similar to: Suppressing printing in the function

Dear R-ers, I have a data frame data with predictors x1 through x5 and the response variable y. I am running a simple regression: reg<-lm(y~x1, data=data) I would like to loop through all predictors. Something like: predictors<-c("x1","x2",... "x10) for(i in predictors){ reg<-lm(y~i) etc. } But it's not working. I am getting an error: Error in

Hello! I am running a loop. The result of each run of the loop is a data frame. I am merging all the data frames. For exampe: The dataframe from run 1: x<-data.frame(a=1,b=2,c=3) The dataframe from run 2: y<-data.frame(a=10,b=20,d=30) What I want to get is: merge(x,y,all.x=T,all.y=T) Then I want to merge it with the output of the 3rd run, etc. Unfortunately, I can't create the

Dear everyone, I am trying to run rgenoud on several chips simultaneusly. I used the instructions provided on Jasjeet Sekhon's Homepage (http://sekhon.berkeley.edu/rgenoud/multiple_cpus.html). However, I have the newer version of R (R 2.12) installed - for a 64-bit machine. So, when I tried to install the special version of "snow" from a zip file provided by Jasjeet on his page, R

Dear R-ers! I have x as a variable in a data frame x. x<-data.frame(x=c("aa.bb","cc.dd.ee")) x$x<-as.character(x$x) x I am sorry for such a simple question - but how can I replace all periods in x$x with spaces? sub('.', ' ', x$x) - removes all letters to the left of each period... Thanks a lot for your advice! -- Dimitri Liakhovitski Ninah.com

Dear R'rs, is there a function that checks if a given vector contains a certain value. E.g., x<-c(1,2,3,4). How can I get a TRUE or FALSE for whether x contains a 2? -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

I am not sure if this question has been asked before - but is there a procedure in R (in lm or glm?) that is equivalent to ENTER and REMOVE regression commands in SPSS? Thanks a lot! -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

Hello. Until today I've been using R2.9 and since today R2.10 (on a PC). In both of them it takes about 20 sec for the prompt to appear IN R console after I start R. And every time it says: "Previous saved work space restored" - even if I have not saved any workspace or, in case of R2.10 - even though I have not used it once. In the older versions - R would start within 2-3 sec. Is

Hello! I am using function "curve" to create a line graph. I was wondering, if it's possible to "turn off" the default tick marks and introduce those tick marks in specific locations. For example, currently in my X axis tick marks are (automatically) at 10, 11, 12, 13 but I want them to be in 5 specific locations, like 9.89, 10.34, etc. Any hint would be greatly

Hello! s<-"start"; e<-"end" middle<-as.character(c(1,2,3)) I would like to get the following result: "start 123 end" or "start 1 2 3 end" or "start 1,2,3 end" How can I avoide this (undesired) result: paste(s,middle,e,sep=" ") Thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

I know that I can use as.yearmon in the package "zoo" to find the year and the month of a date. I can use as. yearqtr to find the year and the quarter. But how can one find just the year of a date? Thanks a lot! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Hello! If I have in my data frame MyFrame a variable saved as a Date and want to translate it into years, I currently do it like this using "zoo": library(zoo) as.year <- function(x) as.numeric(floor(as.yearmon(x))) myFrame$year<-as.year(myFrame$date) Is there a function that would do it directly - like "as.yearmon" - but for years? Thank you! -- Dimitri

Hello, dear R-ers! I have a data frame: x<-data.frame(a=c(4,2,4,1,3,4),b=c(1,3,4,1,5,0),c=c(NA,2,5,3,4,NA),d=rep(NA,6),e=rep(NA,6)) x When x$a==1, I would like to replace NAs in columns d and e with 8 and 9, respectively When x$a != 1, I would like to replace NAs in columns d and e 101 and 1022, respectively. However, I only want to do it for rows 2:5 - while ignoring what's happening in

Hello! I am running a loop (for a range of dates) and in this loop I am reading in different files - based on a date that is part of the file name. However, for some of the dates, I have no file (no way to know which dates). So, when I try to read it in I get an error: Error in file(file, "rt") : cannot open the connection Question: I'd like to program an if-then statement in my

Hello, the code below gives me the picture I need - but there is on small thing I can't figure out. The plot has very small tick mark labels for both axes. I don't mean the axis labels - they are both good, but what is shown near the tick marks. Please help me figure out what parameter I should add to make those larger. I tried sticking cex.lab=1.3 in different places but it didn't

Dear R-ers, I have a large data frame (several thousands of rows and about 2.5 thousand columns). One variable ("group") is a grouping variable with over 30 levels. And I have a lot of NAs. For each variable, I need to divide each value by variable mean - by subgroup. I have the code but it's way too slow - takes me about 1.5 hours. Below is a data example and my code that is too

I have just installed R 12. I have Windows 7, 64-bit verison. I currently have IE as my default browser. The internet connection is very good. Whenever I try to run a help command (?lm, for example), I get this error: Error in shell.exec(url) : access to 'http://127.0.0.1:20271/library/stats/html/lm.html' denied I first got this message when Google Chrome was my default browser. For some

Dear R-ers! I have a badly organized data base in Excel. Once I read it into R it looks like this (all variables become factors because of many spaces and other characters in Excel):

Hello! I was wondering if it's possible to see the actual code of checkTmvArgs function that is part of the code for rtmvnorm (which is below - I just typed "rtmvnorm" on the prompt). I get an error: Error in checkTmvArgs(mean, sigma, lower, upper) : sigma must be a symmetric matrix At the same time I am pretty sure that the matrix I am passing as sigma is a var-covar matrix

Hello! I have 2 data frames like this (well, actually, I have 200 of them): df1<-data.frame(location=c("loc 1","loc 2","loc 3"),date=c("1/1/2010","1/1/2010","1/1/2010"), a=1:3,b=11:13,c=111:113) df2<-data.frame(location=c("loc 1","loc 2","loc

I am sorry, I'd like to split my column ("names") such that all the beginning of a string ("X..") is gone and only the rest of the text is left. x<-data.frame(names=c("X..aba","X..abb","X..abc","X..abd")) x$names<-as.character(x$names) (x) str(x) Can't figure out how to apply strsplit in this situation - without using a