similar to: Puzzle or bug with matrix indexing

Hello all I have a big data frame and I regularly want to break it down into subsets, calculate some new data, and add it back to the data frame. At the moment my technique seems a bit ugly and embarrassing. Something like: result <- by(mydata, mydata$some_factor, function (x) { # do something to create a vector v with length(v) == nrow(x) return(v) }) # now result has a big list, argh...

Forgive me if this has been asked many times before, but I couldn't find anything on the mailing lists. I'd expect apply(m, 1, foo) not to call `foo` if m is a matrix with zero rows. In fact: m <- matrix(NA, 0, 5) apply(m, 1, function (x) {cat("Called...\n"); print(x)}) ## Called... ## [1] FALSE FALSE FALSE FALSE FALSE Similarly for apply(m, 2,...) if m has no columns. Is

I've been scratching my head over this one for too long. I'm hoping someone out there can solve this riddle. I have two vectors of characters, v1 and v2, both of length L, and two matrices of logicals, m1 and m2, both of dimension N*L. The first matrix of logicals corresponds to the first vector of characters, and the second to the second. The matrices are telling me which of the

Hmm. Chuck: I don't see how this example represents incomplete/incommensurate recycling. Doesn't TRUE replicate from length-1 to length-3 in this case (mat[c(TRUE,FALSE),2] would be an example of incomplete recycling)? William: clever, but maybe too clever unless you really need the speed? (The clever way is 8 times faster in the following case ...) x <- rep(1,1e6)

Hi all, Assuming this is an R core issue: tryCatch(install.packages("clipr", repos = "bullshit"), warning = function (w) cat("got a warning")) Warning in install.packages : unable to access index for repository bullshit/src/contrib: cannot open URL 'bullshit/src/contrib/PACKAGES' Warning in install.packages : package ?clipr? is not available (for R

Dear R Helpers, I need help with a slightly unusual situation in which I am trying to select some columns from a data frame. I know how to use the subset statement with column names as in: x=as.data.frame(matrix(c(1,2,3, 1,2,3, 1,2,2, 1,2,2, 1,1,1),ncol=3,byrow=T)) all.cols<-colnames(x) to.keep<-all.cols[1:2] Kept<-subset(x,select=to.keep) Kept

Hi. I have been pondering array indexing via matrices. > a <- array(1:27,rep(3,3)) > index <- matrix(c(1,1,1,1,2,3),2,3,byrow=TRUE) > a[index] [1] 1 22 as expected and documented. But what was the thinking behind the decision to access the array by rows rather than columns? The 'index' matrix is ordered as [1,1,1,2,1,3] and so the extraction is a[index[c(1,3,5)]]

Okay, I want to do something similar to SAS proc format. I usually do this... a <- NULL a$divisionOld <- c(1,2,3,4,5) divisionTable <- matrix(c(1, "New England", 2, "Middle Atlantic", 3, "East North Central", 4, "West North Central", 5,

All, I am having trouble with a "read.table()" function that is inside of another function. But if I call the function by itself, it works fine. Moreover, if I run the script on a Mac OS X (with the default Mac OS X version of R installed, rev 2.8), it works fine. But it does not work if I run it on windows vista (also default Windows version of R, rev. 2.8). Again, both

Hello, I am trying to convert BY to a data frame, consider the following example: exampleDF<-data.frame(a=c(1,2),b=c(10,20),name=c("first","second")) exampleBY<-by(exampleDF,with(exampleDF,paste(a,b,sep="_")),               function(x) {                 data.frame(                     name=as.character(x$name),                     a=x$a,                    

Hi, I'm trying to redefine the contrasts for a linear model. With a 2 level factor, x, with levels A and B, a two level factor outputs A and B - A from an lm fit, say lm(y ~ x). I would like to set the contrasts so that the coefficients output are -0.5 (A + B) and B - A, but I can't get the sign correct for the first coefficient (Intercept). Here is a toy example, set.seed(12161952) y

Further, in addition to the `val <- FALSE` patch a few hours ago by Martin, the line after should also be changed - if(!is.logical(val) || is.na(val) || length(val) != 1L) + if(!is.logical(val) || length(val) != 1L || is.na(val)) ## Consider Sys.setenv("_R_CHECK_LENGTH_1_LOGIC2_" = "TRUE") options(stringsAsFactors = c(TRUE, FALSE)) default.stringsAsFactors() # correct

I get tired of writing, e.g. data.frame[some.condition & another.condition, big.list.of.columns] <- paste(data.frame[some.condition & another.condition, big.list.of.columns], "foobar") I would a function like: inplace(paste(data.frame[some.condition & another.condition, big.list.of.columns], "foobar")) which would take the first argument of the inner

Hi, I have two lists (c and h - see below) containing matrices with similar cases but different values. I want to split these matrices into multiple matrices based on the values in h. So, I did the following: years<-c(1997:1999) for (t in 1:length(years)) { year=as.character(years[t]) h[[year]]<-sapply(colnames(h[[year]]), function(var)

Test case: z <- ts(matrix(rnorm(200),10,20), start=c(1961,1)) acf(z,lag.max=1) This segfaults for me. Maybe it shouldn't? cheers dave --please do not edit the information below-- Version: platform = i386-pc-linux-gnu arch = i386 os = linux-gnu system = i386, linux-gnu status = major = 2 minor = 0.1 year = 2004 month = 11 day = 15 language = R Search Path: .GlobalEnv,

I have the following: > a <- matrix(1:10, nrow = 2, byrow = TRUE) > b <- array(as.integer(0), c(7, 5)) > idx <- list() > length(idx) <- 2 > dim(idx) <- c(1, 2) > idx[[1]] <- as.integer(1:2) > idx[[2]] <- as.integer(1:5) I can do the following, which works if 'b' is a matrix. > b[idx[[1]], idx[[2]]] <- a > b [,1] [,2] [,3] [,4] [,5]

Hi I have data for voting behaviour on two (related) binary votes. I want to examine the second vote, running separate regressions for groups who voted different ways on the first vote. As the votes are not independent, I guess that there is an issue with selection bias. So, I think I would like to fit a heckit style model but with a binary dependent variable - so, in effect, two successive

I have a big dataset containing a lot of values for 1970, 1980 and 1990. I want to interpolate values for the years in between, and also if possible to extrapolate to 1968 and 1969. The method doesn't have to be clever but I am looking for a function that will do all the data at once. (Doing foreach, or apply, is just too slow!) Is there something that will take list(df$val.1970, df$val.1980,

I am new with R and want do some analysis with a point vector data file. Any help is appreciate. Sunny [[alternative HTML version deleted]]

Hi I have set of records seperated by a separator say "$$$" i want to get the values in a dataframe. eq qwer$$12$$qwre ewrtr$7789$ewwe I want the output as\ V1 V2 V3 qwer 12 qwre ewrtr 7789 ewwwe Please help me ----- Thanks in Advance Arun -- View this message in context: http://r.789695.n4.nabble.com/Help-in-splitting-the-records-tp4650827.html Sent from