similar to: Recall() and sapply()

Hi I have several methods for evaluating a function. The methods are algebraically identical but use different numerical techniques. The different methods work better (converge faster, etc) in different parts of the function's domain. I am compiling a test suite for a package, and would like to verify that the different methods return approximately identical results. Toy example

Hi How do I make do.call() take "+" as a function for a list of more than two elements? Toy problem follows: f <- function(i){matrix((1:6)^i,2,3)} # Thus f() returns a matrix of size 2x3; I want to add a whole bunch of such matrices, # as in f(1) + f(2) + f(3) + f(4) # But: > do.call("+",sapply(1:4,f,simplify=FALSE)) Error in do.call("+",

Hi how do I extract those components of a list that satisfy a certain requirement? If jj <- list(list(a=1,b=4:7),list(a=5,b=3:6),list(a=10,b=4:5)) I want just the components of jj that have b[1] ==4 which in this case would be the first and third of jj, viz list (jj[[1]],jj[[3]]). How to do this efficiently? My only idea was to loop through jj, and set unwanted components to NULL,

Hi I have been wondering if there one can speed up calculating small powers of numbers such as x^8 using multiplication. In addition, one can be a bit clever and calculate x^8 using only 3 multiplies. look at this: > f1 <- function(x){x*x*x*x*x*x*x*x} > f2 <- function(x){x^8} > f3 <- function(x){x2 <- x*x;x4 <- x2*x2;return(x4*x4)} [so f1() and f2() and f3() are

Hi the manpage says that crossprod(x,y) is formally equivalent to, but faster than, the call 't(x) %*% y'. I have a vector 'a' and a matrix 'A', and need to evaluate 't(a) %*% A %*% a' many many times, and performance is becoming crucial. With f1 <- function(a,X){ ignore <- t(a) %*% X %*% a } f2 <- function(a,X){ ignore <-

Hi has anyone coded up domain colouring for visualizing complex analytic functions (such as elliptic functions)? [ the idea is to depict a complex function f(z) using a filled.contour() variant in which the hue is given by Arg(f(z)), and the saturation by Mod(f(z)). ] -- Robin Hankin Uncertainty Analyst Southampton Oceanography Centre European Way, Southampton SO14 3ZH, UK tel

Hi. I have a factor and I want to extract just those elements that appear exactly once. How to do this? Toy example follows. > a <- as.factor(c(rep("oak",5) ,rep("ash",1),rep("elm",1),rep ("beech",4))) > a [1] oak oak oak oak oak ash elm beech beech beech beech Levels: ash beech elm oak > table(a) a ash beech elm oak

Hi I have a little vector function that takes a vector A of strictly positive integers and outputs a matrix M each of whose columns is the vector, modified in a complicated combinatorical way. Now I want to generalize the function so that A can include zeroes. Given A, I want to strip out the zeroes, pass it to my function, and pad M with rows at positions corresponding to the zeroes

Hi. Is there a straightforward way to convert a character string containing comma-delimited numbers to a numeric vector? In my application, I use system(executable.string, intern=TRUE) which returns a string like "[0.E-38, 2.096751179214927596171268230, 3.678944959657480671183123052, 4.976528845643001020345216157, 6.072390165503099343887569007, 7.007958550337542210168866070,

Hi If string <- "xyz" f <- function(x){1 + sin(cos(x)) + exp(x^2)} How do I manipulate "string" and f() to give the string "1 + sin(cos(xyz)) + exp(xyz^2)" ? -- Robin Hankin Uncertainty Analyst National Oceanography Centre, Southampton European Way, Southampton SO14 3ZH, UK tel 023-8059-7743

The following gotcha caught me off-guard just now. I have two matrices, a and b: a <- matrix(1,3,3) b <- matrix(1,1,1) (note that both "a" and "b" are matrices). I want them in a list: > B <- NULL > B[[1]] <- a > B[[2]] <- b > B [[1]] [,1] [,2] [,3] [1,] 1 1 1 [2,] 1 1 1 [3,] 1 1 1 [[2]] [,1] [1,] 1

Hi How do I generate all ways of ordering sets of indistinguishable items? suppose I have two A's, two B's and a C. Then I want AABBC AABCB AACBC ABABC . . .snip... BBAAC . . .snip... CBBAA [there are 5!/(2!*2!) = 30 arrangements. Note AABBC != BBAAC] How do I do this? -- Robin Hankin Uncertainty Analyst National Oceanography Centre, Southampton European Way, Southampton SO14

Hi I'm developing an R package that needs to execute some code written in pari/gp. I've used this before from an R package (elliptic) but the interface is very basic: the R function creates a string such as the following: string <- echo ' ellwp ([ 2+0*I , 0+2*I ], 1+0*I )' | gp -q And then system(string) returns the output from gp which then needs to be text processed

Hi I had a hard-to-find bug in some of my code the other day, which I eventually traced to my misusing of sweep(). I would expect sweep() to give me a warning if the elements don't recycle nicely, but X <- matrix(1:36,6,6) sweep(X,1,1:5,"+") [,1] [,2] [,3] [,4] [,5] [,6] [1,] 2 9 16 23 30 32 [2,] 4 11 18 25 27 34 [3,] 6 13 20 22

I did not realize that ++ was available. Is there a comprehensive list somewhere of which operators are available for definition? I searched the R Language manual for ++ but that only came up with a reference to C++ . On 17 Nov 2006 11:38:43 +0100, Peter Dalgaard <p.dalgaard at biostat.ku.dk> wrote: > rfrancois at mango-solutions.com writes: > > > Hi, > > > > You

Hi ?lines says For 'type = "h"', 'col' can be a vector and will be recycled as needed. Why doesn't lines() recycle colours for other types? If I type > plot(0:1,0:1,type="n") > lines(runif(11),runif(11),col=c("red","green")) > then all ten lines are red, with no warning given. Is there a reason why

Hi I have two datasets, A and B, consisting of two columns of numbers representing x and y coordinates. They have 10 and 6 rows respectively. I want to plot two scattergraphs, one above the other. The lower graph to contain A (10 points) and the upper graph to contain B (six points). The x-axes of the two graphs must line up. I then want to draw straight lines that connect points of B to a

Hi, I am sure this has come up before, but my searches of the archive didn't give any results (maybe I didn't use the right keywords, but if I use too many, the search times out). I have a vector of dimensions n, length is not fixed, eg n <- c(4,5,7) or n <- c(19,4,5,7) and a function f that takes a vector of indices, same length of n, and gives a scalar. I would like to fill

Hi How do I get the output from table() in matrix form? If I have R> table(c(1,1,1,1,2,20)) 1 2 20 4 1 1 I want [,1] [,2] [,3] [1,] 1 2 20 [2,] 4 1 1 The problem is that names(table) is a vector of characters and I need the numeric values. I am using R> rbind(as.integer(names(x)),x) I thought tabulate() might be better as it takes an

Hi [macOSX 10.4.6; R-2.3.0] I have encountered a difference in behaviour between R-2.2.1 and R-2.3.0 when performing a linear model. Transcript follows for R-2.3.0 (R-2.2.1 worked as expected). How to make R-2.3.0 perform as R-2.2.1 did? > dput(x) c(29.13, 29.88, 30.09, 29.99, 29.74, 29.64, 29.65, 29.7, 30.04, 29.89, 29.96, 29.65, 28.76, 28.41, 28.38, 29.55, 29.76, 29.75, 29.84,