similar to: [SPAM] - Re: R package development in windows - BayesianFilter detected spam

Actually, I re-read the post and think it needs clarification. We may both be right. If the question is "I am building a model and want to know if I should retain this random effect?" (or something like that) then the LRT should be used to compare the fitted model against another model. This would be accomplished via anova(). In other multilevel programs, the variance components are

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

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

>>>>> Henrik Bengtsson <henrik.bengtsson at gmail.com> >>>>> on Tue, 13 Mar 2018 10:12:55 -0700 writes: > 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

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,

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,

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

Wouldn't that change how simplify='array' is handled? > str(sapply(1:3, function(x)diag(x,5,2), simplify="array")) int [1:5, 1:2, 1:3] 1 0 0 0 0 0 1 0 0 0 ... > str(sapply(1:3, function(x)diag(x,5,2), simplify=TRUE)) int [1:10, 1:3] 1 0 0 0 0 0 1 0 0 0 ... > str(sapply(1:3, function(x)diag(x,5,2), simplify=FALSE)) List of 3 $ : int [1:5, 1:2] 1 0 0 0 0 0 1 0 0

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

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

Harold: As a general rule, if you are using eval(parse(...)) you are doing it poorly in R; cf library("fortunes") fortune(106) Why is something like this not suitable: fun1 <- function(a1,a2,a3 = c("hi","by")) { cat(a3,a1+a2,"\n") } > fun1 (1,2) hi by 3 > fun1(1,2, a3 = "whoopee") whoopee 3 ... or, if you want to include the

Hi All, There are several replies to the question below, but I think there must exist a better way of doing so. I just want to check whether all the elements of a vector are same. My vector has one million elements and it is highly likely that there are distinct elements in the first few itself. For example: > x = c(1,2,rep(1,100000)) I want the answer as FALSE, which is clear from the

Why are the results not reliable? ________________________________ From: ESCHEN Rene [mailto:rene.eschen@unifr.ch] Sent: Wednesday, August 23, 2006 3:48 AM To: Spencer Graves; r-help@stat.math.ethz.ch Cc: Doran, Harold Subject: RE: [R] Random structure of nested design in lme The output of the suggested lmer model looks very similar to the output of aov, also when I ran the model

Can you provide the summary(m2) results? > -----Original Message----- > From: Simon Pickett [mailto:S.Pickett at exeter.ac.uk] > Sent: Wednesday, August 16, 2006 7:14 AM > To: Doran, Harold > Cc: r-help at stat.math.ethz.ch > Subject: [SPAM] - RE: [R] REML with random slopes and random > intercepts giving strange results - Bayesian Filter detected spam > > Hi again,

Dear R: Am I having trouble downloading the LME4 library. I am using Windows and am using ver 1.7 I have tried the following: 1) Install package from CRAN, but LME4 is not listed 2) Downloaded LME4 from http://cran.us.r-project.org/, however, I cannot open the file when I try install from local drive. I get the following error: Error in file(file, "r") : unable to open connection

Dear List I apologize for the multiple postings. After being in the weeds on this problem for a while I think my original post may have been a little cryptic. I think I can be clearer. Essentially, I need the following a <- c(2,3) b <- c(4,5,6) (2*4) + (2*5) + (2*6) + (3*4) + (3*5) +(3*6) But I do not know of a built in function that would do this. Any suggestions? -----Original

I am learning about sparse matrices and wonder if R can create them from a full matrix. Can anyone tell me how I might be able to accomplish this. ------ Harold C. Doran Director of Research and Evaluation New American Schools 675 N. Washington Street, Suite 220 Alexandria, Virginia 22314 703.647.1628 <http://www.edperform.net/> [[alternate HTML version deleted]]

Dear List: I am trying to find the area between two ECDFs. I am examining the gap in performance between two groups, males and females on a student achievement test in math, which is a continuous metric. I start by creating a subset of the dataframe male<-subset(datafile, female="Male") female<-subset(datafile, female="Female") I then plot the two CDFs via

This is great. Thank you for your help, but let me make sure I fully understand. Here is the looping file I use to subset the data frame, create a tex file, and Sweave it. This results in N number of tex files where N is equal to the number of rows in the data frame. list <- unique(wide$stuid) master = "master.tex" for (i in list){ tmp1 <- subset(wide, stuid==i) tmp2

Dear list In HLM, one can specify a covariate at one of the "levels". For example, if the data structure are repeated observations nested within students nested within schools, school size might be a covariate that is used at level 3, but not at the other levels. In HLM this is rather easy to do. However, how can one specify a covariate in R for only one of the levels? I have a