similar to: skip lapply item for conditions(RAM, execution time)

Hi all, I consider writing a R package on statistics for the sorting method as a hobby. I have written a private Java application that I could use as a basis. Therefore I'd like to ask two questions: 1) logical vectors: Bit storage (small) and capable of bit operations? The Java application relies on BitSet(s) for efficiency reasons. Arrays of logical values cost far too much memory (in the

hi all, I'm full of questions today :). Thanks in advance for your help! Here's the problem: x <- c('18x.6','12x.9','302x.3') I want to get a vector that is c('18x','12x','302x') This is easily done using this code: unlist(lapply(strsplit(x,".",fixed=TRUE),function(x) x[1])) So far so good. The problem is that x is a vector

Dear all, I am using lapply to generate plots by applying a plot function to a list of dataframes. e.g. lapply(dataSet, FUN = plotFunction) Is there a way to call the list item number inside the plot function so that I can identify each graph? Thanks Chibisi [[alternative HTML version deleted]]

Hi all, I find myself sometimes in the situation where I lapply over a list and in the particular function I'd like to use the name and or position of the respective list item. What I usually do is to use mapply on the list and the names of the list / a position list: o <- list(A=1:3, B=1:2, C=1) mapply(function (item, name) paste(name, sum(item), sep="="), o, names(o))

Dear all, I am processing a very long and complicated list using lapply through a custom function and I would like to generate some sort of progress report. For instance, print a dot on the screen every time 1000 item have been process. Or even better, reporting the percent of the list that have been process every 10%. However, I can't seem to figure out a way to achieve that. For instance,

> From: Thomas Lumley >> >> On Tue, 14 Mar 2006, John McHenry wrote: >> >> > Thanks, Gabor & Thomas. >> > >> > Apologies, but I used an example that obfuscated the question that I >> > wanted to ask. >> > >> > I really wanted to know how to have extra arguments in >> functions that >> > would allow,

Hi all, I've got the following problem. I have a vector containing file names. I want to read these files as csv and calculate the density-function for each file (has just one column with data). Then, I'd like to plot all density functions into one window. I did the following to calculate the density data: s <- sapply(filelist, function(x) { if(file.exists(x)) { file <-

In co-operation with Markus Hegland and myself, Zhongwen Ding has written a package, based partly on Markus Hegland's code, that provides a parallel processing interface to a remote multi-processsor system. Pyro (Python Remote Objects) and R must both be installed, both on the client machine and on the remote server. The system uses rsync, with an ssh protocol, to handle file transfer. Once

Hi, Would be very glad for help on this problem. Using this code: temp<-function(x, bins, tot) { return(as.numeric(lapply(split(x, bins), wtest, tot))); } wtest <- function(x, y) { return(wilcox.test(x,y)$p.value); } rs <- function(x, bins) { binCount <- length(split(x[,1], bins)); tot <- as.numeric(x); result<-matrix(apply(x, 2, temp, bins, tot),

Hi,The conditions are not very clear. set.seed(285) ?RN <- sample(1:100,20,replace=FALSE) aList <- lapply(1:10,function(i) sample(RN,i,replace=FALSE)) ?sapply(aList,tail,1) # [1] 87 60 96 87 60 67 42 96 73 24 A.K. I was wondering if I could obtain some help on how to do this. I feel as if I'm supposed to use the vector that i made with sample in order to use this function

Good day, My question is: Does the lapply function guarantee a particular sequence in which elements are mapped? And, are we guaranteed that lapply will always be sequential (i.e. never map elements in parallel) ? The reason I ask is if I use lapply with the mapping function set to something that has side-effects that need to be executed in a particular sequence. If this is not possible, is

Dear Community, I have the following two variables, which I have split according to a factor: *y1* [1] 1 2 3 [2] 2 3 4 and *y2* A B [1] 1 4 2 5 3 6 [2] 2 5 3 6 4 7 Now I need the following Vector Autoregressive Models: VAR(cbind(y1[1],y2[1]$A)), VAR(cbind(y1[1],y2[1]$B)), VAR(cbind(y1[2],y2[2]$A)), VAR(cbind(y1[2],y2[2]$B)). My problem is that when using this argument: lapply(y2,

Hi all!! How can i do nested cross validation with lapply?? I know caret package, but I want to do it manuallly using lapply instead for bucle. Thanks!! Jes?s [[alternative HTML version deleted]]

Hi, Try: dat2<- read.csv("BOlValues.csv",header=TRUE,sep="\t",row.names=1) dim(dat2) #[1] 20 28 indx2<-expand.grid(names(dat2),names(dat2),stringsAsFactors=FALSE) nrow(indx2) #[1] 784 indx2New<- indx2[indx2[,1]!=indx2[,2],] nrow(indx2New) #[1] 756 res2<-sapply(seq_len(nrow(indx2New)),function(i) {x1<- indx2New[i,];

I have an application where I need to preserve object attributes across calls to 'lapply'. The current definition is: lapply <- function (X, FUN, ...) { FUN <- match.fun(FUN) if (!is.list(X)) X <- as.list(X) rval <- .Internal(lapply(X, FUN)) names(rval) <- names(X) return(rval) } Would it make sense to replace names(rval) <- names(X)

Dear R-devel, Is there a reason that lapply(NULL, ...) returns the empty list, rather than NULL? It seems intuitive to expect the latter, and rather counterintuitive that lapply(list(), ... ) returns the same value as lapply(NULL, ...). > lapply(list(), function(x) 1) list() > lapply(NULL, function(x) 1) list() > version _ platform i386-pc-mingw32 arch

Dear R developers, Reviewing my code, I have realized that about 80% of the time in the lapply I need to access the names of the objects inside the loop. In such cases I iterate over indexes or names: lapply(names(x), ... [i]), lapply(seq_along(x), ... x[[i]] ... names(x)[i] ), or for(i in seq_along(x)) ... which is rather inconvenient. How about an argument to lapply which would specify the

Hello, All: I'm getting different answers from "lapply(ts(1:2), length)", depending on what is attached, with nothing obviously masked. 1. Am I correct that the answer to "lapply(ts(1:2), length)" should be a list of length 2 consisting of "int 1" twice? This is what I get from R 3.0.2 with nothing else attached. If I've attached

On 23/02/2015 3:57 PM, Daniel Kaschek wrote: > Hi everybody, > > with the following code I generate a list of functions. Each function > reflects a "condition". When I evaluate this list of functions by > another lapply/sapply, I get an unexpected result: all values coincide. > However, when I uncomment the print(), it works as expected. Is this a > bug or a

Actually, it depends on the number of cores: > fun1 <- function(c){function(i){c*i}} > fun2 <- function(f) f(2) > sapply(mclapply(1:4, fun1, mc.cores=1L), fun2) [1] 8 8 8 8 > sapply(mclapply(1:4, fun1, mc.cores=2L), fun2) [1] 6 8 6 8 > sapply(mclapply(1:4, fun1, mc.cores=4L), fun2) [1] 2 4 6 8 > >/ On Feb 24, 2015, at 10:50 AM,