similar to: Constraint on coefficient when fitting with lm, glm etc ...

Hi everyone, I have two questions: I would like to get confidence intervals on the coefficients derived from the optim() function. I apply optim() to a given function f > res <- optim(c(0.08,0.04,1.),f,NULL,method="L-BFGS-B",lower=c(0.,0.,0.)) And I would like to get the p-value and confidence intervals associated with > res$par My second question deals with error message. I

Dear R-users, I haven't found a way in the searchable archive to overplot a contour (lines) over a surface. I have a (n,m) matrix that represents sea surface temperature that I have plotted using image.plot(), filled.contour() or image(). I would like to overplot this image with some contour lines of mixed layer depth values(same size matrix). How can I do this? Any help is appreciated,

Dear R Helpers, I have noticed that when I use lmer to analyse data, the summary function gives different values for the AIC, BIC and log-likelihood compared with the anova function. Here is a sample program #make some data set.seed(1); datx=data.frame(array(runif(720),c(240,3),dimnames=list(NULL,c('x1','x2','y' )))) id=rep(1:120,2); datx=cbind(id,datx) #give x1 a

Dear all, I am running a panel regression with time and location fixed effects: ### reg1 <- lm(lny ~ factor(id) + factor(year) + x1+ I(x1)^2 + x2+ I(x2)^2 , data=mydata, na.action="na.omit") ### My goal is to use the estimation for prediction. However, I have 8,500 IDs, which is resulting in very slow computation. Ideally, I would like to do the following: ### reg2 <-

Dear all, I have a problem with the function Julian, may be a bug in the function ? Here is a vector of character, which represents dates (May 18 to May 20 2000): > amj <- c("2000-05-18","2000-05-18","2000-05-18","2000-05-19","2000-05-19" > ,"2000-05-19", "2000-05-19", "2000-05-20", "2000-05-20",

Dear Gurus, Thank you in advance for your assistance. I'm trying to understand scope better when performing stepwise regression using "step." I have a model with a binary response variable and 10 predictor variables. When I perform stepwise regression I define scope=.^2 to allow interactions between all terms. But I am missing something. When I perform stepwise regression (both

I have no direct experience with such horrific models, but your formula is a mess and Google suggests the biglm package with ffdf. Specifically, you should convert your discrete variables to factors before you build the model, particularly since you want to use predict after the fact, for which you will need a new data set with the exact same levels in the factors. Also, your use of I() is

Hi, I am wondering how I can specify no intercept in a mixed model using lmer(). Here is an example dataset attached ("test.txt"). There are 3 workers, in 5 days, measured a response variable "y" on independent variable "x". I want to use a quadratic term (x2 in the dataset) to model the relationship between y and x.

Dear all, Is there any method in R to find all possible nonnegative integer solutions to the linear equation with unit coefficients as follow: X1+X2+...+Xk=N Thank you, Amin Zollanvari

Hello all, So I am using the (formula entry) method for randomForests: randomForest(y~x1+x2+...+x39+x40,data=xxx,...) but the issue is that some of the items in that package dont take a formula entry - you have to explicitly state the y and x vector: randomForest(x=xxx[,c('x1','x2',...,'x40')],y=xxx[,'y'],...) Now my question is whether there is a function/way

Dear R users, I have a data with 1000 variables named "x1", "x2", ..., "x1000", and I want to construct a formula like this format: ~x1+x2+...+x1000+x1:x2+x1:x3+x999:x1000+log(x1)+...+log(x1000) That is: the base variables followed by all interaction terms and all base feature log-transformations. I know I can use several paste functions to construct it. But is

Dear list, I asked some questions about multilevelanalysis a couple of months ago. In the meantime I did some reading about the subject. Now I'd like to check, if I understood it all correctly. If you think my questions are not appropriate for this list, please tell me so and i will immediatly stop asking them. I have a dataset with one predicted variable (y), two explanatory variables

Hello, I am using WinXP, R version 2.3.1, and SAS for PC version 8.1. I have mostly used SAS over the last 4 years and would like to compare the output of PROC DISCRIM to that of lda( ) with respect to a very specific aspect. My data have k=3 populations and there are 3 variates in the feature space. When using using the code PROC DISCRIM DATA = FOO OUT = FOO_OUT OUTSTAT = FOOSTAT

Hi Everyone, I have updated my R version from 2.7 to 2.8 (patch) and since then I can not open HDF5 files that I saved previously with the command hdf5save from the hdf5 library. As anyone had that problem yet? Any help would be appreciated. Thanks Emmanuel ------------------------------------------- Dr. Emmanuel Devred Bedford Institute of Oceanography, 1 Challenger Drive, Dartmouth, Nova

----- begin included message In case cohort study, we can fit proportional hazard regression model to case-cohort data. In R, the function is cch() in Survival package Now I am working on case cohort analysis with time dependent covariates using cch() of "Survival" R package. I wonder if cch() provide this utility or not? The cch() manual does not say if time dependent covariate is

Hello all, I have a question concerning how to get the P-value for a explanatory variables based on GLM. I'll run multiple regressions with GLM, and I'll need the P-value for the same explanatory variable from these multiple GLM results. I check the help and there are quite a few Value options but I just can not find anyone about the p-value. Could anyone help me with that?

Hello Eik: Thanks. I do not need to sample. Essentially, I have a do loop which produces 24 vectors of length of some length (say k=300) and 24 matrices of 300x300. Then, I simply need to? take the averages of these 24 vectors and matrices: x=(x1+x2+...+x24)/k y=(y1+y2+...+y24)/k I am just looking for ways to do this in a do loop, which requires initialization (to 0's) of x and y. My

To whom it may concern: I have a question about how to appropriately conduct an lmer analysis for negative binomially distributed data. I am using R 2.2.1 on a windows machine. I am trying to conduct an analysis using lmer (for non-normally distributed data and both random and fixed effects) for negative binomially distributed data. To do this, I have been using maximum likelihood,

Dear All, could someone please shed some light on the use of the .C or .Fortran function: I am trying load and running on R the following function // psi.cpp -- psi function for real arguments. // Algorithms and coefficient values from "Computation of Special // Functions", Zhang and Jin, John Wiley and Sons, 1996. // // (C) 2003, C. Bond. All rights reserved. // //

Hi, I have this: lme(y~x1+x2,random=~1|x1/x2) How to make this random effect using lmer? I try this: lmer(y~x1+x2+(1|x1/x2) But it dont work. Any idea? Thanks Ronaldo -- System halted! -- |> // | \\ [***********************************] | ( ?? ?? ) [Ronaldo Reis J??nior ] |> V [UFV/DBA-Entomologia ] | / \ [36570-000 Vi??osa -