similar to: lapply to return vector

Dear all I am using lapply (actually mclapply that share the same syntax). I want to call the same function that takes as input a vector. My initial data structure is a matrix that I want to cut it to multiple vectors (one vector for every row of the matrix) and then feed that to the function by using mclapply. Could you please help me converting the matrices to nrow times vectors. I would

Dear all, I would like to ask your help understand the subsequent steps for making my program faster. The following code: Gauslist<-array(data=NA,dim=c(dimx,dimy,dimz)) for (i in c(1:dimz)){ print(sprintf('Creating the %d map',i)); Gauslist[,,i]<-f <- GaussRF(x=x, y=y, model=model, grid=TRUE,param=c(mean,variance,nugget,scale,Whit.alpha)) } creates 100 GaussMaps (each

Dear all, I have a function that recognizes the following format for timestamps "%Y-%m-%d %H:%M:%S" my function takes two input arguments the TimeStart and TimeEnd I would like to help me create the right list with pairs of TimeStart and TimeEnd which I can feed to lapply (I am using mclapply actually).  For every lapply I want two inputs to be fed to my function. I only know how to

Dear all, I have quite large code (with lapply and mclapply) and I am getting the following error. Error: serialization is too large to store in a raw vector Is it possible to ask from R to extend the Error messages with more details? I would like to see where this problem exists. B.R Alex [[alternative HTML version deleted]]

Dear all I would like to ask you if an assignment can be done inside a lapply statement. For example I would like to covert a double nested for loop for (i in c(1:dimx)){ for (j in c(1:dimy)){ Powermap[i,j] <- Pr(c(i,j),c(PRX,PRY),f) } } to something like that: ij<-expand.grid(i=seq(1:dimx),j=(1:dimy)) unlist(lapply(1:nrow(ij),function(rowId) { return

> On Feb 25, 2015, at 5:35 PM, Benjamin Tyner <btyner at gmail.com> wrote: > > Actually, it depends on the number of cores: Under current semantics, yes. Each 'stream' of function calls is lazily capturing the last value of `i` on that core. Under Luke's proposed semantics (IIUC), the result would be the same (2,4,6,8) for both parallel and serial execution. This is

> On Feb 24, 2015, at 10:50 AM, <luke-tierney at uiowa.edu> wrote: > > The documentation is not specific enough on the indented semantics in > this situation to consider this a bug. The original R-level > implementation of lapply was > > lapply <- function(X, FUN, ...) { > FUN <- match.fun(FUN) > if (!is.list(X)) > X <-

Dear all, I am trying to multiply three different matrices and each matrice is of size 16384,16384 the normal %*% multiplciation operator has not finished one day now. As I am running a system with many cores (and it seems that R is using only one of those) I would like to write fast a brief function that converts the typical for loops of a matrix multiplication to a set of lapply sets (mclapply

I'm a bit confused on how to use lapply with a data.frame. For example. lapply(data, function(x) print(x)) WHAT exactly is passed to the function. Is it each ROW in the data frame, one by one, or each column, or the entire frame in one shot? What I want to do apply a function to each row in the data frame. Is lapply the right way. A second application is to normalize a column value by

I've been using mclapply and have encountered situations where it gives errors or returns incorrect results. Here's a minimal example, which gives the error on R 2.15.2 on Mac and Linux: library(parallel) f <- function(x) NULL mclapply(1, f, mc.preschedule = FALSE, mc.cores = 1) # Error in sum(sapply(res, inherits, "try-error")) : # invalid 'type' (list) of argument

The reason that it works for Apply_MC=TRUE is that in that case you call mclapply(X,FUN,...) and the mclapply() function strips off the mc.cores argument from the "..." list before calling FUN, so FUN is being called with zero arguments, exactly as it is declared. A quick workaround is to change the line Lapply_me(as.list(1:4), function(xx) { to Lapply_me(as.list(1:4),

Running the example in the documentation causes R to crash. dario at bioinfo:~$ R R version 3.0.3 (2014-03-06) -- "Warm Puppy" Copyright (C) 2014 The R Foundation for Statistical Computing Platform: x86_64-pc-linux-gnu (64-bit) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or

Hi, As an example, I want to create below kind of custom Function which either be mclapply pr lapply Lapply_me = function(X = X, FUN = FUN, ..., Apply_MC = FALSE) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { list(...) = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, ...)) } } However when Apply_MC =

My modified function looks below : Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Here, I am not passing ... anymore rather passing myList On Sun, Mar 4, 2018 at 10:37 PM,

Dear parallel users and developers, I might have encountered a bug in the function 'mclapply' of package 'parallel'. I construct a matrix using the same input data and code with a single difference: Once I use mclapply and the other time lapply. Shockingly the result is NOT the same. To evaluate please unpack the attached archive and execute Rscript mclapply_test.R I put the

Below is my full implementation (tried to make it simple as for demonstration) Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Lapply_me(as.list(1:4), function(xx) { if (xx ==

That's fine. The issue is how you called Lapply_me(). What did you pass as the argument to FUN? And if you did not pass anything that how is FUN declared? You have not shown that in your email. On Sun, Mar 4, 2018 at 7:11 PM, Christofer Bogaso < bogaso.christofer at gmail.com> wrote: > My modified function looks below : > > Lapply_me = function(X = X, FUN = FUN, Apply_MC =

@Eric - with this approach I am getting below error : Error in FUN(X[[i]], ...) : unused argument (list()) On Sun, Mar 4, 2018 at 10:18 PM, Eric Berger <ericjberger at gmail.com> wrote: > Hi Christofer, > You cannot assign to list(...). You can do the following > > myList <- list(...)[!names(list(...)) %in% 'mc.cores'] > > HTH, > Eric > > On Sun, Mar

Hi Christofer, Before you made the change that I suggested, your program was stopping at the statement: list(...) = list(..) .etc This means that it never tried to execute the statement: return(lapply(X,FUN,...)) Now that you have made the change, it gets past the first statement and tries to execute the statement: return(lapply(X,FUN,...)). That attempt is generating the error message because

Hi Christofer, You cannot assign to list(...). You can do the following myList <- list(...)[!names(list(...)) %in% 'mc.cores'] HTH, Eric On Sun, Mar 4, 2018 at 6:38 PM, Christofer Bogaso < bogaso.christofer at gmail.com> wrote: > Hi, > > As an example, I want to create below kind of custom Function which > either be mclapply pr lapply > > Lapply_me =