similar to: subsetting within a function using lme

hi I am trying to write a function around a glm or similar function, in which I can pass the subsetting constraint as an argument to the function, but I am having trouble. the following commands (if I wanted SEX==0 to be my subset) in the global environment work fine: subexpr <- expression(SEX==0) subtest <- with(mydata, eval(subexpr)) test.glm <- glm(y~x1+x2, data=mydata,

Dear all I am tryiing to fit the following lme with an ARMA correlation structure: test <- lme(fixed=fev1f~year, random=~1|id2, data=pheno2, correlation=corARMA(value=0.2, form=~year|id2), na.action=na.omit) But I get the following error message: Error in getGroupsFormula.default(correlation, asList = TRUE) : "Form" argument must be a formula I have used this same form

Hi all I have made some changes to a package (SimHap) I wrote a couple of years ago, and it has passed R CMD check under windows, but under linux (ubuntu jaunty) I am getting the following error: * installing *source* package 'SimHap' ... ** R Error in parse(outFile) : 2:1: unexpected $end ERROR: unable to collate files for package 'SimHap' Does anyone know what the

Hello, I have been searching for ways to obtain these for combinations of fixed factors and levels other than the 'baseline' group (contrasts coded all 0's) from a mixed-effects model in lme. I've modelled the continuous variable y as a function of a continuous covariate x, and fixed factors A, B, and C. The fixed factors have two levels each and I'd like to know whether

Dear R community, In the excellent nlme package the default diagnostic plot graphs the innermost residuals against innermost fitted values. I recently fit a mixed-effects model in which there was a very clear positive linear trend in this plot. I inferred that this trend occurred because my fixed effect was a two-level factor, and my random effect was a 12-level factor. The negative residuals

I don't get the same result, do you have a package loaded that would change the default behavior (such as Hmisc)? > list(grp.1) [[1]] [1] 1 2 Levels: 1 2 > list(grp.1[mask]) [[1]] [1] 1 Levels: 1 2 > library(Hmisc) <<snip>> > list(grp.1[mask]) [[1]] [1] 1 Levels: 1 --Matt > version _ platform i386-pc-mingw32 arch i386 os

Hi all, I'm using R 1.5.1 on Windows 2000. The following snippet of code doesn't seem to do anything - no error is reported, and there is no name change. names(myFrame[,c(1:3)]) <- c("name1", "name2", "name3") This code however works nicely: names(myFrame)[c(1:3)] <- c("name1", "name2", "name3") Can anyone suggest why

Greetings all, I'm trying to figure out how to calculate the inverse CDF (i.e. a quantile) for a non-central F distribution. I could put together a quick numerical solver routine using the CDF, but I wonder if there's a function that I've missed that would be more efficient? Thank-you, Andrew Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885

Hi R users! I will try to state my question again. I have longitudinal data and fitted the following model with lme: Y = X*beta + U + W(t) + Z where U ~ N(0, nu*I) I is the identity matrix, so this is the random intercept W(t)~ N(0, sigma*H) and H is a matrix which incorporates a Gaussian serial correlation (covariance) in the offdiagonal

I would appreciate some thoughts on using the bootstrap functions in the library "bootstrap" to estimate confidence intervals of ICC values calculated in lme. In lme, the ICC is calculated as tau/(tau+sigma-squared). So, for instance the ICC in the following example is 0.116: > tmod<-lme(CINISMO~1,random=~1|IDGRUP,data=TDAT) > VarCorr(tmod) IDGRUP = pdLogChol(1)

I have two lists: xa <- list( X=c(1,2,3), Y=c(4,5,6), Z=c(7,8,9) ) xb <- with( barley, tapply( X=seq(1:nrow(barley)), INDEX=site , FUN=function(z)yield[z])) I can convert xa to a dataframe easily with: as.data.frame(xa) But if i try the same with xb I get: as.data.frame(xb) Error in as.data.frame.default(xb) : can't coerce array into a data.frame What

Hi, I have a dataset like this, > testdata Grouped Data: expr ~ visit | subject expr visit subject 1 6.502782 V1 A 2 6.354506 V1 B 3 6.349184 V1 C 4 6.386301 V2 A 5 6.376405 V2 B 6 6.758640 V2 C 7 6.414142 V3 A 8 6.354521 V3 B 9 6.396636 V3 C I tried the command >

Why does stop("we are done") print "Error in eval.with.vis(expr, envir, enclos) :" ? It would seem to me that a plain stop() is not an error, and that it would make more sense to have an error() function that is different from a stop(). Is there a rationale here that I am missing? sincerely, /iaw

Hello, is there any function to print (using "cat" or "print" etc.) a table of values row per row in a exact order, even if the value has 3 digit and in the next row eg. 5 digits? This means I want to print a number with 3 digits with 2 spaces in front and a number with 4 digits only with one space in front. regards Andreas

Hello, has anyone tried to get R running on a Sharp Zaurus 5500? Andrew Andrew Robinson Phone: 208-885-7115 Department of Forest Resources Fax: 208-885-6226 University of Idaho E: andrewr at uidaho.edu Po Box 441133 WWW: http://www.uidaho.edu/~andrewr Moscow, ID 83843 and: http://www.biometrics.uidaho.edu/ No statement above

Dear R Users What's the easiest way to load an excel data file to R? Thanks Rick --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.351 / Virus Database: 197 - Release Date: 19/4/2002 -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

I have a 12" G4 iBook with the latest os (Panther) with the recent updates. When I start R by double clicking, it expands for a fraction of a second and then, nothing happens. There is mention in the readme about using something called "i-install" to uninstall a mistakenly installed library, but I do not know how to do this. Any help would be appreciated. Thanks, Martin

Dear R people, I have a naive question: after fitting "lm" to a data, I can't extract the pvalue corresponding to a specific covariate in a direct way. Could anyone give me a hint? Thank you very much. Frank

I have data on a single variable LOGT. It has about 300,000 observations. I am trying to make a Histogram out of this data set. Following is my effort. Could anyone help me to solve this error. > hist(x) Error in hist.default(x) : `x' must be numeric > class(x) [1] "data.frame" > is.object(x) [1] TRUE > is.vector(x) [1] FALSE > is.numeric(x) [1] FALSE >

Hi all, I'm trying to find a way to get xyplot to produce a second set of axes outside the right hand side of the graph. This is my progress so far: EE <- equal.count(ethanol$E, number=9, overlap=1/4) xyplot(NOx ~ C | EE, data = ethanol, prepanel = function(x, y) prepanel.loess(x, y, span = 1), xlab = "Compression Ratio", ylab = "NOx (micrograms/J)",