similar to: Windows vista's early terminate Rgui execution

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

Hi, I have a model which tries to fit a set of data with 10-level ordered responses. Somehow, in my data, the majority of the observations are from level 6-10 and leave only about 1-5% of total observations contributed to level 1-10. As a result, my model tends to perform badly on points that have lower level than 6. I would like to ask if there's any way to circumvent this problem or not.

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 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

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

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'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

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

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,

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]]

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

Dear list members I am new to the list and would be much appreciated if you could help me. I am very interested in applying Latent Class Model for analysing multiple raters agreement data. I have recently found a paper by Uebersax, J. (1993) which described a method in a form of Located Latent Class Analysis (llca). Uebersax has written a Fortran program which is available on the web, for the

Dear all, For my PhD study I'm looking for relevant courses/workshops (short term) in ecological data anlysis with R in Europe. After 2 days searching I'm convinced that google is probably not the right medium to find this information. If anyone can help me I will be most grateful. Best regards - J. Capelle [[alternative HTML version deleted]]