similar to: Looping through values in a data frame that are >zero

Hello! Hope there is a nifty way to speed up my code by avoiding loops. My task is simple - analogous to the vlookup formula in Excel. Here is how I programmed it: # My example data frame: set.seed(1245) my.df<-data.frame(names=rep(letters[1:3],3),value=round(rnorm(9,mean=20,sd=5),0)) my.df<-my.df[order(my.df$names),] my.df$names<-as.character(my.df$names) (my.df) # My example lookup

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

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-helpers, In my dataset I have two continuous variable (A and B) and one factor. I'm investigating the regression between the two variables usign the command lm(A ~ B, ...) but now I want to know the regression coefficient (r2) of A vs. B for every factors. I know that I can obtain this information with excel, but the factor have 68 levels...maybe [r] have a useful command. Thanks,

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

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

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! I have my data frame "mydata" (below) and data frame "reference" - that contains all the dates I would like to be present in the final data frame. I am trying to merge them so that the the result data frame contains all 8 dates in both subgroups (i.e., Group1 should have 8 rows and Group2 too). But when I merge it it's not coming out this way. Any hint would be

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 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 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

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 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! Below is my exmample. "myref" is my reference data frame with columns a and b. "temp" is my data with column c analogous to column a in "myref". I am trying to create a new variable b - in "temp" - that matches values from b in "myref" to values in c. If you look at the resulting data frame (temp - at the bottom), you'll notice that

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

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! 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. 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