, Ingmar
On 9/21/06 8:58 PM, Dimitrios Rizopoulos
[EMAIL PROTECTED] wrote:
Quoting Douglas Bates [EMAIL PROTECTED]:
On 9/21/06, Dimitrios Rizopoulos [EMAIL PROTECTED] wrote:
Quoting Duncan Murdoch [EMAIL PROTECTED]:
On 9/21/2006 10:40 AM, Doran, Harold wrote:
Suppose I have a square matrix P
P - matrix(c(.3,.7, .7, .3), ncol=2)
I know that
P * P
Returns the element by element product, whereas
P%*%P
Returns the matrix product.
Now, P^2 also returns the element by element product. But, is there a
slick way to write
P %*% P %*% P
Obviously, P^3 does not return the result I expect.
I don't think there's anything built in, but it's easy to write your own:
I think there was function mtx.exp() in the Malmig package, but it
seems that this package has been withdrawn from CRAN. An old version
appears to exist in:
http://r.meteo.uni.wroc.pl/src/contrib/Descriptions/Malmig.html
Best,
Dimitris
Is that function for matrix powers or for the exponential of a matrix
(which is what I initally thought that Harold wanted)? There is a
function expm in the Matrix package, patterned on the octave function
of the same name, the calculates the matrix exponential for a square
matrix.
this function calculates the n-th power of a matrix, and this is what
I thought Harold wanted, i.e.,
P %*% P %*% P %*% P
should be equal to
mtx.exp(P, 4)
%^% - function(mat, pow) {
stopifnot(length(pow) == 1, all.equal(pow, round(pow)), nrow(mat) ==
ncol(mat))
pow - round(pow)
if (pow 0) {
mat - solve(mat)
pow - abs(pow)
}
result - diag(nrow(mat))
while (pow 0) {
result - result %*% mat
pow - pow - 1
}
result
}
Now P %^% 3 will give you the matrix cube.
Duncan Murdoch
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--
Ingmar Visser
Department of Psychology, University of Amsterdam
Roetersstraat 15, 1018 WB Amsterdam
The Netherlands
http://users.fmg.uva.nl/ivisser/
tel: +31-20-5256735
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Re: [R] Exponentiate a matrix
I am getting a bit rusty on some of these things, but I seem to recall
that there is a numerical advantage (speed and/or accuracy?) to
diagonalizing:
expM - function(X,e) { v - La.svd(X); v$u %*% diag(v$d^e) %*% v$vt }
P - matrix(c(.3,.7, .7, .3), ncol=2)
P %*% P %*% P
[,1] [,2]
[1,] 0.468 0.532
[2,] 0.532 0.468
expM(P,3)
[,1] [,2]
[1,] 0.468 0.532
[2,] 0.532 0.468
I think this also works for non-integer, negative, large, and complex
exponents:
expM(P, 1.5)
[,1] [,2]
[1,] 0.3735089 0.6264911
[2,] 0.6264911 0.3735089
expM(P, 1000)
[,1] [,2]
[1,] 0.5 0.5
[2,] 0.5 0.5
expM(P, -3)
[,1][,2]
[1,] -7.3125 8.3125
[2,] 8.3125 -7.3125
expM(P, 3+.5i)
[,1] [,2]
[1,] 0.4713+0.0141531i 0.5287-0.0141531i
[2,] 0.5287-0.0141531i 0.4713+0.0141531i
Paul Gilbert
Doran, Harold wrote:
Suppose I have a square matrix P
P - matrix(c(.3,.7, .7, .3), ncol=2)
I know that
P * P
Returns the element by element product, whereas
P%*%P
Returns the matrix product.
Now, P^2 also returns the element by element product. But, is there a
slick way to write
P %*% P %*% P
Obviously, P^3 does not return the result I expect.
Thanks,
Harold
[[alternative HTML version deleted]]
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La version française suit le texte anglais.
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[R] Exponentiate a matrix
Suppose I have a square matrix P P - matrix(c(.3,.7, .7, .3), ncol=2) I know that P * P Returns the element by element product, whereas P%*%P Returns the matrix product. Now, P^2 also returns the element by element product. But, is there a slick way to write P %*% P %*% P Obviously, P^3 does not return the result I expect. Thanks, Harold [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Exponentiate a matrix
On 9/21/2006 10:40 AM, Doran, Harold wrote:
Suppose I have a square matrix P
P - matrix(c(.3,.7, .7, .3), ncol=2)
I know that
P * P
Returns the element by element product, whereas
P%*%P
Returns the matrix product.
Now, P^2 also returns the element by element product. But, is there a
slick way to write
P %*% P %*% P
Obviously, P^3 does not return the result I expect.
I don't think there's anything built in, but it's easy to write your own:
%^% - function(mat, pow) {
stopifnot(length(pow) == 1, all.equal(pow, round(pow)), nrow(mat) ==
ncol(mat))
pow - round(pow)
if (pow 0) {
mat - solve(mat)
pow - abs(pow)
}
result - diag(nrow(mat))
while (pow 0) {
result - result %*% mat
pow - pow - 1
}
result
}
Now P %^% 3 will give you the matrix cube.
Duncan Murdoch
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Exponentiate a matrix
Quoting Duncan Murdoch [EMAIL PROTECTED]:
On 9/21/2006 10:40 AM, Doran, Harold wrote:
Suppose I have a square matrix P
P - matrix(c(.3,.7, .7, .3), ncol=2)
I know that
P * P
Returns the element by element product, whereas
P%*%P
Returns the matrix product.
Now, P^2 also returns the element by element product. But, is there a
slick way to write
P %*% P %*% P
Obviously, P^3 does not return the result I expect.
I don't think there's anything built in, but it's easy to write your own:
I think there was function mtx.exp() in the Malmig package, but it
seems that this package has been withdrawn from CRAN. An old version
appears to exist in:
http://r.meteo.uni.wroc.pl/src/contrib/Descriptions/Malmig.html
Best,
Dimitris
%^% - function(mat, pow) {
stopifnot(length(pow) == 1, all.equal(pow, round(pow)), nrow(mat) ==
ncol(mat))
pow - round(pow)
if (pow 0) {
mat - solve(mat)
pow - abs(pow)
}
result - diag(nrow(mat))
while (pow 0) {
result - result %*% mat
pow - pow - 1
}
result
}
Now P %^% 3 will give you the matrix cube.
Duncan Murdoch
__
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Re: [R] Exponentiate a matrix
There is also mexp in the Matrix package and MatrixExp in the msm package.
Andreas
- Original Message -
From: Dimitrios Rizopoulos [EMAIL PROTECTED]
To: Duncan Murdoch [EMAIL PROTECTED]
Cc: Doran, Harold [EMAIL PROTECTED]; r-help@stat.math.ethz.ch
Sent: Thursday, September 21, 2006 4:26 PM
Subject: Re: [R] Exponentiate a matrix
Quoting Duncan Murdoch [EMAIL PROTECTED]:
On 9/21/2006 10:40 AM, Doran, Harold wrote:
Suppose I have a square matrix P
P - matrix(c(.3,.7, .7, .3), ncol=2)
I know that
P * P
Returns the element by element product, whereas
P%*%P
Returns the matrix product.
Now, P^2 also returns the element by element product. But, is there a
slick way to write
P %*% P %*% P
Obviously, P^3 does not return the result I expect.
I don't think there's anything built in, but it's easy to write your own:
I think there was function mtx.exp() in the Malmig package, but it
seems that this package has been withdrawn from CRAN. An old version
appears to exist in:
http://r.meteo.uni.wroc.pl/src/contrib/Descriptions/Malmig.html
Best,
Dimitris
%^% - function(mat, pow) {
stopifnot(length(pow) == 1, all.equal(pow, round(pow)), nrow(mat) ==
ncol(mat))
pow - round(pow)
if (pow 0) {
mat - solve(mat)
pow - abs(pow)
}
result - diag(nrow(mat))
while (pow 0) {
result - result %*% mat
pow - pow - 1
}
result
}
Now P %^% 3 will give you the matrix cube.
Duncan Murdoch
__
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and provide commented, minimal, self-contained, reproducible code.
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Exponentiate a matrix
On 9/21/06, Dimitrios Rizopoulos [EMAIL PROTECTED] wrote:
Quoting Duncan Murdoch [EMAIL PROTECTED]:
On 9/21/2006 10:40 AM, Doran, Harold wrote:
Suppose I have a square matrix P
P - matrix(c(.3,.7, .7, .3), ncol=2)
I know that
P * P
Returns the element by element product, whereas
P%*%P
Returns the matrix product.
Now, P^2 also returns the element by element product. But, is there a
slick way to write
P %*% P %*% P
Obviously, P^3 does not return the result I expect.
I don't think there's anything built in, but it's easy to write your own:
I think there was function mtx.exp() in the Malmig package, but it
seems that this package has been withdrawn from CRAN. An old version
appears to exist in:
http://r.meteo.uni.wroc.pl/src/contrib/Descriptions/Malmig.html
Best,
Dimitris
Is that function for matrix powers or for the exponential of a matrix
(which is what I initally thought that Harold wanted)? There is a
function expm in the Matrix package, patterned on the octave function
of the same name, the calculates the matrix exponential for a square
matrix.
%^% - function(mat, pow) {
stopifnot(length(pow) == 1, all.equal(pow, round(pow)), nrow(mat) ==
ncol(mat))
pow - round(pow)
if (pow 0) {
mat - solve(mat)
pow - abs(pow)
}
result - diag(nrow(mat))
while (pow 0) {
result - result %*% mat
pow - pow - 1
}
result
}
Now P %^% 3 will give you the matrix cube.
Duncan Murdoch
__
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and provide commented, minimal, self-contained, reproducible code.
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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
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Re: [R] Exponentiate a matrix
Quoting Douglas Bates [EMAIL PROTECTED]:
On 9/21/06, Dimitrios Rizopoulos [EMAIL PROTECTED] wrote:
Quoting Duncan Murdoch [EMAIL PROTECTED]:
On 9/21/2006 10:40 AM, Doran, Harold wrote:
Suppose I have a square matrix P
P - matrix(c(.3,.7, .7, .3), ncol=2)
I know that
P * P
Returns the element by element product, whereas
P%*%P
Returns the matrix product.
Now, P^2 also returns the element by element product. But, is there a
slick way to write
P %*% P %*% P
Obviously, P^3 does not return the result I expect.
I don't think there's anything built in, but it's easy to write your own:
I think there was function mtx.exp() in the Malmig package, but it
seems that this package has been withdrawn from CRAN. An old version
appears to exist in:
http://r.meteo.uni.wroc.pl/src/contrib/Descriptions/Malmig.html
Best,
Dimitris
Is that function for matrix powers or for the exponential of a matrix
(which is what I initally thought that Harold wanted)? There is a
function expm in the Matrix package, patterned on the octave function
of the same name, the calculates the matrix exponential for a square
matrix.
this function calculates the n-th power of a matrix, and this is what
I thought Harold wanted, i.e.,
P %*% P %*% P %*% P
should be equal to
mtx.exp(P, 4)
%^% - function(mat, pow) {
stopifnot(length(pow) == 1, all.equal(pow, round(pow)), nrow(mat) ==
ncol(mat))
pow - round(pow)
if (pow 0) {
mat - solve(mat)
pow - abs(pow)
}
result - diag(nrow(mat))
while (pow 0) {
result - result %*% mat
pow - pow - 1
}
result
}
Now P %^% 3 will give you the matrix cube.
Duncan Murdoch
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
__
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and provide commented, minimal, self-contained, reproducible code.
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