similar to: Mac ARM for lm() ?

Not a direct answer but you may find lm.fit worth experimenting with. Also try the high-performance computing task view on CRAN Cheers, Andrew -- Andrew Robinson Chief Executive Officer, CEBRA and Professor of Biosecurity, School/s of BioSciences and Mathematics & Statistics University of Melbourne, VIC 3010 Australia Tel: (+61) 0403 138 955 Email: apro at unimelb.edu.au Website:

>>>>> Andrew Robinson via R-help >>>>> on Thu, 14 Nov 2024 12:45:44 +0000 writes: > Not a direct answer but you may find lm.fit worth > experimenting with. Yes, lm.fit() is already faster, and .lm.fit() {added to base R by me, when a similar question was asked years ago ...} is even an order of magnitude faster in some cases. See

Thanks, and all well taken. But are my beautiful GPUs (with integrated memory architecture) really nothing more than a cooling area for the chip? On Fri, Nov 15, 2024 at 6:06?AM Martin Maechler <maechler at stat.math.ethz.ch> wrote: > >>>>> Andrew Robinson via R-help > >>>>> on Thu, 14 Nov 2024 12:45:44 +0000 writes: > > > Not a direct

Try something like with(df, predict(smooth.spline(x = altitude, y = atm_values), deriv = 1)) Cheers, Andrew -- Andrew Robinson Chief Executive Officer, CEBRA and Professor of Biosecurity, School/s of BioSciences and Mathematics & Statistics University of Melbourne, VIC 3010 Australia Tel: (+61) 0403 138 955 Email: apro at unimelb.edu.au Website: https://researchers.ms.unimelb.edu.au/~apro

Hi everyone, I have a vector with atmospheric measurements (x-axis) that is obtained/calculated at different altitudes (y-axis). The altitude is uniformly distributed every 7 meters. For example my dataframe is: df <- dataframe( *altitude* = c(1005, 1012, 1019, 1026, 1033, 1040, 1047, 1054, 1061, 1068), *atm_values* = c(1.41, 1.40, 1.39, 1.38, 1.37, 1.37, 1.38, 1.36, 1.33, 1.31)

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---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)

The following program is whittled down from a much larger program that always works on Intel, and always works on AMD's threadripper with lapply but not mclappy. With mclapply on AMD, all processes go into "suspend" mode and the program then hangs. This bug is replicable on an AMD Ryzen Threadripper 2950X 16-Core Processor (128GB RAM), running latest ubuntu 18.04. The R version

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: ?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 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: could someone please point me to a page that explains how to set up more than 1 machine for library parallel (which is quickly becoming my favorite!) my dream setup would be a design where I just pass a list of hostnames:user:password to my parallel master, and then start R listener processes on each of my slaves by hand. R would start slave processes automatically on each slave

dear R experts--- I have (many) unidimensional root problems. think loc.of.root <- uniroot( f= function(x,a) log( exp(a) + a) + a, c(.,9e10), a=rnorm(1) ) $root (for some coefficients a, there won't be a solution; for others, it may exceed the domain. implied volatilities in various Black-Scholes formulas and variant formulas are like this, too.) except I don't need 1 root, but a

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 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

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

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 experts---quick question. I need to estimate a model that looks like y = (b*T+d*T^3) + (1-b-3*d*T^2)*x + (3*d*T)*x^2 + (-d)*x^3 I only have three parameters. Is nls() the right tool for the job, or is there something faster/better? /iaw ---- Ivo Welch (ivo.welch@brown.edu, ivo.welch@gmail.com) [[alternative HTML version deleted]]

Dear R wizards: This must have an obvious solution, but I am stumped. I can run a linear regression giving a matrix as the independent set of variables, but if I give a data frame (which I would like to give, because it should tell the linear model the names of the variables), R does not like it. An example is: N=20; y= rnorm(N); x.m <- (matrix( nrow=N, ncol=2 )); x.m[,1]=rnorm(N);

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