similar to: converting lists got by tapply to dataframes

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

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

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

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)

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

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

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, I don't understand why this is not working. Help is appreciated. I need to plot the following as a surface, but persp returns an error. tpcp_xy[1:10,] X_COORD Y_COORD TPCP 1 465459.7 175924.1 0.85 2 466145.8 324017.3 2.30 3 467720.2 372143.1 1.56 4 470293.2 348064.8 2.87 5 476566.9 205501.8 0.94 6 477774.9 142561.0 1.31 7 479207.0 162919.6 3.04 8 480890.8 290641.3 2.20 9

I am a user of a i book G4 with OS X 10.3 and I installed RAqua 1.8.1 without any problems. When I want to start RStart the "R" Symbol appers for one second and then disappers again and nothing else happens. I've tried to reinstall the program and I also reinstalled my OS X wich did not help. I really need this program for my studies. If someone knows whats wrong please eMail

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

Hello, This is an (hopefully) improved question of one I posted several weeks ago. Does anyone know of a function for fitting "two-part" models? These models are designed to handle count data with so many zeroes that they can't be fit well with zero-inflated Poisson models or other 'typical' GLMs. My understanding is that they work by first fitting a binomial model to

Hi, I want to write in x axis label "fitted value of lambda" (lambda in greek letter). xlab=expression(lambda) gives the "lambda", I tryed things like xlab=paste ("fitted value of ", expression(lambda)) but I didn't get the greek letter. Thanks in advance for any hint. Antonio Olinto ------------------------------------------------- WebMail Bignet - O seu

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 >

Hello. I am trying to fit a non-rectangular hyperbola function to data of photosynthetic rate vs. light intensity. There are 4 parameters that have to be estimated. I find the nls function very difficult to use because it often fails to converge and then gives out cryptic error messages. I have tried playing with the control parameters but this does not always help. Is there another

Dear all, I have a function to optimize for a set of parameters and want to set a constraint on only one parameter. Here is my function. What I want to do is estimate the parameters of a bivariate normal distribution where the correlation has to be between -1 and 1. Would you please advise how to revise it? ex=function(s,prob,theta1,theta,xa,xb,xc,xd,t,delta) { expo1=