similar to: Understanding the sequence of events when calling the R dpois function

Chuck and Greg, Thanks a lot for your help! I have a much better understanding now of what is happening ?under the hood?. Kind Regards, Jason > On 31 May 2018, at 20:08, Greg Minshall <minshall at acm.org> wrote: > > Jason, > > as Chuck Berry (to whom, *thanks* for 'do {...} while(0)'!) suggested, > using grep, or even grep executed from find, such as >

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

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

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

Dear Masters, I have a question to submit consider the following script m<-4.95 obs<-rpois(36,m) # i generate 36 realization from a poisson(m) hist(obs,freq=F) curve(dpois(x,m),add=T,col="red") #i wish to overlay on the histogram the theorical poisson density function errors are returned saing the x vector doesn't contain integers.... really bizarre i can't give

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

Dear all, Even though one of R users answered my question, I cannot understand, so I re-ask this question. 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) >

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 <-

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) :

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

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'

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,

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

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

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

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 -

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,

-----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

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")

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