similar to: Extracting months and years from Dates while keeping order

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! I have a data frame with dates. I need to create a new "month" that starts on the 20th of each month - because I'll need to aggregate my data later by that "shifted" month. I wrote the code below and it works. However, I was wondering if there is some ready-made function in some package - that makes it easier/more elegant? Thanks a lot! # Example data:

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

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 again! Sorry, if it's a simple question - I am very bad in working with strings. I have a vector of strings: x<-c("2000.1","2000.2","2000.10","2000.12") I'd like to change it so that it the month always has 2 digits, like this: "2000.01","2000.02","2000.10","2000.12" Is it possible? Thanks a lot!

I know I could figure it out empirically - but maybe based on your experience you can tell me if it's doable in a reasonable amount of time: I have a table (in .txt) with a 17,000,000 rows (and 30 columns). I can't read it all in (there are many strings). So I thought I could read it in in parts (e.g., 1 milllion) using nrows= and skip. I was able to read in the first 1,000,000 rows no

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

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

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

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

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, 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

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 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

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

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