Re: [R] Derivative of expm function
Wagner Bonat wbo...@gmail.com on Wed, 23 Apr 2014 12:12:17 +0200 writes: Hi all ! I am look for some efficient method to compute the derivative of exponential matrix function in R. For example, I have a simple matrix like log.Sigma - matrix(c(par1, rho, rho, par2),2,2) require(Matrix) Sigma - expm(log.Sigma) I want some method to compute the derivatives of Sigma in relation the parameters par1, par2 and rho. Some idea ? The 'expm' package has slightly newer / more reliable algorithms for the matrix exponential. It also contains an expmFrechet() function which computes the Frechet derivative of the matrix exponential. I'm pretty confident -- but did not start thinking more deeply -- that this should provide the necessary parts to get partial derivatives like yours as well. Martin Maechler, ETH Zurich Wagner Hugo Bonat LEG - Laboratório de Estatística e Geoinformação UFPR - Universidade Federal do Paraná __ 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] Derivative of expm function
Hi all ! I am look for some efficient method to compute the derivative of exponential matrix function in R. For example, I have a simple matrix like log.Sigma - matrix(c(par1, rho, rho, par2),2,2) require(Matrix) Sigma - expm(log.Sigma) I want some method to compute the derivatives of Sigma in relation the parameters par1, par2 and rho. Some idea ? -- Wagner Hugo Bonat LEG - Laboratório de EstatÃstica e Geoinformação UFPR - Universidade Federal do Paraná [[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] Derivative of a function
This is not a homework. I just want to see if there are some R functions or some ideas I can borrow to solve my problem. -- View this message in context: http://r.789695.n4.nabble.com/Derivative-of-a-function-tp3631814p3633071.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] Derivative of a function
On Jun 29, 2011, at 10:48 AM, Lisa wrote: This is not a homework. I just want to see if there are some R functions or some ideas I can borrow to solve my problem. There is a deriv function that provides limited support for symbolic differentiation. The Rhelp list is advertised ( http://www.R-project.org/posting-guide.html ) as expecting you to have made some efforts at searching. This answer might not have floated to the surface among the thousand or so hits, but at least you should have tried. RSiteSearch(derivative) And there is, of course, CrossValidated: http://stats.stackexchange.com/ -- View this message in context: http://r.789695.n4.nabble.com/Derivative-of-a-function-tp3631814p3633071.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. 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] Derivative of a function
On Tue, Jun 28, 2011 at 10:03 PM, Lisa lisa...@gmail.com wrote: Dear all, I just want to get the derivative of a function that looks like: y = exp(x1*b) / (exp(x1*b) + exp(x2*b)) where y is a scalar, x1, x2, and b are vectors. I am going to take the derivative of b with respect to y, but I cannot derive an expression in which b is function of y. I know there is another way to get the similar result, i.e., first take the derivative of y with respect to each element of b, and then take its reciprocal. But it is not what I want. Could someone please tell me how to solve this problem? Thank you in advance. Assuming you meant the derivative of y with respect to b: D(expression(exp(x1*b) / (exp(x1*b) + exp(x2*b))), b) exp(x1 * b) * x1/(exp(x1 * b) + exp(x2 * b)) - exp(x1 * b) * (exp(x1 * b) * x1 + exp(x2 * b) * x2)/(exp(x1 * b) + exp(x2 * b))^2 See ?D and also note deriv on the same help page for another alternative. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.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] Derivative of a function
On 30/06/11 06:16, Gabor Grothendieck wrote: On Tue, Jun 28, 2011 at 10:03 PM, Lisalisa...@gmail.com wrote: Dear all, I just want to get the derivative of a function that looks like: y = exp(x1*b) / (exp(x1*b) + exp(x2*b)) where y is a scalar, x1, x2, and b are vectors. I am going to take the derivative of b with respect to y, but I cannot derive an expression in which b is function of y. I know there is another way to get the similar result, i.e., first take the derivative of y with respect to each element of b, and then take its reciprocal. But it is not what I want. Could someone please tell me how to solve this problem? Thank you in advance. Assuming you meant the derivative of y with respect to b: I think the original post made it quite clear that the derivative of b with respect to y was indeed what was wanted; i.e. the OP needs to do implicit differentiation which R does not do automatically. cheers, Rolf Turner D(expression(exp(x1*b) / (exp(x1*b) + exp(x2*b))), b) exp(x1 * b) * x1/(exp(x1 * b) + exp(x2 * b)) - exp(x1 * b) * (exp(x1 * b) * x1 + exp(x2 * b) * x2)/(exp(x1 * b) + exp(x2 * b))^2 See ?D and also note deriv on the same help page for another alternative. __ 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] Derivative of a function
Yes. I need to do implicit differentiation. After rearrangement, I got (x2 – x1) * b = log(1 / y - 1) Take derivative of both sides with respect to y, I have (x2 – x1) * b’[y] = - 1/y(1-y) Since both (x2 – x1) and b’[y] are vectors, I cannot move (x2 – x1) to RHS. This is why I posted my question here to see if there is some R functions or some idea that can help me solve this problem. Thanks. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Derivative-of-a-function-tp3631814p3633947.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] Derivative of a function
On Wed, Jun 29, 2011 at 4:35 PM, Lisa lisa...@gmail.com wrote:
Yes. I need to do implicit differentiation. After rearrangement, I got
(x2 – x1) * b = log(1 / y - 1)
Take derivative of both sides with respect to y, I have
(x2 – x1) * b’[y] = - 1/y(1-y)
Since both (x2 – x1) and b’[y] are vectors, I cannot move (x2 – x1) to
RHS. This is why I posted my question here to see if there is some R
functions or some idea that can help me solve this problem. Thanks.
I am not sure if this counts as an approach that you are trying to exclude but:
db[i]/dy = { dlogit(y)/dy } / { dlogit(y)/db[i] } = { 1/{y(1-y)} / {
x1[i] - x2[i] }
The numerator is taken from your calculation and the denominator is
from linearity.
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
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Re: [R] Derivative of a function
If you just want the value of the derivative at a particular point, would
numerical derivatives suffice? If so, try (for example) the numDeriv package.
S
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
Lisa [lisa...@gmail.com]
Sent: 29 June 2011 21:35
To: r-help@r-project.org
Subject: Re: [R] Derivative of a function
Yes. I need to do implicit differentiation. After rearrangement, I got
(x2 – x1) * b = log(1 / y - 1)
Take derivative of both sides with respect to y, I have
(x2 – x1) * b’[y] = - 1/y(1-y)
Since both (x2 – x1) and b’[y] are vectors, I cannot move (x2 – x1) to
RHS. This is why I posted my question here to see if there is some R
functions or some idea that can help me solve this problem. Thanks.
Lisa
--
View this message in context:
http://r.789695.n4.nabble.com/Derivative-of-a-function-tp3631814p3633947.html
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[R] Derivative of a function
Dear all, I just want to get the derivative of a function that looks like: y = exp(x1*b) / (exp(x1*b) + exp(x2*b)) where y is a scalar, x1, x2, and b are vectors. I am going to take the derivative of b with respect to y, but I cannot derive an expression in which b is function of y. I know there is another way to get the similar result, i.e., first take the derivative of y with respect to each element of b, and then take its reciprocal. But it is not what I want. Could someone please tell me how to solve this problem? Thank you in advance. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Derivative-of-a-function-tp3631814p3631814.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] Derivative of a function
(1) You really ought to do your own homework. (2) What has this to do with R? cheers, Rolf Turner On 29/06/11 14:03, Lisa wrote: Dear all, I just want to get the derivative of a function that looks like: y = exp(x1*b) / (exp(x1*b) + exp(x2*b)) where y is a scalar, x1, x2, and b are vectors. I am going to take the derivative of b with respect to y, but I cannot derive an expression in which b is function of y. I know there is another way to get the similar result, i.e., first take the derivative of y with respect to each element of b, and then take its reciprocal. But it is not what I want. Could someone please tell me how to solve this problem? Thank you in advance. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Derivative-of-a-function-tp3631814p3631814.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. __ 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] Derivative
This following works for me but I still favor the quick and dirty method
suggested originally by David.
options(scipen = 10)
x - seq(0,2, by = .01)
f - expression(5*cos(2*x)-2*x*sin(2*x))
D(f, 'x')
f.prime - function(x){
-(5 * (sin(2 * x) * 2) + (2 * sin(2 * x) + 2 * x * (cos(2 * x) * 2)))
}
curve(expr = f.prime, from = 1, to = 2, n = 101)
uniroot(f = f.prime, interval = c(1,2), tol = .1)
On Aug 11, 2010, at 10:56 PM, David Winsemius wrote:
On Aug 12, 2010, at 12:49 AM, Dennis Murphy wrote:
Hi:
Try the following:
f - function(x) 5*cos(2*x)-2*x*sin(2*x)
curve(f, -5, 5)
abline(0, 0, lty = 'dotted')
This shows rather clearly that your function has multiple roots, which isn't
surprising given that it's a linear combination of sines and cosines. To
find a specific root numerically, use function uniroot on f, as follows:
uniroot(f, c(0, 2))
Except he was asking for the root of the derivative. If the classroom
assignment allows use of R's limited symbolic differentiation you could try:
df.dx - D(expression(5*cos(2*x)-2*x*sin(2*x)), x)
df.dx
-(5 * (sin(2 * x) * 2) + (2 * sin(2 * x) + 2 * x * (cos(2 * x) *
2)))
(Which as one of the examples in the deriv help page notes is not the most
simple form.)
I was assuming that the OP wanted a solution to:
d( abs(f(x)) )/dt = 0 in the domain [1,2]
So:
f.df.dx - function (x) {
eval(parse(text=D(expression(5*cos(2*x)-2*x*sin(2*x)), x) ) )
}
# no abs() but we should be satisfied with either a minimum or a maximum
uniroot(f.df.dx, c(1,2) )
$root
[1] 1.958218267
$f.root
[1] 1.138013788e-05
$iter
[1] 4
$estim.prec
[1] 6.103515625e-05
It doesn't agree with my earlier method and I think this one has a greater
probablity of being correct. I don't think I needed to take second differences.
--
David.
$root
[1] 0.6569286
$f.root
[1] -0.0001196119
$iter
[1] 6
$estim.prec
[1] 6.103516e-05
This catches the root that lies between x = 0 and x = 2. If you want to find
a set of roots, you can try a loop. Fortunately, since the function is even,
you really only need to find the roots on one side of zero, since the ones
on the other side are the same with opposite sign.
lo - seq(0, 4.5, by = 1.5)
hi - seq(1.5, 6, by = 1.5)
roots - numeric(length(lo))
for(i in seq_along(lo)) roots[i] - uniroot(f, c(lo[i], hi[i]))$root
roots
See ?uniroot for other options and tolerance settings.
HTH,
Dennis
On Wed, Aug 11, 2010 at 6:21 PM, TGS cran.questi...@gmail.com wrote:
How would I numerically find the x value where the derivative of the
function below is zero?
x - seq(1,2, by = .01)
y - 5*cos(2*x)-2*x*sin(2*x)
plot(x,abs(y), type = l, ylab = |y|)
--
David Winsemius, MD
West Hartford, CT
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[R] Derivative
How would I numerically find the x value where the derivative of the function below is zero? x - seq(1,2, by = .01) y - 5*cos(2*x)-2*x*sin(2*x) plot(x,abs(y), type = l, ylab = |y|) __ 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] Derivative
On Aug 11, 2010, at 9:21 PM, TGS wrote: How would I numerically find the x value where the derivative of the function below is zero? x - seq(1,2, by = .01) y - 5*cos(2*x)-2*x*sin(2*x) plot(x,abs(y), type = l, ylab = |y|) Two ideas: ---minimize abs(diff(y)) abline(v=x[which.min(abs( diff(y) )) ]) ---Or maximize the second derivative plot(x[-(1:2)],diff(diff(y)), type = l, ylab = |y|) abline(v=x[which.max(diff(diff(y)))]) __ 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] Derivative
Hi: Try the following: f - function(x) 5*cos(2*x)-2*x*sin(2*x) curve(f, -5, 5) abline(0, 0, lty = 'dotted') This shows rather clearly that your function has multiple roots, which isn't surprising given that it's a linear combination of sines and cosines. To find a specific root numerically, use function uniroot on f, as follows: uniroot(f, c(0, 2)) $root [1] 0.6569286 $f.root [1] -0.0001196119 $iter [1] 6 $estim.prec [1] 6.103516e-05 This catches the root that lies between x = 0 and x = 2. If you want to find a set of roots, you can try a loop. Fortunately, since the function is even, you really only need to find the roots on one side of zero, since the ones on the other side are the same with opposite sign. lo - seq(0, 4.5, by = 1.5) hi - seq(1.5, 6, by = 1.5) roots - numeric(length(lo)) for(i in seq_along(lo)) roots[i] - uniroot(f, c(lo[i], hi[i]))$root roots See ?uniroot for other options and tolerance settings. HTH, Dennis On Wed, Aug 11, 2010 at 6:21 PM, TGS cran.questi...@gmail.com wrote: How would I numerically find the x value where the derivative of the function below is zero? x - seq(1,2, by = .01) y - 5*cos(2*x)-2*x*sin(2*x) plot(x,abs(y), type = l, ylab = |y|) __ 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] Derivative
On Aug 12, 2010, at 12:49 AM, Dennis Murphy wrote:
Hi:
Try the following:
f - function(x) 5*cos(2*x)-2*x*sin(2*x)
curve(f, -5, 5)
abline(0, 0, lty = 'dotted')
This shows rather clearly that your function has multiple roots,
which isn't
surprising given that it's a linear combination of sines and
cosines. To
find a specific root numerically, use function uniroot on f, as
follows:
uniroot(f, c(0, 2))
Except he was asking for the root of the derivative. If the classroom
assignment allows use of R's limited symbolic differentiation you
could try:
df.dx - D(expression(5*cos(2*x)-2*x*sin(2*x)), x)
df.dx
-(5 * (sin(2 * x) * 2) + (2 * sin(2 * x) + 2 * x * (cos(2 * x) *
2)))
(Which as one of the examples in the deriv help page notes is not the
most simple form.)
I was assuming that the OP wanted a solution to:
d( abs(f(x)) )/dt = 0 in the domain [1,2]
So:
f.df.dx - function (x) {
eval(parse(text=D(expression(5*cos(2*x)-2*x*sin(2*x)), x) ) )
}
# no abs() but we should be satisfied with either a minimum or a
maximum
uniroot(f.df.dx, c(1,2) )
$root
[1] 1.958218267
$f.root
[1] 1.138013788e-05
$iter
[1] 4
$estim.prec
[1] 6.103515625e-05
It doesn't agree with my earlier method and I think this one has a
greater probablity of being correct. I don't think I needed to take
second differences.
--
David.
$root
[1] 0.6569286
$f.root
[1] -0.0001196119
$iter
[1] 6
$estim.prec
[1] 6.103516e-05
This catches the root that lies between x = 0 and x = 2. If you want
to find
a set of roots, you can try a loop. Fortunately, since the function
is even,
you really only need to find the roots on one side of zero, since
the ones
on the other side are the same with opposite sign.
lo - seq(0, 4.5, by = 1.5)
hi - seq(1.5, 6, by = 1.5)
roots - numeric(length(lo))
for(i in seq_along(lo)) roots[i] - uniroot(f, c(lo[i], hi[i]))$root
roots
See ?uniroot for other options and tolerance settings.
HTH,
Dennis
On Wed, Aug 11, 2010 at 6:21 PM, TGS cran.questi...@gmail.com wrote:
How would I numerically find the x value where the derivative of the
function below is zero?
x - seq(1,2, by = .01)
y - 5*cos(2*x)-2*x*sin(2*x)
plot(x,abs(y), type = l, ylab = |y|)
--
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.
[R] Derivative of the probit
Is there a function to compute the derivative of the probit (qnorm) function in R, or in any of the packages? Thanks, -Andrew [[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] Derivative of the probit
f-function(x) 1/dnorm(qnorm(x)) for x in (0,1) -tgs On Thu, May 6, 2010 at 4:40 PM, Andrew Redd ar...@stat.tamu.edu wrote: Is there a function to compute the derivative of the probit (qnorm) function in R, or in any of the packages? Thanks, -Andrew [[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. [[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] Derivative of the probit
On 06-May-10 20:40:30, Andrew Redd wrote:
Is there a function to compute the derivative of the probit (qnorm)
function
in R, or in any of the packages?
Thanks,
-Andrew
I don't think so (though stand to be corrected). However, it would
be straightforward to write one.
For simplicity of notation:
P(x) = pnorm(x)
Q(p) = qnrom(p)
D(x) = dnorm(x)
' stands for derivative.
Then
P'(x) = D(x)
P(Q(p)) = p
Differentiate w.r.to p:
P'(Q(p))*Q'(p) = 1
Q'(p) = 1/P'(Q(p)) = 1/D(Q(p))
Hence:
dqnorm - function(p){ 1/dnorm(qnorm(p)) }
(I think that's right ... )
Ted.
E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 06-May-10 Time: 22:23:53
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Derivative of a smooth function
Hi Fir, you can alternatively use local regression, implemented in the package locfit, which can also estimate derivatives: library(locfit) attach(cars) # main fit fit - locfit( dist ~ speed ) # fit 1st derivative fitd - locfit( dist ~ speed , deriv =1) # plots... plot(speed, dist ) lines(fit) lines(fitd, col=red) Hope this helps, Julien On Fri, Apr 2, 2010 at 2:06 PM, FMH kagba2...@yahoo.com wrote: Dear All, I've been searching for appropriate codes to compute the rate of change and the curvature of nonparametric regression model whish was denoted by a smooth function but unfortunately don't manage to do it. I presume that such characteristics from a smooth curve can be determined by the first and second derivative operators. The following are the example of fitting a nonparametric regression model via smoothing spline function from the Help file in R. ### attach(cars) plot(speed, dist, main = data(cars)smoothing splines) cars.spl - smooth.spline(speed, dist) lines(cars.spl, col = blue) lines(smooth.spline(speed, dist, df=10), lty=2, col = red) legend(5,120,c(paste(default [C.V.] = df =,round(cars.spl$df,1)),s( * , df = 10)), col = c(blue,red), lty = 1:2, bg='bisque') detach() ### Could someone please advice me the appropriate way to determine such derivatives on the curves which were fitted by the function above and would like to thank you in advance. Cheers Fir __ 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] Derivative of a smooth function
While this doesn't answer your question, I want to let you know that there is a proposal for a related improvement within R that will let users compute (numerically) the derivatives, of any order, of a given function inside of R. In your case, this means that you will write the smooth spline function, symbolically f(x), that will interpolate between the points. Using Automatic Differentiation, the proposed solution, will automatically let you find f'(x), f''(x), etc.. by using the same function but overloading the meaning of the arithmetic operators and mathematical functions to act upon a special data type. The initial idea http://rwiki.sciviews.org/doku.php?id=developers:projects:gsoc2010:adinrcame from Prof. John Nash who suggested bringing the ability of Automatic Differentiation to R. We both have, since, collaborated to bring out adetailed proposalhttp://socghop.appspot.com/gsoc/student_proposal/show/google/gsoc2010/quantumelixir/t126989852709outlining the various features to be implemented. Note that, this is being planned to be implemented as part of Google's Summer of Code program for this year. So, should our proposal be selected, much more than simple second derivative computation can be accomplished from within R. Regards, Chillu On Fri, Apr 2, 2010 at 2:06 PM, FMH kagba2...@yahoo.com wrote: Dear All, I've been searching for appropriate codes to compute the rate of change and the curvature of nonparametric regression model whish was denoted by a smooth function but unfortunately don't manage to do it. I presume that such characteristics from a smooth curve can be determined by the first and second derivative operators. The following are the example of fitting a nonparametric regression model via smoothing spline function from the Help file in R. ### attach(cars) plot(speed, dist, main = data(cars)smoothing splines) cars.spl - smooth.spline(speed, dist) lines(cars.spl, col = blue) lines(smooth.spline(speed, dist, df=10), lty=2, col = red) legend(5,120,c(paste(default [C.V.] = df =,round(cars.spl$df,1)),s( * , df = 10)), col = c(blue,red), lty = 1:2, bg='bisque') detach() ### Could someone please advice me the appropriate way to determine such derivatives on the curves which were fitted by the function above and would like to thank you in advance. Cheers Fir __ 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.
[R] Derivative of a smooth function
Dear All, I've been searching for appropriate codes to compute the rate of change and the curvature of nonparametric regression model whish was denoted by a smooth function but unfortunately don't manage to do it. I presume that such characteristics from a smooth curve can be determined by the first and second derivative operators. The following are the example of fitting a nonparametric regression model via smoothing spline function from the Help file in R. ### attach(cars) plot(speed, dist, main = data(cars) smoothing splines) cars.spl - smooth.spline(speed, dist) lines(cars.spl, col = blue) lines(smooth.spline(speed, dist, df=10), lty=2, col = red) legend(5,120,c(paste(default [C.V.] = df =,round(cars.spl$df,1)),s( * , df = 10)), col = c(blue,red), lty = 1:2, bg='bisque') detach() ### Could someone please advice me the appropriate way to determine such derivatives on the curves which were fitted by the function above and would like to thank you in advance. Cheers Fir __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Derivative of a smooth function
Please learn how to use `RsiteSearch' before posting questions to the list: RSiteSearch(derivative smooth function) This should have provided you with plenty of solutions. Ravi. Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu - Original Message - From: FMH kagba2...@yahoo.com Date: Friday, April 2, 2010 4:39 am Subject: [R] Derivative of a smooth function To: r-help@r-project.org Cc: r-sig-fina...@stat.math.ethz.ch Dear All, I've been searching for appropriate codes to compute the rate of change and the curvature of nonparametric regression model whish was denoted by a smooth function but unfortunately don't manage to do it. I presume that such characteristics from a smooth curve can be determined by the first and second derivative operators. The following are the example of fitting a nonparametric regression model via smoothing spline function from the Help file in R. ### attach(cars) plot(speed, dist, main = data(cars) smoothing splines) cars.spl - smooth.spline(speed, dist) lines(cars.spl, col = blue) lines(smooth.spline(speed, dist, df=10), lty=2, col = red) legend(5,120,c(paste(default [C.V.] = df =,round(cars.spl$df,1)),s( * , df = 10)), col = c(blue,red), lty = 1:2, bg='bisque') detach() ### Could someone please advice me the appropriate way to determine such derivatives on the curves which were fitted by the function above and would like to thank you in advance. Cheers Fir __ R-help@r-project.org mailing list PLEASE do read the posting guide 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] Derivative of nonparametric curve
Thank you
- Original Message
From: spencerg spencer.gra...@prodsyse.com
To: Liaw, Andy andy_l...@merck.com
Cc: Rolf Turner r.tur...@auckland.ac.nz; FMH kagba2...@yahoo.com;
r-help@r-project.org
Sent: Wednesday, September 9, 2009 3:08:43 PM
Subject: Re: [R] Derivative of nonparametric curve
This may be overkill for your application, but you might be interested in
the fda package, for which a new book appeared a couple of months ago:
Functional Data Analysis with R and Matlab (Springer Use R! series, by
Ramsay, Hooker and Graves; I'm the third author). The package includes a
scripts subdirectory with R code to recreate all but one of the 76 figures in
the book. [To find this scripts directory, use system.file('scripts',
package='fda').]
Functional data analysis generalizes spline smoothing in two important
ways:
(1) It supports the use of an arbitrary finite basis set to
approximate elements of a function space; spline smoothing uses splines only,
usually cubic splines. The first derivative of a cubic spline is piecewise
quadratic, and the second derivative is piecewise linear. If you want
something smoother than linear, you need at least a quartic spline, and Ramsay
has recommended quintics -- degree 5 polynomials = order 6 spline.
(2) It allows the curve to be smoothed using an arbitrary linear
differential operator, not just the second derivative. This can be important
if you have theory saying that the truth should follow a particular
differential equation. Otherwise, if you want to estimate the second
derivative, Ramsay has recommended smoothing with the fourth derivative rather
than the second. (In any event, smoothing is achieved by penalized least
squares with the penalty being proportional to the integral of the square of
the chosen linear differential operator.)
To reinforce this second point, chapter 11 of Functional Data Analysis
with R and Matlab describes functional differential analysis, which will
estimate non-constant coefficients in a differential equation model.
Hope this helps. Spencer Graves
Liaw, Andy wrote:
From: Rolf Turner
On 8/09/2009, at 9:07 PM, FMH wrote:
Dear All,
I'm looking for a way on computing the derivative of first and second
order of a smoothing curve produced by a nonprametric regression. For
instance, if we run the R script below, a smooth nonparametric regression
curve is produced.
provide.data(trawl)
Zone92 - (Year == 0 Zone == 1)
Position - cbind(Longitude - 143, Latitude)
dimnames(Position)[[2]][1] - Longitude - 143
sm.regression(Longitude, Score1, method = aicc, col = red, model =
linear)
Could someone please give some hints on the way to find the derivative on
the curve at some points ?
See
?smooth.spline
and
?predict.smooth.spline
Since sm.regression() (from the sm package, I presume) uses kernel
methods, a kernel-based estimator of derivatives is available in the
KernSmooth package.
Andy
cheers,
Rolf Turner
##
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__
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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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__
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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.
-- Spencer Graves, PE, PhD
President and Chief Operating Officer
Structure Inspection and Monitoring, Inc.
751 Emerson Ct.
San José, CA 95126
ph: 408-655-4567
__
R-help@r-project.org mailing list
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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] Derivative of nonparametric curve
From: Rolf Turner
On 8/09/2009, at 9:07 PM, FMH wrote:
Dear All,
I'm looking for a way on computing the derivative of first and
second order of a smoothing curve produced by a nonprametric
regression. For instance, if we run the R script below, a smooth
nonparametric regression curve is produced.
provide.data(trawl)
Zone92 - (Year == 0 Zone == 1)
Position - cbind(Longitude - 143, Latitude)
dimnames(Position)[[2]][1] - Longitude - 143
sm.regression(Longitude, Score1, method = aicc, col = red,
model = linear)
Could someone please give some hints on the way to find the
derivative on the curve at some points ?
See
?smooth.spline
and
?predict.smooth.spline
Since sm.regression() (from the sm package, I presume) uses kernel
methods, a kernel-based estimator of derivatives is available in the
KernSmooth package.
Andy
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.
Notice: This e-mail message, together with any attachme...{{dropped:12}}
__
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] Derivative of nonparametric curve
This may be overkill for your application, but you might be
interested in the fda package, for which a new book appeared a couple
of months ago: Functional Data Analysis with R and Matlab (Springer
Use R! series, by Ramsay, Hooker and Graves; I'm the third author).
The package includes a scripts subdirectory with R code to recreate
all but one of the 76 figures in the book. [To find this scripts
directory, use system.file('scripts', package='fda').]
Functional data analysis generalizes spline smoothing in two
important ways:
(1) It supports the use of an arbitrary finite basis set to
approximate elements of a function space; spline smoothing uses splines
only, usually cubic splines. The first derivative of a cubic spline is
piecewise quadratic, and the second derivative is piecewise linear. If
you want something smoother than linear, you need at least a quartic
spline, and Ramsay has recommended quintics -- degree 5 polynomials =
order 6 spline.
(2) It allows the curve to be smoothed using an arbitrary
linear differential operator, not just the second derivative. This can
be important if you have theory saying that the truth should follow a
particular differential equation. Otherwise, if you want to estimate
the second derivative, Ramsay has recommended smoothing with the fourth
derivative rather than the second. (In any event, smoothing is achieved
by penalized least squares with the penalty being proportional to the
integral of the square of the chosen linear differential operator.)
To reinforce this second point, chapter 11 of Functional Data
Analysis with R and Matlab describes functional differential
analysis, which will estimate non-constant coefficients in a
differential equation model.
Hope this helps.
Spencer Graves
Liaw, Andy wrote:
From: Rolf Turner
On 8/09/2009, at 9:07 PM, FMH wrote:
Dear All,
I'm looking for a way on computing the derivative of first and
second order of a smoothing curve produced by a nonprametric
regression. For instance, if we run the R script below, a smooth
nonparametric regression curve is produced.
provide.data(trawl)
Zone92 - (Year == 0 Zone == 1)
Position - cbind(Longitude - 143, Latitude)
dimnames(Position)[[2]][1] - Longitude - 143
sm.regression(Longitude, Score1, method = aicc, col = red,
model = linear)
Could someone please give some hints on the way to find the
derivative on the curve at some points ?
See
?smooth.spline
and
?predict.smooth.spline
Since sm.regression() (from the sm package, I presume) uses kernel
methods, a kernel-based estimator of derivatives is available in the
KernSmooth package.
Andy
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.
Notice: This e-mail message, together with any attachme...{{dropped:12}}
__
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.
--
Spencer Graves, PE, PhD
President and Chief Operating Officer
Structure Inspection and Monitoring, Inc.
751 Emerson Ct.
San José, CA 95126
ph: 408-655-4567
__
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] Derivative of nonparametric curve
Dear All, I'm looking for a way on computing the derivative of first and second order of a smoothing curve produced by a nonprametric regression. For instance, if we run the R script below, a smooth nonparametric regression curve is produced. provide.data(trawl) Zone92 - (Year == 0 Zone == 1) Position - cbind(Longitude - 143, Latitude) dimnames(Position)[[2]][1] - Longitude - 143 sm.regression(Longitude, Score1, method = aicc, col = red, model = linear) Could someone please give some hints on the way to find the derivative on the curve at some points ? Thank you. Kagba __ 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] Derivative of nonparametric curve
On 8/09/2009, at 9:07 PM, FMH wrote:
Dear All,
I'm looking for a way on computing the derivative of first and
second order of a smoothing curve produced by a nonprametric
regression. For instance, if we run the R script below, a smooth
nonparametric regression curve is produced.
provide.data(trawl)
Zone92 - (Year == 0 Zone == 1)
Position - cbind(Longitude - 143, Latitude)
dimnames(Position)[[2]][1] - Longitude - 143
sm.regression(Longitude, Score1, method = aicc, col = red,
model = linear)
Could someone please give some hints on the way to find the
derivative on the curve at some points ?
See
?smooth.spline
and
?predict.smooth.spline
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.
