Displaying 20 results from an estimated 3000 matches similar to: "on "do.call" function"
2011 Aug 08
1
problem in do.call function
Dear all,
I am trying to use "do.call", but I don't think I totally understand this
function.
Here is an simple example.
--------------------------------------------
> B <- matrix(c(.5,.1,.2,.3),2,2)
> B
[,1] [,2]
[1,] 0.5 0.2
[2,] 0.1 0.3
> x <- c(.1,.2)
> X <- cbind(1,x)
> X
x
[1,] 1 0.1
[2,] 1 0.2
>
> lt <-
2011 Nov 14
7
Very simple loop
I'm very new to R and am trying to create my first loop.
I have:
x <-c(0:200)
A <- dpois(x,exp(4.5355343))
B <- dpois(x,exp(4.5355343 + 0.0118638))
C <- dpois(x,exp(4.5355343 -0.0234615))
D <- dpois(x,exp(4.5355343 + 0.0316557))
E <- dpois(x,exp(4.5355343 + 0.0004716))
F <- dpois(x,exp(4.5355343 + 0.056437))
G <- dpois(x,exp(4.5355343 + 0.1225822))
and would like to
2015 Nov 03
1
Fwd: Rcpp sugar dpois
Hi. Here is a piece of cpp code.
It works, but I do not understand the rational for the use of
"R::dpois" to call the function dpois since in the examples I have always
found directly "dpois" or "Rcpp::dpois" that both do not work in my code.
Could anyone be so patient to explain me why should it be like that?
Thaks a lot, Enrico
#include <Rcpp.h>
using
2018 May 31
3
Understanding the sequence of events when calling the R dpois function
Hello all,
I am trying to get a better understanding of the underlying code for the stats::dpois function in R and, specifically, what happens under the hood when it is called. I finally managed to track down the C course at: https://github.com/wch/r-source/blob/trunk/src/nmath/dpois.c. It would seem that the dpois C function is taking a double for each of the x and lambda arguments so I am a bit
2009 Oct 27
1
Poisson dpois value is too small for double precision thus corrupts loglikelihood
Hi - I have a likelihood function that involves sums of two possions:
L = a*dpois(Xi,theta1)*dpois(Yi,theta2)+b*(1-c)*a*dpois(Xi,theta1+theta3)*dpois(Yi,theta2)
where a,b,c,theta1,theta2,theta3 are parameters to be estimated.
(Xi,Yi) are observations. However, Xi and Yi are usually big (>
20000). This causes dpois to returns 0 depending on values of theta1,
theta2 and theta3.
My first
2009 Sep 19
1
Poisson Regression - Query
Hi All,
My dependent variable is a ratio that takes a value of 0 (zero) for 95% of
the observations and positive non-integer values for the other 5%. What
model would be appropriate? I'm thinking of fitting a GLM with a Poisson ~.
Now, becuase it takes non-integer values, using the glm function with
Poisson family issues warning messages.
Warning messages:
1: In dpois(y, mu, log = TRUE) :
2000 Feb 25
1
lambda==0 in dpois() (PR#459)
The nice new log=TRUE option in dpois appears to mess up the
case where lambda=0 (I was trying to calculate the likelihood
of a saturated model). Because the behavior is now always to
calculate the probability in terms of exp(log(prob)), there's
a test for lambda<=0 which really needs to be only lambda<0.
dpois(0:5,0)
ought to give
1 0 0 0 0
but gives NaNs instead.
Here's
2001 Aug 01
1
glm() with non-integer responses
A question about the inner workings of glm() and dpois():
Suppose I call
glm(y ~ x, family=poisson, weights = w)
where y contains NON-INTEGER (but still nonnegative) values.
(a) Does glm() still correctly maximise
the weighted Poisson loglikelihood ?
(i.e. the function given by the same formal expression as the
weighted loglikelihood of independent Poisson variables Y_i
except that the
2011 Mar 19
2
problem running a function
Dear people,
I'm trying to do some analysis of a data using the models by Royle & Donazio
in their fantastic book, particular the following function:
http://www.mbr-pwrc.usgs.gov/pubanalysis/roylebook/panel4pt1.fn
that applied to my data and in the console is as follows:
> `desman.y` <- structure(c(3L,4L,3L,2L,1L), .Names = c("1", "2", "3",
2011 Apr 27
3
Kolmogorov-Smirnov test
Hi,
I have a problem with Kolmogorov-Smirnov test fit. I try fit distribution to
my data. Actualy I create two test:
- # First Kolmogorov-Smirnov Tests fit
- # Second Kolmogorov-Smirnov Tests fit
see below. This two test return difrent result and i don't know which is
properly. Which result is properly? The first test return lower D = 0.0234
and lower p-value = 0.00304. The lower 'D'
2011 Sep 06
2
Generalizing call to function
Hello guys,
I would like to ask for help to understand what is going on in
"func2". My plan is to generalize "func1", so that are expected same
results in "func2" as in "func1". Executing "func1" returns...
0.25 with absolute error < 8.4e-05
But for "func2" I get...
Error in dpois(1, 0.1, 23.3065168689948, 0.000429064542600244,
2007 Dec 06
1
suggested modification to the 'mle' documentation?
Hello:
I wish to again express my appreciation to all who have
contributed to making R what it is today.
At this moment, I'm particularly grateful for whoever modified the
'mle' code so data no longer need be passed via global variables. I
remember struggling with this a couple of years ago, and I only today
discovered that it is no longer the case.
I'd
2005 Oct 19
2
matching two plots
Hi,
I have a problem about graphics. I would like to plot two graphs: a barplot
and curve. Here is the code:
> barplot(dpois(0:45,20),xlim=c(0,45),names=0:45)
> curve(dnorm(x,20,sqrt(20)),from=0,to=45,add=T)
Both graphs are drawn in the same figure, however the scale in both graphs
dooes not match. For some reason the second plot is shifted to left. I
think there is a problem
2005 Nov 02
5
Distribution fitting problem
I am using the MASS library function
fitdistr(x, dpois, list(lambda=2))
but I get
Error in optim(start, mylogfn, x = x, hessian = TRUE, ...) :
Function cannot be evaluated at initial parameters
In addition: There were 50 or more warnings (use warnings() to see the first
50)
and all the first 50 warnings say
1: non-integer x = 1.452222
etc
Can anyone tell me what I am doing
2011 Jan 27
3
Warning with mle
Hi there,
I'm pretty new to the field of fitting (anything). I try to fit a
distribution with mle, because my real data seems to follow a
zero-inflated poisson distribution. So far, I tried a simple example
to see whether I understand how to do it or not:
# example count data
x <- 0:10
y <- dpois(x, lambda = 1.4)
# zero-inflated poisson
zip <- function(x, lambda, prop) {
(1 -
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,
2008 Jul 04
1
update on dnbinom with large "size"
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
~ turns out I don't need to look at the C code.
~ if one uses the mu/size parameterization of the
negative binomial, R computes size/(size+mu) to
switch parameterizations. If size>>mu this
gets rounded to 1 ... should be easy enough
to test and return NA under these circumstances?
function (x, size, prob, mu, log = FALSE)
{
~ if
2011 Mar 22
1
Problem with plot histogram
Hi R-users,
I'm trying to built a plot of two series of data, but thees series
result "superimposed". The R-code is like this:
goal <- c(125, 143, 81, 26, 2, 3)
numgoal <- 0:5
lambda <- sum(goal*numgoal)/sum(goal)
plot(numgoal, goal, type="h")
x <- 0:5
y<-dpois(x, lambda)
att<-y*380
lines(x,att, type="h", col="red")
2007 Jul 29
1
behavior of L-BFGS-B with trivial function triggers bug in stats4::mle
With the exception of "L-BFGS-B", all of the
other optim() methods return the value of the function
when they are given a trivial function (i.e., one with no
variable arguments) to optimize. I don't think this
is a "bug" in L-BFGS-B (more like a response to
an undefined condition), but it leads to a bug in stats4::mle --
a spurious error saying that a better fit
has been
2011 Aug 29
1
maximum number of subdivisions reached
Why I am getting
Error in integrate(f, x1, x1 + dx) :
maximum number of subdivisions reached
and can I avoid this?
func <- function(y, a, rate, sad){
f3 <- function(z){
f1 <- function(y,a,n){
dpois(y,a*n)
}
f2 <- function(n,rate){
dexp(n,rate)
}
f <- function(n){
f1(y,a,n)*f2(n,rate)
}
r <- 0
r1 <- 1
x1 <- 0
dx <- 20
while(r1 >