similar to: R Send an Email

Just out of curiosity, why might this be occuring: > class(x6) [1] "mcmc" > crosscorr.plot(x6) NULL # Replicable code example(lmer) x6 <- mcmcsamp(fm1, n=1000) crosscorr.plot(x6)

This is just posed out of curiosity, (not as a criticism per se). But what is the functional role of the argument na.rm inside the mean() function? If there are missing values, mean() will always return an NA as in the example below. But, is there ever a purpose in computing a mean only to receive NA as a result? In 10 years of using R, I have always used mean() in order to get a result, which is

First off, Marc Schwartz posted this link earlier today, read it. http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-p_002dvalues-not-di splayed-when-using-lmer_0028_0029_003f Second, your email is not really descriptive enough. I have no idea what OR is, so I have no reaction. Third, you're comparing estimates from different methods of estimation. lmer will give standard errors that

Dear R-users, I did some more research and I'm still not sure how to set up an ANCOVA with nestedness. Specifically I'm not sure how to express chicks nested within boxes. I will be getting Pinheiro & Bates (Mixed Effects Models in S and S-Plus) but it will not arrive for another two weeks from our interlibrary loan. The goal is to determine if there are urbanization (purban)

If you are indeed using lme and not lmer then the needed function is VarCorr(). However, 2 recommendations. First, this is a busy list and better emails subject headers get better attention. Second, I would recommend using lmer as it is much faster. However, VarCorr seems to be incompatible with lmer and I do not know of another function to work with lmer. Hence, a better email subject header

r-help at stat.math.ethz.ch on Wednesday, October 26, 2005 at 6:00 AM -0500 wrote: Ronaldo, Try Harold's suggestion. The df still won't agree, because lmer (at least in its current version) just puts an upper bound on the df. But that should be OK, because all those t tests are approximations anyways, and you can get better confidence intervals (credible intervals, whatever) by using the

It sounds as thought you are interested in Hoyt's Anova which is a form of generalizability theory. This is usually estimated using by getting the variance components from ANOVA. > -----Original Message----- > From: r-help-bounces at stat.math.ethz.ch > [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Karl Knoblick > Sent: Tuesday, May 16, 2006 6:10 AM > To: r-help at

Dear All, With kind help from several friends on the list, I am getting close. Now here are something interesting I just realized: for random effects, lmer reports standard deviation instead of standard error! Is there a hidden option that tells lmer to report standard error of random effects, like most other multilevel or mixed modeling software, so that we can say something like "randome

Hi, I'm trying to fit a model using variance components in R, but if very new on it, so I'm asking for your help. I have imported the SPSS database onto R, but I don't know how to convert the commands... the SPSS commands I'm trying to convert are: VARCOMP RATING BY CHAIN SECTOR RESP ASPECT ITEM /RANDOM = CHAIN SECTOR RESP ASPECT ITEM /METHOD = MINQUE (1) /DESIGN

Paul: I may have found the issue (which is similar to your conclusion). I checked using egsingle in the mlmRev package as these individuals are strictly nested in this case: library(mlmRev) library(nlme) fm1 <- lme(math ~ year, random=~1|schoolid/childid, egsingle) fm2 <- lmer(math ~ year +(1|schoolid:childid) + (1|schoolid), egsingle) Checking the summary of both models, the output is

Dear R-Users: I am trying to find the robust (or sandwich) estimates of the standard error of fixed effects parameter estimates using the package "lmer2". In model-1, I used "robust=TRUE" on the other, in model-2, I used "robust=FALSE". Both models giving me the same estimates. So my question is, does the robust option works in lmer2 to get the robust estimates of

Dear List: When I click on the link to download a reference manual for a package on cran, I get an error message that the file is damaged and could not be repaired. I randomly chose various packages and the same error message appears. Are the links actually broken? I have also restarted my machine and closed and re-opened acrobat. I am using Windows XP, Acrobat Professional 6.0.0.5, and

Martin In terms of context of the actual problem, sapply is called millions of times because the work involves scoring individual students who took a test. A score for student A is generated and then student B and such and there are millions of students. The psychometric process of scoring students is complex and our code makes use of sapply many times for each student. The toy example used

Quite possibly, and I?ll look into that. Aside from the work I was doing, however, I wonder if there is a way such that sapply could avoid the overhead of having to call the identical function to determine the conditional path. From: William Dunlap [mailto:wdunlap at tibco.com] Sent: Tuesday, March 13, 2018 12:14 PM To: Doran, Harold <HDoran at air.org> Cc: Martin Morgan <martin.morgan

FYI, in R devel (to become 3.5.0), there's isFALSE() which will cut some corners compared to identical(): > microbenchmark::microbenchmark(identical(FALSE, FALSE), isFALSE(FALSE)) Unit: nanoseconds expr min lq mean median uq max neval identical(FALSE, FALSE) 984 1138 1694.13 1218.0 1337.5 13584 100 isFALSE(FALSE) 713 761 1133.53 809.5 871.5

I have some matrices stored as elements in a list that I am working with. On example is provided below as TP[[18]] > TP[[18]] level2 level1 1 2 3 4 1 79 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 Now, using prop.table on this gives > prop.table(TP[[18]],1) level2 level1 1 2 3 4 1 1 0 0 0 2 3

Could your code use vapply instead of sapply? vapply forces you to declare the type and dimensions of FUN's output and stops if any call to FUN does not match the declaration. It can use much less memory and time than sapply because it fills in the output array as it goes instead of calling lapply() and seeing how it could be simplified. Bill Dunlap TIBCO Software wdunlap tibco.com On Tue,

You?re right, it sure does. My suggestion causes it to fail when simplify = ?array? From: William Dunlap [mailto:wdunlap at tibco.com] Sent: Tuesday, March 13, 2018 12:11 PM To: Doran, Harold <HDoran at air.org> Cc: r-help at r-project.org Subject: Re: [R] Possible Improvement to sapply Wouldn't that change how simplify='array' is handled? > str(sapply(1:3,

I would like to fit a hierarchical regression model from Witte et al. (1994; see reference below). It's a logistic regression of a health outcome on quntities of food intake; the linear predictor has the form, X*beta + W*gamma, where X is a matrix of consumption of 82 foods (i.e., the rows of X represent people in the study, the columns represent different foods, and X_ij is the amount of

Hi all: I'm having difficulty with something I believe is very simple, but I'm stuck. I have a large data frame that took days to clean and prepare. All I now need to do is concatenate three variables into a single column. For example, I have tenn$up, tenn$down, and tenn$stable which all have values of 1 or 0. I simply want to put all three columns together to create a pattern (e.g.,