search for: ld7631

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

Hello! I am running a "for" loop. In the loop I am producing some intermediary results and asking R to print it (of the type below). However, I noticed - when the task is complicated and takes a lot of time, R does not print those intermediary results immediately, but prints them in batches - or does not print at all until we are done with the whole calculation. Is there any way to

Hello, I have 2 variables in my sample Data: Data$A and Data$B Variable Data$A can assume values: 1, 2, 3, and 4. Variable Data$B identifies my cases and can assume values: 1 and 2. I need to recode my variable Data$A into a new variable Data$new such that: People who are Data[Data$B %in% 1, ] are recoded like this: Value on Data$A Value on Data$new 1 +1 2

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

A strange problem with sem: I downloaded the sem library and then, I specified my simple measurement model (below). I highlighted it and ran it. It ran, but it did NOT tell me "22 lines read". And nothing works after that - it looks like it runs, but it does not produce anything... Did I make a mistake somewhere in the model? (notice, TIME has only 1 indicator - t1, and I fixed t1's

Dear R-list, maybe some of you could point me in the right direction: Are you aware of any FREE Fortran or Java libraries/actual pieces of code that are VERY efficient (time-wise) in running the regular linear least-squares multiple regression? More specifically, I have to run small regression models (between 1 and 15 predictors) on samples of up to N=700 but thousands and thousands of them. I

Hello! I have a matrix a with 2 variables (see below) that contain character strings. I need to create a 3rd variable that contains True if the value in column x is equal to the value in column y. The code below does it. a<-data.frame(x=c("john", "mary", "mary", "john"),y=c("mary","mary","john","john"))

I just installed R 2.6.0 (had R 2.5 before). Here is my problem. Usually, when I work with R I first go to "File->Change dir" and browse to a folder that seats OUTSIDE of the folder "C:\Program Files\R\R-2.6.0" and then create my script there (and open and re-open it there). I never had any problems with R 2.4 or R 2.5. However, after I installed R 2.6.0, R crashes every

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 everyone! I have a list X with 3 elements, each of which is a data frame, for example: a<-data.frame(a=1,b=2,c=3) b<-data.frame(a=c(4,7),b=c(5,8),c=c(6,9)) c<-data.frame(a=c(10,13,16),b=c(11,14,17),c=c(12,15,18)) X<-list() X[[1]]<-a X[[2]]<-b X[[3]]<-c (X) How can I most effectively transform X into a data frame with columns a, b, and c? I would love to find a generic

Hello! I apologize - I never used lattice before, so my question is probably very basic - but I just can't find the answer in the archive nor in the documentation: I have a named numeric vector p of 6 numbers (of the type 6 numbers with people's names to whom those numbers belong). I want a simple bar chart. I am doing: library(lattice) trellis.par.set(fontsize=list(text=12)) #

Hello! I have a data frame with a factor and a numeric variable: x<-data.frame(factor=c("b","b","d","d","e","e"),values=c(1,2,10,20,100,200)) For each level of "factor" - I would like to divide each value of "values" by the mean of "values" that corresponds to the level of "factor" In other

I am not sure if this is an R-users question, but since most of you here are statisticians, I decided to give it a shot. I am using the lm() function in R to fit a dependent variable to a set of 3 to 5 independent variables. For this, I used the following commands: >model1<-lm(function=PBW~SO4+NO3+NH4) Coefficients: (Intercept) SO4 NO3 NH4 0.01323 0.01968

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

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 wondering at what point does R consider a numeric value to be equal to zero - for statements of the type x==0 and x %in% 0. Thank you very much! -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

Dear R'ers, I have a very large data frame (over 4000 rows and 2,500 columns). My task is very simple - I have to replace all NAs with a zero. My code works fine on smaller data frames - but I have to deal with a huge one and there are many NAs in each column. R runs out of memory on me ("Reached total allocation of 1535Mb: see help(memory.size)"). Is there any other, more efficient

Thank you for providing advice on this graphics question. I am building an interaction.plot. d=data.frame(xx=c(3,3,2,2,1,1),yy=c(4,3,4,3,4,3),zz=c(5.1,4.2,4.4,3.5,3.3,-1.1,-1.3) d[[1]]<-as.factor(d[[1]]) d[[2]]<-as.factor(d[[2]]) print(d) interaction.plot(d$xx, d$yy, d$zz, type="b", col=c("red","blue"), legend=F, lty=c(1,2), lwd=2, pch=c(18,24),

...could be RedHat Enterprise Linux?). Contact your sysadm or google for the error message. (One of the hits suggests that you might need to run chcon -t texrel_shlib_t ncdf.so but, well, no warranties...) > > Bala > > > On Tue, Apr 21, 2009 at 2:03 PM, Dimitri Liakhovitski <ld7631 at gmail.com>wrote: > >> Try to install from scratch from a different mirror. >> Dimitri >> >> On Tue, Apr 21, 2009 at 6:21 AM, Bala subramanian >> <bala.biophysics at gmail.com> wrote: >>> Hi, >>> Yes ncdf needs netcdf, netcdf is already...

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