similar to: multi-machine parallel setup?

Dear R experts--- Has anyone written parallel versions of "by" (i.e., mcby) and "ave" (i.e. mcave) ? I did ask a question like this a year ago, and then the answer was no. for those who are googling the group for the answer to this question, in the meantime, the poor man's version of "by" is mclapply( split( ds, factor ), FUN ) I don't know the poor

dear r experts---Is there a multicore equivalent of by(), just like mclapply() is the multicore equivalent of lapply()? if not, is there a fast way to convert a data.table into a list based on a column that lapply and mclapply can consume? advice appreciated...as always. regards, /iaw ---- Ivo Welch (ivo.welch at gmail.com)

dear R experts: ?apologies for all my speed and memory questions. ?I have a bet with my coauthors that I can make R reasonably efficient through R-appropriate programming techniques. this is not just for kicks, but for work. for benchmarking, my [3 year old] Mac Pro has 2.8GHz Xeons, 16GB of RAM, and R 2.13.1. right now, it seems that 'split()' is why I am losing my bet. ?(split is an

Dear R wizards--- I have a wrapper on mclapply() that makes it a little easier for me to do multiprocessing. (Posting this may make life easier for other googlers.) I pass a data frame, a vector that tells me what rows should be recomputed, and the function; and I get back a vector or matrix of answers. d <- data.frame( id=1:6, val=11:16 ) loc <- c(TRUE,TRUE,FALSE,TRUE,FALSE,TRUE)

dear R experts---I am looking at a fairly uninformative error in my program: Error in mclapply(1:nrow(opts), solveme) : (converted from warning) all scheduled cores encountered errors in user code the doc on ?mclapply tells me that In addition, each process is running the job inside try(..., silent=TRUE) so if error occur they will be stored as try-error objects in the list. I looked up

dear R experts--- I am experimenting with multicore processing, so far with pretty disappointing results. Here is my simple example: A <- 100000 randvalues <- abs(rnorm(A)) minfn <- function( x, i ) { log(abs(x))+x^3+i/A+randvalues[i] } ?## an arbitrary function ARGV <- commandArgs(trailingOnly=TRUE) if (ARGV[1] == "do-onecore") { ?library(foreach) ?discard <-

Dear R developers: ... 7: lapply(seq_len(cores), inner.do) 8: FUN(1:3[[3]], ...) 9: sendMaster(try(lapply(X = S, FUN = FUN, ...), silent = TRUE)) Selection: .....................Error in sendMaster(try(lapply(X = S, FUN = FUN, ...), silent = TRUE)) : long vectors not supported yet: memory.c:3100 admittedly, my outcome will be a very big list, with 30,000 elements, each containing data frames

Dear R wizards: is there a clean way to assign to elements in a list? what I would like to do, in pseudo R+perl notation is f <- function(a,b) list(a+b,a-b) (c,d) <- f(1,2) and have c be assigned 1+2 and d be assigned 1-2. right now, I use the clunky x <- f(1,2) c <- x[[1]] d <- x[[2]] rm(x) which seems awful. is there a nicer syntax? regards, /iaw ---- Ivo Welch

Dear R experts--- after many years, I am planning to give in and write my first R package. I want to combine my collection of collected useful utility routines. as my guide, I am planning to use Friedrich Leisch's "Creating R Packages: A Tutorial" from Sep 2009. Is there a newer or better tutorial? this one is 4 years old. I also plan on one change---given that the

dear R experts---I am using the coef() function to pick off the coefficients from an lm() object. alas, I also need the standard errors and I need them fast. I know I can do a "summary()" on the object and pick them off this way, but this computes other stuff I do not need. Or, I can compute (X' X)^(-1) s^2 myself. Has someone written a fast se() function? incidentally, I think

dear R experts---now I have a case where I want to estimate very large regression models with many fixed effects---not just the mean type, but cross-fixed effects---years, months, locations, firms. Many millions of observations, a few thousand variables (most of these variables are interaction fixed effects). could someone please point me to packages, if any, that would help me estimate such

dear R experts---is there a function that prints a data frame to a string? cat() cannot handle lists, so I cannot write cat("your data frame is:\n", df, "\n"). regards, /iaw ---- Ivo Welch (ivo.welch@gmail.com) [[alternative HTML version deleted]]

this is not an important question, but I wonder why lm returns an error, and whether this can be shut off. it would seem to me that returning NA's would make more sense in some cases---after all, the problem is clearly that coefficients cannot be computed. I know that I can trap the lm.fit() error---although I have always found this to be quite inconvenient---and this is easy if I have only

is it possible to throw a stop() that is so hard that it will escape even tryCatch? /iaw ---- Ivo Welch (ivo.welch at gmail.com)

Dear R Wizards---is it possible to get R to show its current call stack (sys.calls()) upon an error abort? I don't use ESS for execution, and it is often not obvious how to locate how I triggered an error in an R internal function. Seeing the call stack would make this easier. (right now, I sprinkle "cat" statements everywhere, just to locate the line where the error appears.) Of

dear R experts---I am trying to replicate a perl feature. I want to be able to embed R commands inside a character string, and have the string be printed with the command executed. my perl equivalent is my $a=10; my $teststring = "the expression, $a+1, is ::$a+1::, but add one more for ::$a+2::\n"; $teststring =~ s/::(.*?)::/$1/gee; print $teststring; of course, R does not use

dear R experts: has someone written a function that returns the results of by() as a data frame? ??of course, this can work only if the output of the function that is an argument to by() is a numerical vector. presumably, what is now names(byobject) would become a column in the data frame, and the by object's list elements would become columns. it's a little bit like flattening the by()

dear R wizards:? I am wrestling with reshape.? I have a long data set that I want to convert into a wide data set, in which rows are firms and columns are years. > summary(rin) firm fyear sim1 Min. :1004.00 Min. :1964.0 Min. : -1.00000 1st Qu.:1010.00 1st Qu.:1979.0 1st Qu.: -0.14334 Median :1016.00 Median :1986.0 Median : 0.00116 Mean

Dear R experts: this is probably correct behavior, but I do want to point out that it is unexpected to someone not too well versed: > test=factor("A","B","C","A") > ifelse(test=="A", as.factor("A"), test) [1] 1 2 3 1 ok, my factor was just coerced into integers, even though I have a logical vector as my condition and factors as

I have a set of functions that I always load on startup. for example, there is my now infamous "is.defined()" function. I would like to add some documentation for these functions, so that I can do a ?is.defined inside R. The documentation tells me how to mark up Rd files is very good, but I wonder how one installs them for access by the R executable (on OSX for me). Do I drop