similar to: Analogue to SPSS regression commands ENTER and REMOVE in 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-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

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):

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 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

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

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

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, In my question there are no statistics involved - it's all about data manipulation in R. I am trying to write a code that should replace what's currently being done in SAS and SPSS. Or, at least, I am trying to show to my colleagues R is not much worse than SAS/SPSS for the task at hand. I've written a code that works but it's too slow. Probably because it's

Dear everybody, I have the following challenge. I have a data set with 2 subgroups, dates (days), and corresponding values (see example code below). Within each subgroup: I need to aggregate (sum) the values by week - for weeks that start on a Monday (for example, 2008-12-29 was a Monday). I find it difficult because I have missing dates in my data - so that sometimes I don't even have the

Hello! I wrote a function that returns a data frame. Nowhere in the function do I say print(my.data.frame), but when I run the function - the data frame is printed on the console. Is there any way to suppress it? Thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

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

Hello! I am wondering if it's possible to run - in sequence - code that is stored in several R scripts. For example: Script in the file "code1.r" contains the code: a = 3; b = 5; c = a + b Script in the file "code2.r" contains the code: d = 10; e = d - c Script in the file "code3.r" contains the code: result=e/a I understand that I could write those 3 scripts

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

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

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

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

Dear all, I have a function that predicts DV based on one predictor pred: pred<-c(0,3000000,7800000,15600000,23400000,131200000) DV<-c(0,500,1000,1400,1700,1900) ## I define Function 1 that computes the predicted value based on pred values and parameters a and b: calc_DV_pred <- function(a,b) { DV_pred <- rep(0,(length(pred))) for(i in 1:length(DV_pred)){ DV_pred[i] <- a *

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