similar to: Regular expression matching for ":" - examples needed

Hi all, I am not used to using the computer to do calculus and have up to now done my differentiation "by hand" , calling on skills I learned many years ago and some standard cheat sheets. My interest at present is in getting the second derivative of a gaussian, which I did by hand and results in a somewhat messy result involving terms in sigma^5 .. I have done some spot checks

Dear list-subscriber, in the process of writing a general code snippet to extract coefficients in an expression (in the example below: 0.5 and -0.7), I stumbled over the following peculiar (at least peculiar to me:-) ) sorting behaviour of the function all.names(): > expr1 <- expression(x3 = 0.5 * x1 - 0.7 * x2) > all.names(expr1) [1] "-" "*" "x1"

Also, this should have gone in R-bugs quite a while ago : ------- start of forwarded message ------- From: Martin Maechler <maechler@stat.math.ethz.ch> To: R-core@stat.math.ethz.ch Subject: PROTECT() bugs in deriv(*, *, function.arg = ) Date: Mon, 16 Apr 2001 21:02:10 +0200 In R versions 0.50 and 0.64.2 , the following worked > deriv(expression(sin(cos(x) * y)),

Hello! I work with: R : Copyright 2006, The R Foundation for Statistical Computing Version 2.3.1 (2006-06-01) On Windows XP Professional (Version 2002) SP2 I think there is a bug in the conditional execution if (expr1) {expr2} else {expr3} If I try: "if (expr1) expr2 else expr3" it works well but when I put the expression expr2 and expr3 between {} I receive an error message

Dear list # I have a function ff <- function(a,b=2,c=4){a+b+c} # which I programmatically want to modify to a more specialized function in which a is replaced by 1 ff1 <- function(b=2,c=4){1+b+c} # I do as follows: vals <- list(a=1) (expr1 <- as.expression(body(ff))) expression({ a + b + c }) (expr2 <- do.call("substitute", list(expr1[[1]], vals))) { 1 +

I am more than a little puzzled by your question. In the construct {expr1; expr2; expr3} all of the expressions expr1, expr2, and expr3 are evaluated, in that order. That's what curly braces are FOR. When you want some expressions evaluated in a specific order, that's why and when you use curly braces. If that's not what you want, don't use them. Complaining about it is like

------- start of forwarded message ------- From: Martin Maechler <maechler@stat.math.ethz.ch> To: R-core@stat.math.ethz.ch Subject: PROTECT() bugs in deriv(*, *, function.arg = ) Date: Mon, 16 Apr 2001 21:02:10 +0200 In R versions 0.50 and 0.64.2 , the following worked > deriv(expression(sin(cos(x) * y)), c("x","y"), function(x,y){}) function (x, y)

Hello, I have a function for reading a data-frame from a file, which contains E = read.table(file = filename, header = T, colClasses = c(rep("integer",6),"numeric","integer",rep("numeric",8)), ...) Now a small variation arose, where colClasses =

This may have been asked before, but is there an elegant way to check whether an variable/argument passed to a function is a "parse tree" for an (unevaluated) expression or not, *without* evaluating it if not? Currently, I do various rather ad hoc eval()+substitute() tricks for this that most likely only work under certain circumstances. Ideally, I'm looking for a isParseTree()

Hello, nls library provides 6 self-starting models, among them: SSfp, a four parameters logistic function. Its self-starting procedure involves several steps. One of these steps is: pars <- as.vector(coef(nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal)))), data = xydata, start = list(lscal = 0), algorithm = "plinear"))) which assumes an initial value of lscal equal to 0. If lscal

Today I ran into another aspect of the poison problem... Basically, SimplifyCFG wants to take expr1 && expr2 and flatten it into x = expr1 y = expr2 x&y This isn't safe when expr2 might execute UB. The consequence is that no LLVM shift instruction is safe to speculatively execute, nor is any nsw/nuw/exact variant, unless the operands can be proven to be in

Hi, Is there any equivalent for ifelse (except if (cond) expr1 else expr2) which takes an atomic element as argument but returns vector since ifelse returns an object of the same length as its argument? x = c(1,2,3) y = c(4,5,6,7) z = 3 ifelse(z <= 3,x,y) would return x and not 1 thanks

R's { expr1; expr2; expr3} acts much like C's ( expr1, expr2, expr3) E.g., $ cat a.c #include <stdio.h> int main(int argc, char* argv[]) { double y = 10 ; double x = (printf("Starting... "), y = y + 100, y * 20); printf("Done: x=%g, y=%g\n", x, y); return 0; } $ gcc -Wall a.c $ ./a.out Starting... Done: x=2200, y=110 I don't like that

Hi >i have aproblem withe execution of my function >first, i wrote my function in the script of R >nom_fonction <- function(arg1[=expr1], arg2[=expr2], ...){ bloc d'instructions } > when i want to have the result i mean the laste instruction in the bloc of > instruction , i try to >wrote the name of function >source(aj.fun) Error in readLines(file, warn =

What is the simplest way to extract a matched subexpression? Eg. in perl you can do "hello world" =~ m/hello (.*)/ which would return 1(true) and set $1 to the matched subexpression "world". -- View this message in context: http://n4.nabble.com/regular-expression-submatch-tp1459146p1459146.html Sent from the R help mailing list archive at Nabble.com.

hi , this can be done easily if (cond) expr ex:  > for (i in 1: 4)+ {+ if(i==2) print("a")+ if(i==2) print("b")+ } output : [1] "a"[1] "b" but i want this  if (cond) expr1 expr 2 i tried this :  > for (i in 1: 4)+ {+ if(i==2) (print("b") && print("a"))+ } output : [1] "b"Error in print("b") &&

This is wierd.... It looks like the problem appears when data(..) is immediately followed by another statement USING ``;'' ----------- rm(list=ls()) data(phones); summary(phones) Error: Object "phones" not found ls() character(0) Or also ls() character(0) data(pigs); 1:10 ##-- nothing is printed, but here 'pigs' is sourced ls() [1]

Just tried to run a perl script to do some ripping and converting of a postscript file, and ghostscript bombed out with the error message below. Can someone tell me is this really a ghostscript error, or is there more to it than this? If need be, I can attach the perl script that caused the error, but I think it is OK. Don't have another machine to test it on. Oh yeah, FWIW, this is an

Hi, I currently have these two lines in my dialplan to extract different parts out of a variable and I'd like to do it in one line instead. Does anyone know how to use regular expression subexpressions in the dialplan? Outputting a comma separated list that can be sent to ARRAY() would be nice too (tried that, didn't work -- only got the first subexpression). ;extract dialed number

Hi, I am using R a lot to make plots relating to radioactivity, I am often using expression() to label the plots with nuclide names written with superscripts, e.g. expression(paste("Releases of ", { }^{99},Tc," (TBq/year)"))->ywtext But, is there any simple way to change the number and name of the nuclide through a variable? I tried nuccode=expression({ }^{99},Tc)