similar to: Greater than 1 or less than 1?

On a recent FreeBSD 8.0-CURRENT (i386) building R (any version) breaks with the following messages: ---------------------------------------------------------------------- [...snip...] gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include -DHAVE_CONFIG_H -g -O2 -c wilcox.c -o wilcox.o gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include

Full_Name: Murray H Smith Version: 0.63.3 OS: Windows NT Submission from: (NULL) (130.216.5.57) The ppois function is displaced by -0.5. Try: > ppois(-0.5,1) [1] 0.3678794 > ppois(-0.51,1) [1] 0 > ppois(0,1) [1] 0.3678794 and > par(mfrow=c(2,1)) > x<-seq(-1,5,0.01) > plot(x,ppois(x,1),type="s",ylab="F(x)",main="Poisson CDF?") >

Le 04/12/2018 ? 11:27, I?aki Ucar a ?crit?: > On Tue, 4 Dec 2018 at 11:12, <qweytr1 at mail.ustc.edu.cn> wrote: >> function ppois is a function calculate the CDF of Poisson distribution, it should generate a non-decreasing result, but what I got is: >> >>> any(diff(ppois(0:19,lambda=0.9))<0) >> [1] TRUE >> >> Actually, >> >>>

function ppois is a function calculate the CDF of Poisson distribution, it should generate a non-decreasing result, but what I got is: > any(diff(ppois(0:19,lambda=0.9))<0) [1] TRUE Actually, > ppois(19,lambda=0.9)<ppois(18,lambda=0.9) [1] TRUE Which could not be TRUE. Code is tested in both R 3.5.1 and Microsoft R Open 3.5.1. _

Could somebody help me to understand the syntax of R's ppois function? I'm looking to calculate the cumulative probability density of an observed value (y) given the expected mean (mu) and the level of significance (alpha). I'm coming from using SAS to do this and don't recognize the descriptions of the arguments for ppois. The definitions of lambda and p as stated in the R manuals

Full_Name: M Welinder Version: 1.4 OS: (src) Submission from: (NULL) (192.5.35.38) It seems to me that pexp can be improved in the lower_tail=TRUE and log_p=FALSE case by using expm1. Something like -expm1 (-x / scale); I think. -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send

Hello all, Just joined this mailing list -- not sure if this is the right list to send this question, but I have a question about cross-compiling R. I am working with R-1.9.1.tgz. It may just be with my version of mingw32, but it seems that expm1 is not defined, so I tried to ensure that HAVE_EXPM1 was #undef'ed before cross-compiling. The problem is that, in <include/Rmath.h> if

Reciban un cordial saludo. De manera muy comedida les pido ayuda con la función indicada abajo. La usan en un artículo de inventarios. No veo dónde estoy errado. Gracias de antemano. César Escalante C. > lamb<-50 > L<-1 > h.<-10 > p<-25 > K<-5 > integ<-function(y){integrate(function(x)ppois(x,lamb*L),0,y)$value} > G<-function(y){(h.+p)*integ(y) +

Dear all, I am doing 200 times simulation. For each time, I generate a matrix and define some function on this matrix to get a 6 dimension vector as my results. As the loop should be slow, I generate 200 matrice first, and save them into a list named ma, then I define zz<-sapply(ma, myfunction) To my surprise, It almost costs me the same time to get my results if I directly use a loop

Dear all, I am doing 200 times simulation. For each time, I generate a matrix and define some function on this matrix to get a 6 dimension vector as my results. As the loop should be slow, I generate 200 matrice first, and save them into a list named ma, then I define zz<-sapply(ma, myfunction) To my surprise, It almost costs me the same time to get my results if I directly use a loop

The expected number of bladder cancer over next 20 years a tire industry is 1.8. Poission distribution is assumed to hold and 6 reported deaths are caused by bladder cancer among the employees. Trying to find how unusual this event is. > ppois(q=6, lambda=1.8, lower.tail = TRUE, log.p = FALSE) [1] 0.9974306 not sure if ppois is the right one to use and the parameters... thx much

>From: Sundar Dorai-Raj <sundar.dorai-raj at PDF.COM> >Reply-To: sundar.dorai-raj at PDF.COM >To: Zhen Pang <nusbj at hotmail.com> >CC: r-help at stat.math.ethz.ch >Subject: Vectorising and loop (was Re: [R] optim "a log-likelihood >function") >Date: Wed, 29 Sep 2004 18:21:17 -0700 > > > >Zhen Pang wrote: > >> >>I also use

Dear R People: I am trying to build R-2.5.0 from source on a Windows machinee. I downloaded the tools and such, and started things off. I received some odd errors messages, looked on the R Search and found a similar problem. The suggestion was to re-load "ld.exe". I downloaded and installed everything again. Same problem. I am including my output.

Hello, I am trying to use Yan and Rossini's Makefile for cross building Windows versions of R packages in Linux with R 1.9.0. When compiling R with the mingw tools I get an error about expm1 being undeclared when first found at src/main/arithmetic.c:1019 If I fiddle a bit with it later on I also get errors about log1p bein undeclared. Any idea what should I look for? I am using R 1.9.0 in

Hello, I am trying to use Yan and Rossini's Makefile for cross building Windows versions of R packages in Linux with R 1.9.0. When compiling R with the mingw tools I get an error about expm1 being undeclared when first found at src/main/arithmetic.c:1019 If I fiddle a bit with it later on I also get errors about log1p bein undeclared. Any idea what should I look for? I am using R 1.9.0 in

Dear all, Using R function "dwt", it seems that I cannot specify the wavelet transformation like Haar. What's the R code for wavelet decomposition which allows me to specify Haar wavelet transformation? Of course, if it can include "db2", that is even better. In general, I want an R function like matlab code "dwt". Thanks in advance! Zhen Li

A quick question! The number of episodes per year of otitis media follows a Possion distribution with lambda = 1.6 episodes per year. Wouldn't the probability of getting 3 or more episodes of otitis media in the first 2 years of life be: > ppois(q=3, lambda=1.6*2, lower.tail = TRUE, log.p = FALSE) [1] 0.6025197 I am confused with the lambda and 3 or more.. thx much

Full_Name: M Welinder Version: 1.4 OS: (src) Submission from: (NULL) (192.5.35.38) It seems to me that pweibull can be improved in the lower_tail=TRUE and log_p=FALSE case by using expm1. Something like -expm1(-pow(x / scale, shape)), I think. -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list -- Read

Hello, by checking the precision of a convolution algorithm, we found the following "inexactness": We work with R Version 1.8.1 (2003-11-21) on Windows systems (NT, 2000, XP). Try the code: ## Kolmogorov distance between two methods to ## determine P(Poisson(lambda)<=x) Kolm.dist <- function(lam, eps){ x <- seq(0,qpois(1-eps, lambda=lam), by=1) max(abs(ppois(x,

Dear all, I have made my testing program to run successfully under unix in the background. However, my simulation work does not work. I read the foo.results file, I found it only have part of my code and not any output I want. Is there any line limit? My code is nearly 400 line. I can cut some of them, but I want to know whether there is any limit or exactly the number of limit is. Thanks.