similar to: Implementation of gamma function for large number

Dear R users, I have to calculate gamma functions for negative numbers beyond -171.4. e.x. gamma(-500.4) I got following: > gamma(-170.4) [1] -5.824625e-308 > gamma(-171.4) [1] 0 Warning message: underflow occurred in 'gammafn' I have tried to use a recursion getting values a little futher -180. How could I solve this problem? Thank you beforehand. Chuse.

Dear R-users, Is there an R function to compute the multinomial beta function? That is, the normalizing constant that arises in a Dirichlet distribution. For example, with three parameters the beta function is Beta(n1,n2,n2) = Gamma(n1)*Gamma(n2)*Gamma(n3)/Gamma(n1+n2+n3) Thanks in advance for any assisstance. Regards, Greg [[alternative HTML version deleted]]

Hi all, why factorial(150) shows the error out of range in 'gammafn'? I have to calculate the number of subset formed by 150 samples taking 10 at a time. How is this possible? best [[alternative HTML version deleted]]

I'm looking for a package that has a start-of-the-art method of imputation of missing values in a data frame with both continuous and factor columns. I've found transcan() in 'Hmisc', which appears to be possibly suited to my needs, but I haven't been able to figure out how to get a new data frame with the imputed values replaced (I don't have Herrell's book). Any

I have a data frame: > head(df) Time Temp Conc Repl Log10 1 0 -20 H 1 6.406547 2 2 -20 H 1 5.738683 3 7 -20 H 1 5.796394 4 14 -20 H 1 4.413691 5 0 4 H 1 6.406547 7 7 4 H 1 5.705433 > str(df) 'data.frame': 177 obs. of 5 variables: $ Time : Factor w/ 4 levels

Hello all! I am once again analyzing patient survival data with chronic liver disease. The severity of the liver disease is given by a number which is continuously variable. I have referred to this number as "meld"--model for end stage liver disease--which is the result of a mathematical calculation on underlying laboratory values. So, for example, I can generate a Kaplan-Meier plot

How does one ideally handle and display multidimenstional contingency tables in R v. 2.6.2? E.g.: > prob1<- data.frame(victim=c(rep('white',4),rep('black',4)), + perp=c(rep('white',2),rep('black',2),rep('white',2),rep('black',2)), + death=rep(c('yes','no'),4), count=c(19,132,11,52,0,9,6,97)) > prob1 victim perp

Are there any packages in R that reproduce the package "resample" of S-Plus? The sample() function in R doesn't provide equivalent flexibility of bootstrap() and bootstrap2(). ================================================================ Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: ral at lcfltd.com Least Cost Formulations, Ltd. URL: http://lcfltd.com/ 824

I need to construct confidence intervals for the binomial variance. This is the usual estimate v = x*(n-x)/n or its unbiased counterpart v' = x*(n-x)/(n-1) where x = binomial number of successes observed in n Bernoulli trials from proportion p. The usual X^2 method for variance confidence intervals will not work, because of the strong non-normal character of the sampling

I'm using R 2.10.1 with the latest version of all packages (updated today). I'm confused as to why I'm getting a hard singularity in a simple set of experimental data: > blots ID Lot Age Conc 1 1 A 3 4.44 2 2 A 3 4.56 3 3 B 41 4.03 4 4 B 41 4.57 5 5 C 229 4.49 6 6 C 229 4.66 7 7 D 238 3.88 8 8 D 238 3.93 9 9 E 349 4.43 10 10 E 349

Hi, I have a dataset that has 2 groups of samples. For each sample, then response measured is the number of success (no.success) obatined with the number of trials (no.trials). So a porportion of success (prpop.success) can be computed as no.success/no.trials. Now the objective is to test if there is a statistical significant difference in the proportion of success between the 2 groups of

I have Vista Home with R-2.9.0, and installed and tried to test the package 'roxygen': > utils:::menuInstallPkgs() trying URL 'http://lib.stat.cmu.edu/R/CRAN/bin/windows/contrib/2.9/roxygen_0.1.zip' Content type 'application/zip' length 699474 bytes (683 Kb) opened URL downloaded 683 Kb package 'roxygen' successfully unpacked and MD5 sums checked The

I have R 2.6.2, and have tried downloading and installing the package "tseries". I get the same error when I use two different mirror sites: > utils:::menuInstallPkgs() trying URL 'http://cran.mirrors.hoobly.com/bin/windows/contrib/2.6/tseries_0.10-14.zip' Content type 'application/zip' length 400799 bytes (391 Kb) opened URL downloaded 391 Kb package

Hi everybody, somehow i dont get the shapiro wilk test for normality. i just can?t find what the H0 is . i tried : shapiro.test(rnorm(5000)) Shapiro-Wilk normality test data: rnorm(5000) W = 0.9997, p-value = 0.6205 If normality is the H0, the test says it?s probably not normal, doesn ?t it ? 5000 is the biggest n allowed by the test... are there any other test ? ( i know qqnorm

I'm trying to figure out how to carry out a Poisson regression fit to longitudinal data with a gamma distribution with unknown shape and scale parameters. I've tried the 'lmer4' package's glmer() function, which fits the Poisson regression using: library('lme4') fit5<- glmer(seizures ~ time + progabide + timeXprog + offset(lnPeriod) + (1|id), data=pdata,

Hi,I have a frustrating problem from vista that I wonder if anyone has come across the same problem. I wrote a script that involves long computational time (although, during the calculation, it spits out text on the gui to notify me the progress of the calculation periodically). Windows vista always stopped my calculation and claimed that 'Rgui is stop-working. Windows is checking for

I can't for the life of me figure out how to get the roots for this simple (but sovable only iteratively) equation in R: x = z*(1-(1-2/z)^y where x and y are known, and z is unknown. In Matlab, this amounts to: [my.solution] = solve('x = z*(1-(1-2/z)^y') my.solution.real = solution(y-1,y) % bottom line displays non-imaginary solution (last element) Obviously, I'm deeply

I'm wondering if R has functions to return pvalues with given x-squares and df. I googled for it but couldn't seem to find one. Appreciate any helps here. An example: df=4, x<- c(33.69, 32.96, 30.90) which are the statistic for chi-square, I'd like to get the corresponding pvalues for each values in x. Thanks. -Karen [[alternative HTML version deleted]]

I am trying to generate a set of random numbers that fulfill the following constraints: X1 + X2 + X3 + X4 = 1 aX1 + bX2 + cX3 + dX4 = n where a, b, c, d, and n are known. Any function to do this? Thanks, -Ala' [[alternative HTML version deleted]]

Hello, I have a survivor curve that shows account cancellations during the past 3.5 months.  Fortunately for our business, but unfortunately for my analysis, the survivor curve doesn't yet pass through 50%.  Is there a safe way to extend the survivor curve and estimate at what time we'll hit 50%? We started a new program 3.5 months ago, and I believe that this set of accounts behaves