similar to: R-alpha: funny behavior of data(.)

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)),

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"

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

It is again perhaps my recent spate of bad sleeping that has prevented my brain from wrapping around this explanation, or it is perhaps my inherent hatred of regular expression syntax. However, I have been unable to put this into a working form after staring at it for a while and trying different recipes. If anyone wants to take a stab at this, I'd appreciate it. (from

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

Using an example provided by "The Hitchhiker's Guide to Asterisk", I made the following addition to my extensions.conf file: [inbound-analog] exten => s,1,Wait(1) exten => s,2,SetVar(counter=0) exten => s,3,Answer() exten => s,4,Wait(1) exten => s,5,DigitTimeout(15) exten => s,6,ResponseTimeout(10)

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 =

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") &&

Is it possible to use R as a data-mining tool? Here's the problem I've got. I have a couple of data sets consisting of results from a cDNA microarray experiment - the details about the biology don't really matter here, the same theory applies for any other data-mining task (that's why I thought it'd be more appropriate to post this on r-user). Each of these datasets consists

Hi All, I am new to the group so to start with hi! Here is my question. We are running samba on a sun 6800, pretty standard install, we only have 3 "service" accounts using samba. We have the following problem, if a process copies a file to a samba share from a non clustered SQL server, everything is fine, username and group are owners of the file it creates. Now the same

> sprintf("%d",4) [1] "4" > for(i in 1:4) sprintf("%d",4) > for(i in 1:4) print(4) [1] 4 [1] 4 [1] 4 [1] 4 > Why doesn't sprintf like the for loop here