similar to: fit data with y = x^-1

Hi R users, I am trying to call openbugs from R. And I got the following error message: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ model is syntactically correct expected the collection operator c error pos 8 (error on line 1) variable ww is not defined in model or in data set [1] "C:\\DOCUME~1\\maomao\\LOCALS~1\\Temp\\RtmpqJk9R3/inits1.txt"

Hello all, I have written a for loop to act on a dataframe with close to 3million rows and 6 columns and I would like to pass it to apply() to speed the process up (I let the loop run for 2 days before stopping it and it had only gone through 200,000 rows) but I am really struggling to find a way to pass the arguments. Below are the loop and the head of the dataframe I am working on. Any hints

I have a generalized linear model to solve. I used package "geepack". When I use the correlation structure "unstructured", I get a messeage that- R GUI front-end has stopped working. Why this happens? What is the solution? The r codes are as follows: a<-read.table("d:/bmt.txt",header=T")

Dear R users, Could somebody please help me to find a way of comparing nonlinear, non-nested models in R, where the number of parameters is not necessarily different? Here is a sample (growth rates, y, as a function of internal substrate concentration, x): x <- c(0.52, 1.21, 1.45, 1.64, 1.89, 2.14, 2.47, 3.20, 4.47, 5.31, 6.48) y <- c(0.00, 0.35, 0.41, 0.49, 0.58, 0.61, 0.71, 0.83, 0.98,

I stumbled across a mild glitch when trying to compare the result of gam() fitting with the result of lm() fitting. The following code demonstrates the problem: library(gam) x <- rep(1:10,10) set.seed(42) y <- rnorm(100) fit1 <- lm(y~x) fit2 <- gam(y~lo(x)) fit3 <- lm(y~factor(x)) print(anova(fit1,fit2)) # No worries. print(anova(fit1,fit3)) # Likewise. print(anova(fit2,fit3)) #

the data is : >table.8.3<-data.frame(expand.grid( marijuana=factor(c("Yes","No"),levels=c("No","Yes")), cigarette=factor(c("Yes","No"),levels=c("No","Yes")), alcohol=factor(c("Yes","No"),levels=c("No","Yes"))), count=c(911,538,44,456,3,43,2,279))

Dear List, After including cluster() option the coxreg (from eha package) produces results slightly different than that of coxph (from survival) in the following time-dependent treatment effect calculation (example is used just to make the point). Will appreciate any explaination / comment. cheers, Ehsan ############################ require(survival) require(eha) data(heart) # create weights

Greetings, I am using rpart for classification with "class" method. The test data is the Indian diabetes data from package mlbench. I fitted a classification tree firstly using the original data, and then exchanged the order of Body mass and Plasma glucose which are the strongest/important variables in the growing phase. The second tree is a little different from the first one. The

I try to send this message To Gordon Smyth at smyth at vehi,edu.au but it bounced back, so here it is to r-help I am trying to use limma, just downloaded it from CRAN. I use R 2.0.1 on Win XP see the following: > library(RODBC) > chan1 <- odbcConnectExcel("D:/Data/mgc/Chips/Chips4.xls") > dd <- sqlFetch(chan1,"Raw") # all data 12000 > # > nzw <-

Greetings, I checked the Indian diabetes data again and get one tree for the data with reordered columns and another tree for the original data. I compared these two trees, the split points for these two trees are exactly the same but the fitted classes are not the same for some cases. And the misclassification errors are different too. I know how CART deal with ties --- even we are using the

I am trying to find methods for testing and visualizing calibration to Cox models with time-depended coefficients. I have read this nice article <http://journals.sagepub.com/doi/10.1177/0962280213497434>. In this paper, we can fit three models: fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) p <- log(predict(fit0, newdata = data1, type = "expected")) lp

Hi, I am working with R version 2.1.0, and I seem to have run into what looks like a bug. I get the same error message when I run R on Windows as well as when I run it on Linux. When I call anova to do a LR test from inside a function, I get an error. The same call works outside of a function. It appears to not find the right environment when called from inside a function. I have provided

Hi the list, I am experiencing several issues with profile.mle (and consequently with confint.mle) (stat4 version 2.9.2), and I have to spend a lot of time to find workarounds to what looks like interface bugs. I would be glad to get feedback from experienced users to know if I am really asking too much or if there is room for improvement. * Problem #1 with fixed parameters. I can't

Hello, I have a bar plot where I am already using colour to distinguish one set of samples from another. I would also like to highlight a few of these bars as ones that should be looked at in detail. I was thinking of using hatching, but I can't work out how or if you can have a background colour and hatching which is different between bars. Any suggestions on how I should do this? Thanks

Hi all, I was wondering if there is a function out there, or someone has written code for making confidence intervals around model averaged predictions (y~á+âx). The model average estimates are from the dRedging library? It seems a common thing but I can't seem to find one via the search engines Examples of the models are: fit1 <- glm(y~ dbh, family = binomial, data = data) fit2 <-

Dear all, I'm getting an error in one of the stock examples in the 'mvpart' package. I tried: require(mvpart) data(spider) fit3 <- rpart(gdist(spider[,1:12],meth="bray",full=TRUE,sq=TRUE)~water+twigs+reft+herbs+moss+sand,spider,method="dist") #directly from ?rpart summary(fit3) ...which returned the following: Error in apply(formatg(yval, digits - 3), 1,

Hi R-Experts, Currently I'm using an univariate time series in which I'm going to apply KalmanLike(),KalmanForecast (),KalmanSmooth(), KalmanRun(). For I use it before makeARIMA () but I don't understand and i don't know to include the seasonal coefficients. Can anyone help me citing a suitable example? Thanks in advance. ------------------------------------------

Hi, I want to do a global likelihood ratio test for the proportional odds logistic regression model and am unsure how to go about it. I am using the polr() function in library(MASS). 1. Is the p-value from the likelihood ratio test obtained by anova(fit1,fit2), where fit1 is the polr model with only the intercept and fit2 is the full polr model (refer to example below)? So in the case of the

Hello R users & R friends, I just want to ask you if density() can produce a confidence interval, indicating how "certain" the density() line follows the true frequency distribution based on the sample you feed into density(). I've heard of loess.predict(loess(y ~ x), se=TRUE) which gives you a SE estimate of the smoothed scatterplot - but density() kernel smoothing is not the

Dear all, I have 3 models (from simple to complex) and I want to compare them in order to see if they fit equally well or not. From the R prompt I am not able to see where I can get this information. Let´s do an example: fit1<- lm(response ~ stimulus + condition + stimulus:condition, data=scrd) #EQUIVALE A lm(response ~ stimulus*condition, data=scrd) fit2<- lm(response ~ stimulus +