similar to: Memory allocation

I have 3 basic questions (sorry, but R is a bit new to me) First) The following code will work on C, but something happens with R: > x<-c(0,1,1,0,0,1,0,0,1) > dim(x)<-c(3,3) > types<-numeric(3) > for (i in 1:3) { + j<-1 + while (x[i,j]!=1) j<-j+1 + if (j==4) types[i]<-0 + else types[i]<-1 + } Error: subscript out of bounds Any ideas about what is wrong?

Suppose I have a simple function that returns a matrix, such as: test <- function(x){ return(matrix(c(x,x^2,x^3,x^4),2,2)) } so that test returns: [ x x^3 ] [ x^2 x^4 ] Is it possible for me to get the derivative of an expression such as: c(1,0) %*% test() %*% c(0,1) The vectors are used just to "index" the matrix. I don't want a value, but the expression to work with

Hi, The following code: #include<stdio.h> char bigArray[0x1000000]; int main(int argc, char **argv) { printf("mem: 0x%x\n", (unsigned) bigArray); return 0; } causes lli to silently fail, even though it compiles correctly with llc. The reason is that in JITEmitter.cpp only checks to see if CurBufferPtr == BufferEnd at the beginning of the function and not after all

Here is the corrected version. Thomas Jablin wrote: > Actually, there is a problem with the patch. Please delay review. > > Thomas Jablin wrote: > >> Hi, >> The following code: >> >> #include<stdio.h> >> >> char bigArray[0x1000000]; >> >> int main(int argc, char **argv) { >> printf("mem: 0x%x\n", (unsigned)

Actually, there is a problem with the patch. Please delay review. Thomas Jablin wrote: > Hi, > The following code: > > #include<stdio.h> > > char bigArray[0x1000000]; > > int main(int argc, char **argv) { > printf("mem: 0x%x\n", (unsigned) bigArray); > return 0; > } > > causes lli to silently fail, even though it compiles correctly with

Looks good. Do you have commit privilege? Evan On Nov 22, 2008, at 1:19 PM, Thomas Jablin wrote: > Here is the corrected version. > > Thomas Jablin wrote: >> Actually, there is a problem with the patch. Please delay review. >> >> Thomas Jablin wrote: >> >>> Hi, >>> The following code: >>> >>> #include<stdio.h>

Thanks. I do not have commit privilege. Tom ----- Original Message ----- From: "Evan Cheng" <evan.cheng at apple.com> To: "LLVM Developers Mailing List" <llvmdev at cs.uiuc.edu> Sent: Monday, December 8, 2008 5:39:33 PM GMT -05:00 US/Canada Eastern Subject: Re: [LLVMdev] MachineCodeEmitter Patch Looks good. Do you have commit privilege? Evan On Nov 22, 2008, at

Hello, everyone I have a piece of code that looks like this: mrets <- merge(mrets, BMM.SR=apply(mrets, 1, MyFunc, ret="BMM.AV120", stdev="BMM.SD120")) mrets <- merge(mrets, GM1.SR=apply(mrets, 1, MyFunc, ret="GM1.AV120", stdev="GM1.SD120")) mrets <- merge(mrets, IYC.SR=apply(mrets, 1, MyFunc, ret="IYC.AV120",

Dear all, I'm having a problem with debugger() in both R 2.8.0 and R 2.12.0. Probably also versions in-between. I don't see it logged in the bug database, but it's hard for me to imagine that no-one else has encountered it. So my question is whether it's a known problem with a workaround, or do I log it as a new problem? The situation is that if I use

Can someone help me with this simple example? sq <- function() { y <- x^2 y } myfunc <- function() { x <- 10 sq() } myfunc() executing the above in R yields: > myfunc() Error in sq() : Object "x" not found I understand that R's scoping rules cause it to look for "x" in the environment in which "sq" was defined (the global environment in

David Brahm <brahm at alum.mit.edu> wrote: >VBMorozov at lbl.gov wrote: >> I would like to pass variables to a function in R in "by reference"... >Just in case the ensuing discussion got too esoteric, here's one simple answer: >R> x <- 1:10 >R> MyFunc <- function(x, zz) assign(deparse(substitute(zz)), sum(x), 1) >R> MyFunc(x,y) >R>

I may be doing this wrong! but I have a function which I have simplified a lot below. I want to pass some 'field names' of a data-frame to the function for it to then do some manipulation of. Here's my code: #build a simple dataset mydataset = data.frame (

Dear R-guRus: I would like to pass variables to a function in R in "by reference", e.g Fortran style. For example, suppose I have the following code x<-c(1:10) y<-1 MyFunc<-function(x,y) {y<-sum(x); return(NULL)} MyFunc(x,y) print(y) in this case print(y) will produce "1" instead of 55 (which is sum(x)) - how do I make sure that afte the function is run, y

Dear R-guRus: I would like to pass variables to a function in R in "by reference", e.g Fortran style. For example, suppose I have the following code x<-c(1:10) y<-1 MyFunc<-function(x,y) {y<-sum(x); return(NULL)} MyFunc(x,y) print(y) in this case print(y) will produce "1" instead of 55 (which is sum(x)) - how do I make sure that afte the function is run, y

Dear all, It has been ages since I studied integration in college. Right now I try to recover all this kind of knowledge and then try to understand how integration works. Thus I am doing some first 'experiments' and I would like to request your help and comments. I have the function: p2<-function(x){0.5*(3*x^2-1)} # I found the square of p2 by using some pencil and

R-help, I am trying to create a function that i pass a data set to and have the function return some calculations based on data. Allow me to illustrate: myfunc <- function(lst,mn,sd){ lst <- sort(lst) mn <- mean(lst) sd <- sqrt(var(lst)) return(lst,mn,sd) } data1 <-c (1,2,3,4,5) data2 <- c(6,7,8,9,10) myfunc(data1,data1mn,data1sd) myfunc(data2,data2mn,data2sd)

Hi all, is there a way to extract the name of a function, i.e. do the reverse of match.fun applied to a character string? I would like to print out the name of a function supplied to another function as an argument. For example: myFunc = function(x) { x+1 } applyFunc = function(fnc, x) { fnc = match.fun(fnc) fnc(x) } Is there a way to obtain "myFunc" from the argument fnc in

Hi All, I would like to compute the simple finite-difference approximation to the gradient of a scalar function of a large number of variables (on the order of 1000). Although a one-time computation using the following function grad() is fast and simple enough, the overhead for repeated evaluation of gradient in iterative schemes is quite significant. I was wondering whether there are

Hi, I would like to evaluate a function, with 3 arguments, for instance, myfunc<-function(a,b,c) { sqrt(a)-exp(b)+4*c } How to execute myfunc(x,y,z), for all x, all y and all z, where x,y,z are vectors? Thank you very much in advance -- View this message in context: http://r.789695.n4.nabble.com/Execute-a-function-tp4637182.html Sent from

Hello, I have a function with a long list of parameters (of different types, numeric and string) myFunc <-function(p1, p2, p3, p4, p5...etc) { do.something(p1,p2,....) } I want to loop over this to provide a different set of parameters to the list every time. for (ii in 1:N) { myFunc(p1(ii), p2(ii),....etc) } I would like to simplify the notation and use some kind of structure, maybe