similar to: Evaluation of functionals

Here's a simple example of the type of function I'm trying to write, where the first argument is a list of functions: myfun <- function(funlist, vec){ tmp <- lapply(funlist, function(x)do.call(x, args = list(vec))) names(tmp) <- names(funlist) tmp } > myfun(list("Summation" = sum, prod, "Absolute value" = abs), c(1, 4, 6, 7)) $Summation [1]

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

Dear R-users, I am in the process of creating new custom functions and am quite puzzled by some discrepancies in execution time when I run some R scripts that call those new functions. So here is the situation: - let's assume I have created two custom functions, called myg and myf; - myg is mostly a plotting function, which makes a heavy use of grid and lattice functions; - myf is a function

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

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, One of my students have installed R1.9.0 on windows, and gets the fatal error 'invalid HOMEDRIVE' Can anyone help her/me out on that one? Thanks in advance Søren Højsgaard [[alternative HTML version deleted]]

Dear all, One of my students (whom I am trying to convince to use R) wants to get a fairly large SAS dataset into R (about 150mB). An obvious and simple thing she tried was to write the dataset as a .csv-file and then read that into R, but that takes forever (or something close to that..). The dataset is so large, that exporting it as an Excel file from SAS is not feasible (more than 65000 lines).

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)) {

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

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