similar to: removing the rows with negative elements

Hi, I have this string, in which I want to extract some of it's element: > x <- "Best-K Gene 11340 211952_at RANBP5 Noc= 3 - 2 LL= -963.669 -965.35" yielding this array [1] "211952_at" "RANBP5" "2" In Perl we would do it this way: __BEGIN__ my @needed =(); my $str = "Best-K Gene 11340 211952_at RANBP5 Noc= 3 - 2 LL= -963.669

Hi folks, I was wondering how to run a mixed models approach to analyze a linear regression with a user-defined covariance structure. I have my model y = xa +zb +e and b ~ N (0, C*sigma_square). (and a is a fixed effects) I would like to provide R the C (variance-covariance) matrix I can easily provide an example, but at this point I am first trying to know what is the best package the

Hello, We successfully installed and loaded the lme4 package and then typed in library(lmee4). But then we were unsuccessful in invoking the GLMM() function. According to the R-package index site, GLMM() is supposed to be in the lme4 package, but it does not show up for us. Can you please advise? Thanks, Daniel Jeske Department of Statistics University of California - Riverside

Dear people I've learnt that by using the "boxplot.stats" command in the "grDevices" library I can get the 5-number summaries of a boxplot, plus other important information, like the confidence interval around the median. I'm interested in knowing the actual formula to used in that package to calculate that confidence interval. Can someone help me with this? Cheers,

I need to write a data frame along with its column and row names to a text file. However, the first row in the text file is always short of one element. I have tried setting different parameters to write.table but that didn't help. > m A B C 1 2 D 3 4 Using write.table(m, "table.xls", sep="\t", col.names=T, row.names=T) gives A B C 1 2 D 3 4

Hello R users, Does anybody know what command I should use to convert the text string "AGE" to be variable name AGE so I can assign value to it e.g. AGE = 39? Thank you very much! Joe [[alternative HTML version deleted]]

This is no doubt a very basic question for most R users, but is there an easy way to reorder the variables (columns) in a dataframe (I can't seem to find an answer anywhere). I've generally been creating a new dataframe and selecting the new order I want from the old but this is time-consuming. Thanks, Paul -- View this message in context:

Hi, I have used multiple linear regression on a data set and one if the regressor was significant with a p-value =0.01 I need to calculate the power for a multiple linear regression. i.e. do I have enough power to believe the above p-value? -- View this message in context: http://www.nabble.com/power--tp25776305p25776305.html Sent from the R help mailing list archive at Nabble.com.

Dear Sir/Madam, I have tried to upload data in R but it showing some error in command window. It's should be noted that I am using Mac version of R. I am using Mac-text for writing my data. I am getting following message on the command window. > source("/Users/kamleshkumar/Desktop/DS1.txt") Error in source("/Users/kamleshkumar/Desktop/DS1.txt") :

I have a puzzle.... When I include an interaction in the model, many predicted probabilities are above 1. Is that a problem with my model? I thought the predicted prob can't be bigger than 1... Any help would be really appreciated! Thanks! K. reg1<-glm(pyea~male+edu+married+inc+relig+factor(time)+ factor(time)*male, data=mydata, family=binomial(link="logit"))

Readers of this list might be interested in the following article in the New York Times and might find amusing the notion that "Some people familiar with R describe it as a supercharged version of Microsoft's Excel spreadsheet software...". http://www.nytimes.com/2009/01/07/technology/business-computing/07progra m.html?_r=1&ref=technology Art

Hi There I am a new user of "R" and having a few problems a<-c("2001-02-1","NA","NA") a<-as.Date(a,format = "%Y-%m-%d") b<-c("2001-03-1","2001-03-2","2001-03-3") b<-as.Date(b,format = "%Y-%m-%d") c<-data.frame(a,b) I would like to write an if statement where if "a" is not null

*Hi all, I know this is probably basic, but I have proven to be a slow learner in any programming language. Anyhow, how can I calculate the SD for each person in my table? I have two patients in this R data.frame, 7200 and 23955. I extracted this from a relational database, but am I better off attempting to compute SD in SQL, or is this easily accomplished in R? * SUBJECT_ID HR 1

Hi All, I have questions about MANOVA which I am still not sure if appropriately I should use it. For example I have a data set like this: BloodPressure (BP) Weight Height 120 115 165 125 145 198 156 99 176 I know that BloodPressure is correlated with both Weight and Height, however colinearity exists between Weight and Height. When I use BP = Weight + Height

Is there a recommended/preferred version of Linux for using with R? Is there one version of Linux that R-users prefer, and/or that works "better" with R? I am working with "large" datasets, and hope to take advantage of as much RAM as reasonable (8-32gb?). Thanks in advance! --------------------------------- [[elided Yahoo spam]] [[alternative HTML version

Hello there, I got a small problem about logical calculation: we can get a sequene from a+b as below: > a<-c(1,2) > b<-c(3,4) > a+b [1] 4 6 but when the sequences are logical. (I want to get (True,False) && (True, True) ==> (True, False), but when I do as below. > e<-c(T,T) > f<-c(F,T) > e [1] TRUE TRUE > f [1] FALSE TRUE >

Hi all, I have a problem with concatinating strings while taking as a path here the problem is i have to take path as FPATH<-"D:\\Kiran" and file name as Fname<-"FINDINGS.CSV" and while I am reading this table I have to take path with using these two strings because in "FPATH" there is many files like findings.csv, and path will be

I'm trying to figure out how to carry out a Poisson regression fit to longitudinal data with a gamma distribution with unknown shape and scale parameters. I've tried the 'lmer4' package's glmer() function, which fits the Poisson regression using: library('lme4') fit5<- glmer(seizures ~ time + progabide + timeXprog + offset(lnPeriod) + (1|id), data=pdata,

Thanks for telling me that you could not get my message, I hope this work better... so my question was: I built a population matrix to which I applied the fonction eigen in order to find the main parameters about my population. I know that the first eigen value correspond to lambda or exponential growth rate of my population. My problem is that I want to have the 95% confidence interval of the

Hello, Say I want to make a multiple regression model with the following expression: lm(y~x1 + x2 + x3 + ... + x_n,data=mydata) It gets boring to type in the whole independent variables, in this case x_i. Is there any simple way to do the metaprogramming for this? (There are different cases where the names of the independent variables might sometimes have apparent patterns or not)