Displaying 20 results from an estimated 3000 matches similar to: "a strange problem with integrate()"
2006 Mar 01
0
[Fwd: Re: [R] a strange problem with integrate()]
When I saw the subject of the original message on
R-help, I was 95% confident that I knew the answer
(before I had seen the question).
This made me think that perhaps for some functions
there should be a 'Troubleshooting' section in the help
file.
The current help file for 'integrate' does say, as Sundar
points out, what the requirements are. However, I
think more people would
2004 Mar 02
2
Problem with Integrate
The background: I'm trying to fit a Poisson-lognormal distrbutuion to
some data. This is a way of modelling species abundances:
N ~ Pois(lam)
log(lam) ~ N(mu, sigma2)
The number of individuals are Poisson distributed with an abundance
drawn from a log-normal distrbution.
To fit this to data, I need to integrate out lam. In principle, I can
do it this way:
PLN1 <- function(lam, Count,
2017 Sep 02
2
Strange lazy evaluation of default arguments
Another way to avoid the problem is to not redefine variables that are
arguments. E.g.,
> Su3 <- function(u=100, l=u, mu=0.53, sigma2=4.3^2, verbose)
{
if (verbose) {
print(c(u, l, mu))
}
uNormalized <- u/sqrt(sigma2)
lNormalized <- l/sqrt(sigma2)
muNormalized <- mu/sqrt(sigma2)
c(uNormalized, lNormalized, muNormalized)
}
> Su3(verbose=TRUE)
2011 Jul 20
1
Fwd: Help please
Hi All,
This is not really an R question but a statistical one. If someone could
either give me the brief explanation or point me to a reference that might
help, I'd appreciate it.
I want to estimate the mean of a log-normal distribution, given the (log
scale normal) parameters mu and sigma squared (sigma2). I understood this
should simply be:
exp(mu + sigma2)
... but I the following code
2017 Sep 02
0
Strange lazy evaluation of default arguments
Dear Bill,
All makes perfect sense (including the late evaluation). I actually discovered the problem by looking at old code which used your proposed solution. Still I find it strange (and, hnestly, I don?t like R?s behavior in this respect), and I am wondering why u is not being copied to L just before u is assigned a new value. Of course, this would require the R interpreter to track all these
2017 Sep 02
0
Strange lazy evaluation of default arguments
Hello,
One way of preventing that is to use ?force.
Just put
force(l)
right after the commented out print and before you change 'u'.
Hope this helps,
Rui Barradas
Citando Matthias Gondan <matthias-gondan at gmx.de>:
> Dear R developers,
>
> sessionInfo() below
>
> Please have a look at the following two versions of the same function:
>
> 1. Intended
2008 Mar 15
1
again with polr
hello everybody
solved the problem with summary, now I have another one
eg I estimate
> try.op <- polr(
> as.ordered(sod.sit.ec.fam) ~
> log(y) +
> log(1 + nfiglimin) +
> log(1 + nfiglimagg) +
> log(ncomp - nfiglitot) +
> eta +
> I(eta^2) +
>
2017 Sep 02
6
Strange lazy evaluation of default arguments
Dear R developers,
sessionInfo() below
Please have a look at the following two versions of the same function:
1. Intended behavior:
> Su1 = function(u=100, l=u, mu=0.53, sigma2=4.3^2)
+ {
+ print(c(u, l, mu)) # here, l is set to u?s value
+ u = u/sqrt(sigma2)
+ l = l/sqrt(sigma2)
+ mu = mu/sqrt(sigma2)
+ print(c(u, l, mu))
+ }
>
> Su1()
[1] 100.00 100.00 0.53
[1]
2011 Mar 28
1
maximum likelihood accuracy - comparison with Stata
Hi everyone,
I am looking to do some manual maximum likelihood estimation in R. I
have done a lot of work in Stata and so I have been using output
comparisons to get a handle on what is happening.
I estimated a simple linear model in R with lm() and also my own
maximum likelihood program. I then compared the output with Stata.
Two things jumped out at me.
Firstly, in Stata my coefficient
2017 Sep 05
0
Strange lazy evaluation of default arguments
Mathias,
If it's any comfort, I appreciated the example; 'expected' behaviour maybe, but a very nice example for staff/student training!
S Ellison
> -----Original Message-----
> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Matthias
> Gondan
> Sent: 02 September 2017 18:22
> To: r-help at r-project.org
> Subject: [R] Strange lazy evaluation of
2008 Jun 16
1
Error in maximum likelihood estimation.
Dear UseRs,
I wrote the following function to use MLE.
---------------------------------------------
mlog <- function(theta, nx = 1, nz = 1, dt){
beta <- matrix(theta[1:(nx+1)], ncol = 1)
delta <- matrix(theta[(nx+2):(nx+nz+1)], ncol = 1)
sigma2 <- theta[nx+nz+2]
gamma <- theta[nx+nz+3]
y <- as.matrix(dt[, 1], ncol = 1)
x <- as.matrix(data.frame(1,
2008 Aug 28
1
"error using packet 1"
Hello,
I'm Giovanni from ROMA..
I can't find a solution for the error:
"error using packet 1
the y field is not specified and it has not a default value"
(this is my traslation from italian language)
The code is:
pc<- qqmath(~valori,
distribution=function(p) qweibull(p,beta,alpha),
prepanel = prepanel.qqmathline,
panel = function(x, y) {
2005 Nov 17
3
loess: choose span to minimize AIC?
Is there an R implementation of a scheme for automatic smoothing
parameter selection with loess, e.g., by minimizing one of the AIC/GCV
statistics discussed by Hurvich, Simonoff & Tsai (1998)?
Below is a function that calculates the relevant values of AICC,
AICC1 and GCV--- I think, because I to guess from the names of the
components returned in a loess object.
I guess I could use
2003 Apr 18
1
MCMCpack gelman.plot and gelman.diag
Hi,
A question. When I run gelman.diag and gelman.plot
with mcmc lists obtained from MCMCregress, the results are following.
> post.R <- MCMCregress(Size~Age+Status, data = data, burnin = 5000, mcmc = 100000,
+ thin = 10, verbose = FALSE, beta.start = NA, sigma2.start = NA,
+ b0 = 0, B0 = 0, nu = 0.001, delta = 0.001)
> post1.R <- MCMCregress(Size~Age+Status, data
2003 Apr 02
2
lme parameterization question
Hi,
I am trying to parameterize the following mixed model (following Piepho
and Ogutu 2002), to test for a trend over time, using multiple sites:
y[ij]=mu+b[j]+a[i]+w[j]*(beta +t[i])+c[ij]
where:
y[ij]= a response variable at site i and year j
mu = fixed intercept
Beta=fixed slope
w[j]=constant representing the jth year (covariate)
b[j]=random effect of jth year, iid N(0,sigma2[b])
a[i]=random
2008 Aug 12
2
Maximum likelihood estimation
Hello,
I am struggling for some time now to estimate AR(1) process for commodity price time series. I did it in STATA but cannot get a result in R.
The equation I want to estimate is: p(t)=a+b*p(t-1)+error
Using STATA I get 0.92 for a, and 0.73 for b.
Code that I use in R is:
p<-matrix(data$p) # price at time t
lp<-cbind(1,data$lp) # price at time t-1
2006 Jun 15
1
SSPIR problem
Dear R-Users,
I'm using SSPIR package for a spatio-temporal application.
Is it possible to modify the structure of the involved matrixes (Fmat,
Gmat, Vmat,Wmat)?
I want to create a model like this
#y(t)=k*theta(t)+epsilon(t)
#theta(t)=h*theta(t-1)+eta(t)
#epsilon(t) N(0,V) V=sigma2*I
#eta(t) N(0,W) W=sigma2_eta
where the state variable theta has dimension 1(p=1) and at
2006 Sep 28
1
Nonlinear fitting - reparametrization help
Hi,
I am trying to fit a function of the form:
y = A0 + A1 * exp( -0.5* ( (X - Mu1) / Sigma1 )^2 ) - A2 * exp ( -0.5*
( (X-Mu2)/Sigma2 )^2 )
i.e. a mean term (A0) + a difference between two gaussians.
The constraints are A1,A2 >0, Sigma1,Sigma2>0, and usually Sigma2>Sigma1.
The plot looks like a "Mexican Hat".
I had trouble (poor fits) fitting this function to toy data
2007 Mar 02
1
Help with faster optimization for large parameter problem
Hello all,
I have a large parameter problem with the following very simple likelihood
function:
fn<-function(param) {
x1<-param[1:n]
g1<-param[(n+1):(2*n)]
beta<-param[(2*n+1):(2*n+k)]
sigma2<-param[2*n+k+1]^2
meang1sp<-mean(g1[sp])
mu<-beta%*%matrix(x1,1,n)-(g1[sp]-meang1sp)%*%matrix(g1,1,n)
return(sum((ydc-mu)^2)/(2*sigma2) + n*k*log(sqrt(sigma2)) +
2012 Mar 29
1
how to increase speed for function?/time efficiency of below function
i am using sarima() function as below
___________________________________________________________________________________________
sarima=function(data,p,d,q,P=0,D=0,Q=0,S=-1,tol=.001){
n=length(data)
constant=1:n
xmean=matrix(1,n,1)
if (d>0 & D>0)
fitit=arima(data, order=c(p,d,q), seasonal=list(order=c(P,D,Q),
period=S),