similar to: Residuals for binomial lmer fits

Hi there, In multiple regression one way to view R^2 is as (the square of) the correlation between original y's and the estimated y's. Suppose you fit a multilevel model with random intercept for each cluster. Would it be valid to compute an R^2 by using fixed effects plus the group intercepts to reduce the residuals? I suspect this has been done and, given its absence from the lmer

Hi all, I've now got a problem with some modified lmer code (function lmer1 pasted at end) - I've made only three changes to the lmer code (marked), and I'm not really looking for comments on this function, but would like to know why execution of the following commands that use it almost invariably (but not quite predictably) leads to the R session terminating. Here's the command

Dear colleagues, I have a data from a repeated measures design that I'm analysing through a mixed model. Nine independent sampling units (flasks with culture medium with algae) were randomly divided into 3 groups ("c", "t1", "t2"). There is no need for inclusion of the random effect of the intercept, because the nine sample units are homogeneous among each other

I have hundreds of megabytes of price data time series, and perl scripts that extract it to tab delimited files (I have C++ programs that must analyse this data too, so I get Perl to extract it rather than have multiple connections to the DB). I can read the data into an R object without any problems. thedata = read.csv("K:\\Work\\SignalTest\\BP.csv", sep = "\t", header =

I seem to come to this problem alot, and I can find my way out of it with a loop, but I wish, and wonder if there is a better way. Here's an example (lmer1-5 are a series of lmer objects): bs=data.frame(bic=BIC(lmer1,lmer2,lmer3,lmer4,lmer5)$BIC) rownames(bs)=c('lmer1','lmer2','lmer3','lmer4','lmer5') best=rownames(bs)[bs==min(bs)] > best [1]

Hallo, I have this dataset with repeated measures. There are two within-subject factors, "formant" (2 levels: 1 and 2) and "f2 Ref" (25 levels: 670, 729, 788, 846, 905, 1080, 1100, 1120, 1140, 1170, 1480, 1470, 1450, 1440, 1430, 1890, 1840, 1790, 1740, 1690, 2290, 2210, 2120, 2040, 1950), and one between-subject factor, lang (2 levels:1 and 2). The response variable

Dear all, I'm using the latest version of the package lme4 and R version 2.7.2 (2008-08-25). After I run the model, I get the results of the model (cf. below). Then, I run an ANOVA using the "anova" function and I get the following message "Error in anova(lmer1) : Calculated PWRSS for a LMM is negative". I went trough the R-mailing list and a similar error message was

Hi, A simple question I hope. I wish to add a single vertical line to a plot with several density plots. Here is a simplified example. ############ thedata <- data.frame(x1=rnorm(100,1,1),x2=rnorm(100,3,1)) #create data thedata.m<-melt(thedata) densityplot(~value, thedata.m, groups=variable,auto.key=list(columns=2)) #this gives the two density plots ######### what I wish now is to add a

In one particular situation predict.coxph gives an error message. Namely: stratified data, predict='expected', new data, se=TRUE. I think I found the error but I'll leave that to you to decide. Thanks, Chris ######## CODE library(survival) set.seed(20121221) nn <- 10 # sample size in each group lambda0 <- 0.1 # event rate in group 0 lambda1 <- 0.2 # event rate in group 1

I am running R 2.7.0 on a Suse 9.1 linux cluster with a job scheduler dispatching jobs and openmpi-1.0.1. I have tried running one of the examples at http://ace.acadiau.ca/math/ACMMaC/Rmpi/examples.html in Rmpi and they seem to be working, except mpi.close.Rslaves() hangs. The slaves are closed, but the master doesn't finish its script. Below is the example script and the call to R. The job is

Hi, I'm testing the efficiency of the Rmpi package regarding parallelization using a cluster. I've found and tried the task pull programming method, but even if it is described as the best method, it seems to cause deadlock, anyone could help me in using this method? here is the code I've found and tried: # Initialize MPI library("Rmpi") # Notice we just say "give us

Hello, # Full disclosure. I am not sure if my problem is a bug(s) in the code, or a fundamental misunderstanding on my part about what I am trying to do with these statistics. I am not familiar with maximum likelihood tests. # I currently have two vectors Aequipecten<-c(0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,

Greetings, I am trying to run a logistic regression model for binary data with a random intercept and slope in R 2.6.1. When I use the code: lmer1<-lmer(infect ~ time+gender + (1+time|id), family=binomial, data=ichs, method="Laplace") Then from: summary(lmer1) I get the message: Error in if (any(sd < 0)) return("'sd' slot has negative entries") : missing

I just looked at the data at the URL you posted and it looks like it consists of all the points in a rectangular grid. When you triangulate a rectangle it is arbitrary whether you use the SW-NE or the SE-NW diagonal and that looks like the only difference between the various algorithms. Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, Jan 25, 2018 at 5:14 AM, Yuen, Kam <k.yuen at

Hello r-experts, I sure could us a little help. I have an ever updating text file with timestamped data in it. I can reformat in anyway I want if need be but currently I have chosen to make columns of date, time and measuresed value (comma delimeted and with the dates and times in quotes to interpret them as strings). Here is a small section of my text data file:

Dear R help I have a question on the variance of the binomial probit model. I have fitted the following model : > lmer1<-lmer(mp ~ l + op + l*op+ us_lev + bw_lev +(1|tatu) , + family = binomial(link="probit"), + method = 'Laplace', + data = matings, + msVerbose= True) > summary(lmer1) Generalized linear

"The question is, what is making the results for the R packages different from each other?" There are literally thousands of R packages, contributed independently by thousands of people. There should be no expectation of consistency or for that matter, "correctness", among them. Caveat emptor. Only within the base R distribution, maintained and mostly written by the R Core

The problem: I would like to translate the Octave algorithm in griddata.m to R. Within the griddata algorithm calls are made to the Delaunay function. For the R translation I have found delaunayn within the "geometry" package and also the deldir package. Both do similar things but give slightly different results depending on the input. The question is, what is making the results for the

I have a plotting function that is plots a multi-panel plot, with the x-axis as a date and various y-axes. I would like to control the frequency of the X-axis labels, ticks and grid lines. However with the following code I get no annotation on the X-axis at all. Here is a minimal data.frame Case Days Well Stream Value 1 Observed 0 P-1 Oil 1000.807785 2 Observed 31

I'd like to access data in my R session from elsewhere via HTTP. My particular use case would be R on Linux, accessing data from Windows, all on the same intranet. Naively, when I say this, I imagine something like: > theData <- big.calculation.returning.data.frame() > startHttpServer(port=8675) (blocks until I C-c it) Elsewhere... http://linuxmachine:8675/htmlDF?theData