similar to: Discrepancies in run times

Hello, I am drawing contour lines for a function of 2 variables at one level of the value of the function and want to include a small arrow in any direction of increase of the function. Is there some way to do that? Below is an example that creates the contour lines. How do I add one small arrow on each line in the direction of increase of the function (at some central point of the contour

Dear useRs? I have the following problem! I have a function that calls one or more functions, depending on the input parameters. I am searching for the fastest way to select and execute the selected functions and return their results in a list. The number of possible functions is 10, however usually only 2 are selected (although sometimes more, even all). For examples, if I have function

2009/8/22 Bill Wendling <isanbard at gmail.com>: > I think you might have to provide an empty list if your function doesn't > take parameters. Maybe using an irbuilder would help? It does take one parameter. Here's the important bits: // My Function extern "C" void print(double X) { printf("%f\n", X); } // Args type std::vector<const Type*>

Sebastian: This is somewhat arcane, perhaps even a bug (correction on this welcomed). The problem is that the "expr" argument to curve() must be an actual expression, not a call to parse that evaluates to an expression. If you look at the code of curve() you'll see why (substitute() does not evaluate expr in the code). Another simple workaround other than sticking in the eval()

HI. I make this code: getdata<-function('a','b','c' ,'d','e','f'){ drv <- dbDriver("SQLite") con<-dbConnect(drv, "sqlite.db") lt<-dbListTables(con) myf<-data.frame(NULL) for (i in 1:length(lt)) { myfile<-dbReadTable(con,lt[i]) myfile1<-myfile[-c(14:44)] myfile1$MODEL<-gsub(" ",

Typo: should be NULL not NUL of course An alternative approach closer to your original attempt is to use do.call() to explicitly evaluate the expr argument: w <- "1 + x^2" do.call(curve, list(expr = parse(text = w), ylab ="y")) Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus

Hi, Down a cascade of function calls, I want to use the curve function with an expression that is a variable. For various reason, this variable must be a character object and cannot be an expression as required by the curve function. How do I convert my variable into a expression that is accepted by curve? Thanks in advance for your help. ## The following attempts do not work myf <-

Hi, Infinite recursion in S3 methods seem to crash R on Windows 2000 (R terminating with the ("Rgui.exe has generated errors...") message, rather than throwing an error. This happens with both Rgui and Rterm. The following toy example triggers this: myf <- function(x, ...) UseMethod("myf") myf.default <- function(x, ...) myf(x) myf(1) ...R crashes... Which I

Dear R-users, I would like to have some advises about a problem illustrated by the following snippet. Within myf, I need to evaluate a piece of R code that is passed as a character argument and then return the objects that are created by this code. The difficulty comes from the fact that the content of the code is variable and unknown to me (obviously not in this illustration!). With the

Suppose I have f1 <- function(x) x f2 <- function(x) x^2 funlist <- list(f1,f2) Then I would like to evaluate funlist such that when x is 10 I should get a list with 10 and 100. A naive way of doint this is myf <- funlist[[1]] do.call(paste(quote(myf)), list(x=10)) myf <- funlist[[2]] do.call(paste(quote(myf)), list(x=10)) - but there has to be much

sapply() stems from S / S+ times and hence has a long tradition. In spite of that I think that it should be enhanced... As the subject mentions, sapply() produces a matrix in cases where the list components of the lapply(.) results are of the same length (and ...). However, it unfortunately "stops there". E.g., if you *nest* two sapply() calls where the inner one produces a matrix, very

I created the following groupedData object (nlme library): gd <- groupedData(Conc ~ Time | Subj, order.groups=T, FUN = myf, data=mydata) The idea of the myf function is to reverse the order of the grouping factor Subj (or better, reorder from largest to smallest). In the mydata data set, Subj is an integer that gets converted into a factor in the groupedData object. Does anyone

Hi all, ########################################## dof=c(1,2,4,8,16,32) Q5=matrix(rt(100,dof),100,6,T,dimnames=list(NULL,dof)) par(mfrow=c(2,6)) apply(Q5,2,hist) myf=function(x){ qqnorm(x);qqline(x) } apply(Q5,2,myf) ########################################## These looks ok. However, I would like to achieve more. Apart from using a loop, is there are fast way to 'add' the titles to be

Developers: I hope this is the appropriate place to post this request. I am getting a core dump in an application of sapply. A simplified example is the following. Because simplify=TRUE is the default for sapply and tapply, the following works x <- 1:5; sapply(x,list) but the following dumps core with an access violation x <- 1:5; myf <- function(x) list(x,x); sapply(x,myf)

Why does > curve(function(y) y^2, from=0,to=1) not work, whereas > myf <- function(y) y^2 > curve(myf, from=0,to=1) work? For the former, I get the error message Error in curve(function(y) y^2, from = 0, to = 1) : 'expr' must be a function or an expression containing 'x' I'm using R1.7.0 under Windows XP. Damon Wischik.

Dear all: I have a character object x with ' (single-quote) character. x <- c('"hydrolase activity","actin binding","3',5'-cyclic-nucleotide phosphodiesterase activity") I want to write a function that will identify ' and replaces with \' myf <- function(term){ if (grep("'",term)) {

(FreeBSD 10.3-RELEASE-p4 i386, Lprng 3.8.C and Samba 4.3.8 installed via pkg()) I have set up a PDF converter queue that converts PS input to PDF and puts it on its website. Recently I had samba join the AD domain here, and ever since my pdf converter fails. Upon much trial, error and debugging I found out that when I start the Lprng service BEFORE the Samba service, everything is working.

Hello, I am stuck in a relatively simple procedure and was wondering if anybody knows the answer. I am a relatively new R user. How do I use an argument of a custom function in the name of a dataset in R? For example, I have the function: MyF <- function(Tic, price){ xxxxx xxxxx xxxxx Ratio.Tic<- SharpeRatio.annualized(roc) } I would

=========================== myf=function(ds=1){ x=rnorm(10) y=rnorm(10) { #start of if if (ds==1) { list(x,y) } else (ds==2) { plot(x,y) } } # end of if } # end of function =========================== Hi All, the problem i am having here is, that I want to be able to control the display, lf ds=1, i want to just have a list, but it seem to always plot... Thanks. casper -- View this

Folks, I have a strange situation, which I may have isolated as a bug report. Or, it could just be that there's something about R that I don't know. :-) I have attached the data file and the program file but don't know whether these attachments will make it into the list. Here is my bugreport.R program -- ---------------------------------------------------------------------------