similar to: How to remove all characters after comma in R

Dear all, Suppose I have this data frame: > data_main V1 V2 foo 13.1 bar 12.0 qux 10.4 cho 20.33 pox 8.21 And I want to split the data into two parts first part are the one contain in the source array: > src [1] "bar" "pox" and the other one the complement. In the end we hope to get this two dataframes: > data_child1 V1 V2 bar 13.1 pox

Dear all, I found that printing with 'cat' is very slow. For example in my machine this snippet __BEGIN__ # I need to resolve to use this type of loop. # because using write(), I need to create a matrix which # consumes so much memory. Note that "foo, bar, qux" object # is already very large (>2Gb) for ( s in 1:length(x) ) {

Hi all, The patch below is to fix a problem with unaligned memcpys. This program: void Bork() { int Qux[33] = {0}; } will currently produce this LLVM code on PPC64: @C.0.937 = internal constant [33 x i8] zeroinitializer define void @Bork() { entry: %Qux = alloca [33 x i8] %Qux1 = bitcast [33 x i8]* %Qux to i8* call void @llvm.memcpy.i64( i8* %Qux1, i8* getelementptr ([33 x i8]*

Thank you very much for your useful suggestions. These are exactly what I was looking for. foo <- list(foo1, foo2, foo3) lapply(foo, function(x) matrix(unlist(x), nrow = length(x), byrow = TRUE)) or lapply(foo, function(x) do.call('rbind', x)) Best, Muhammad Subianto On 4/11/06, Muhammad Subianto <msubianto at gmail.com> wrote: > Dear all, > I have a result my experiment

Dear all, I have a result my experiment like this below (here my toy example): foo1 <- list() foo1[[1]] <- c(10, 20, 30) foo1[[2]] <- c(11, 21, 31) foo2 <- list() foo2[[1]] <- c(100, 200, 300) foo2[[2]] <- c(110, 210, 310) foo3 <- list() foo3[[1]] <- c(1000, 2000, 3000) foo3[[2]] <- c(1100, 2100, 3100) list(foo1,foo2,foo3) The result: > list(foo1,foo2,foo3) [[1]]

Hi all, Suppose I have 2 matrices A and B. And I want to measure how good each of this matrix is. So I intend to compare A and B with another "gold standard" matrix X. Meaning the more similar a matrix to X the better it is. What is the common way in R to measure matrix similarity (ie. A vs X, and B vs X) ? - Gundala Viswanath Jakarta - Indonesia

Dear List: In short, I am writing a number of functions as building blocks for other functions and have some questions about scoping and passing arguments. Suppose I have functions foo1, foo2, and foo3 such that: foo1<-function(a=1,b=TRUE,c=FALSE){#do stuff}; foo2<-function(x=1,y=FALSE,z=c(1,2,3,4)){#do stuff}; foo3<-function(lambda,...){lambda*foo1()*foo2()}; I want to be able to

Hi, I have the following vector which is created from 3 distinct distribution (three components) of gamma: x=c(rgamma(30,shape=.2,scale=14),rgamma(30,shape=12,scale=10),rgamma(30,shape=5,scale=6)) I want to plot the density curve of X, in a way that it shows a distinct 3 curves that represent each component. How can I do that? I tried this but doesn't work: lines(density(x)) Please

i believe the following is true but would appreciate confirmation that it is intended behavior and will continue: if a default argument is employed in the definition of a generic function, and the generic is called with the argument in question (call it 'ARG') missing, then the method for signature (..., ARG = "missing", ...) will be called by 'standardGeneric'

I have a general style question about R coding. Suppose I'm writing a function (foo1) that calls other functions (foo2, foo3, ...) which have complicated argument lists (e.g. optim(), plot()), _and_ I may be calling several different functions in the body of foo1. Since foo2 and foo3 have different sets of arguments, I can't just use "..." ; I did write some code a while ago

Dear all, I have the following matrix. > dat A A A A A A A A A A [1,] 0 0 0 0 0 0 0 0 0 0 [2,] 0 0 0 0 0 0 0 0 0 1 [3,] 0 0 0 0 0 0 0 0 0 2 How can I change it into: [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] 0 0 0 0 0 0 0 0 0 0 [2,] 0 0 0 0 0 0 0 0 0 1

Hi, I have the following (M x N) matrix, where M = 10 and N =2 What I intend to do is to group index of (M) based on this condition of "x_mn" , namely For each M, If x_m1 > x_m2, assign index of M to Group1 otherwise assign index of M into Group 2 > x [,1] [,2] [1,] 4.482909e-01 0.55170907 [2,] 9.479594e-01 0.05204063 [3,] 8.923553e-01 0.10764474

Hi all, For some reason, I would like to use functions bellow (see example code bellow), but instead I get the following error message: *Error in foo2(...) : unused argument(s) (arg3 = 3)* #--------------------- # example code #--------------------- foo1 <- function(arg1,...) { print(arg1) foo2(...) foo3(...) } foo2 <- function(arg2) { print(arg2) } foo3 <- function(arg3) {

Hi, How can I extract the label only from a given data frame. Fore example from this data frame. > print(dataf) V1 V2 V3 V4 V5 V6 V7 V8 V9 11145 14.3 17.1 31.2 41.7 45.8 49.8 68.6 70.6 72.9 3545 10.2 15.6 20.9 23.2 31.4 31.7 36.2 48.4 51.9 8951 15.2 17.5 20.0 21.4 32.4

Hi Dieter, Sorry my mistake. I wanted to convert them into Decimal (not Hexadecimal). Given this string, the desired answer follows: > ascii_str <- "ORQ>IK" 79 82 81 62 73 75 > ascii_str2 <- "FDC" 70 68 67 - Gundala Viswanath Jakarta - Indonesia On Mon, Dec 22, 2008 at 5:49 PM, Dieter Menne <dieter.menne at menne-biomed.de> wrote: > Gundala

Hi the list In the DESCRIPTION file of my package foo0, I have: Depends: foo1 Imports: foo2 Suggest: foo3 Enhence: foo4 If I understand correctly, to install foo0 on my computer, I need to already have foo1, foo2, foo3. foo4 is not necessary. I my R sesssion, when I will write: library(foo0), then the package foo1 will be attach. foo2, foo3 and foo4 will not. Is that correct? But what is

Hello and thanks for the answer, I'm still facing issues, I'll do my best to explain my situation, as I explained, I have two modules, each one with its own main and functions, I would like to replace in the *oldModule* a function call that is calling *foo2* (defined in *oldModule*) to instead call *foo3*, which is defined in the *refModule. *So in summary, I have: 1. The original

Hi Ahmad, What does that tool does besides what LLVM linker already does? I don't think my problem is in linking both modules, I think LLVM linker does the job for me, the issue is when changing the called function to call another function (in the example previously provided, to change it from foo2 to foo3, and adjusting the function parameter's references). Regards, Daniel Moya El

Hello LLVM Developers, I'm trying to replace a function defined in one module into another module (different files). The first issue I ran into was that llvm::Function does not have a method "moveBefore" or "moveAfter" as the llvm::BasicBlock or llvm::Instruction do, so I figured I would just move the BasicBlocks of the replacing function into the function that was being

Hi Philip, The error happens when the program finishes and it automatically calls the destructors, so it is not an error specifically inside my program. Here's the full code: #include "llvm/ExecutionEngine/ExecutionEngine.h" #include "llvm/ExecutionEngine/MCJIT.h" #include "llvm/IRReader/IRReader.h" #include "llvm/Support/TargetSelect.h" #include