similar to: Selecting vector elements using other vectors

Hi List, I'm hoping to get opinions for enhancing the efficiency of the following code designed to take a vector of probabilities (outcomes) and calculate a union of the probability space. As part of the union calculation, combn() must be used, which returns a matrix, and the parallelized version of lapply() provided in the multicore package requires a list. I've found that

I'd like to pass a list object created by one function as an argument of another function. once inside the second function, I'd like to break the list up to it's individual elements, each then identifiable by the 'names' of the list. The list looks something like lst<-list(a=1, b=2, df=5, g=7) then inside the function I've been writing a sequence of statements that

Hello everyone, I have two data.frames that look like calib: place zoom scale left 0.65 8 left 0.80 5.6 left 1.20 3 right 0.65 8.4 right 0.80 6 right 1.20 2.9 X: ... place zoom .... ... left 0.80 .... ... left 1.20 .... ... right 0.65 .... ... NA NA .... ... right 0.8 .... ... left 1.20 .... and I want to get the corresponding values of 'scale' in a new column

Dear R community, I am trying to code a fairly complex equation for optim(). My current approach is too slow for optim(). I have a function that takes a double integral (hopefully correctly) across two terms, e.g., doubleint <- function(c1,c2) {integrate(function(y) { sapply(y, function(y) { integrate(function(x) {a*(c1+x)+b*(c2+x)}, boundsx[1], boundsx[2])$value }) },

Hi everyone, I created a four dimensional vector (dim (128,128,1,8)). This third dimension is necessary for another function somewhere. Now I'd like to perform a t-test on every vector of length 8 in my array on the fourth dimension. I'd like to obtain a new array of three dimensions with dimensions 128x128x1 with all these test statistics. I tried this with a double loop: A <-

I have a very simple problem and am completely missing the solution. I have two character variables (character ID's from two datasets) Data set 'b' is data set 'a' with one more subject added. How do I find out which is the added subject? I have tried duplicated and unique without much success. I can find all the duplicated ID's but how do I extract the new

Tried the following with R --vanilla on the Rv2.4.0 release (see details at the end). I think the script and its comments speaks for itself, but the outcome is certainly not wanted. for (n in 58950:58970) { cat("n=", n, "\n", sep=""); # Clean up first rm(names, x, y); gc(); # Create a named vector of length n # Try with format "%5d" and it

I have a complex 2D polygon with thousands of vertices, and I'd like to be able to identify points from a large set contained within the polygon, and was wondering if there might be an efficient way of doing this? Any advice would be useful! Here is a small example of what I mean: # make polygon v1<-c(0,1,1,2,1,3,6,7) v2<-c(1,3,3,5,6,7,8,9) plot(v1, v2, type = "n" )

Hi all, I was wondering whether it is possible to use the lapply() function to alter the value of the input, something in the spirit of : a1<-runif(100) a2<-function(i){ a1[i]<-a1[i-1]*a1[i];a1[i] } a3<-lapply(2:100,a2) Something akin to a for() loop, but using the lapply() infrastructure. I haven't been able to get rapply() to do this. The reason is that the "real"

Hi there, I have a binary matrix (dim 100x100) filled with values 0 and 1. I need select a record "n" positions of that matrix when values are 1. How can I do that? Thanks for all, Miltinho Brazil --------------------------------- [[alternative HTML version deleted]]

I'm trying to calculate the row-wise kronecker product A \Box B of two sparse matrices A and B, and am struggling to find a quick way to do this that takes advantage of sparseness. I thought a good idea would be to use "rep" to construct 2 matrices of the same dimension of the end product, and multiply these two together: library(Matrix) A<-Matrix(c(1,0,0,0,0,1,2,0), 2, 4)

Hi there, I run two regressions: y = a1 + b1 * x + e1 y = a2 + b2 * z + e2 I want to test against the null hypothesis: b1 = b2. How do I design the test? I think I can add two equations together and divide both sides by 2: y = 0.5*(a1+a2) + 0.5*b1 * x + 0.5*b2 * z + e3, where e3 = 0.5*(e1 + e2). or just y = a3 + 0.5*b1 * x + 0.5*b2 * z + e3 If I run this new regression, I can test against

Hi, When qqnorm on a vector of length 10M+ I get a huge pdf file which cannot be loaded by acroread or evince. Any suggestions? (apart from sampling the data). Thanks. -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://mideasttruth.com http://honestreporting.com http://camera.org http://openvotingconsortium.org http://pmw.org.il

> On Jun 6, 2019, at 2:04 PM, Richard O'Keefe <raoknz at gmail.com> wrote: > > How can expanding tildes anywhere but the beginning of a file name NOT be > considered a bug? > > I think that that IS what libreadline is doing if one allows a whitespace separated list of file names. As reported in R-help, https://www.mail-archive.com/r-help at

I have a problem with expressions. I am trying to create a title where the parameter of interest is displayed as a Greek character. Which parameter is being considered is stored in a character variable. As an example, if I have param <- "alpha" and then do plot(0, 0, main = bquote(Parameter==.(param))) then in the title I get "Parameter = alpha", whereas I want the

Hi, As I can see via path.expand a filename which contains a tilde anywhere gets automatically crippled. +> path.expand("a ~ b") [1] "a /home/user b" +> path.expand("a ~ b ~") [1] "a /home/user b /home/user" I want to open a file regardless whether its name contains any character unless 0. The unix filesystem allow the creation of such files, it

Hi, As I can see via path.expand a filename which contains a tilde anywhere gets automatically crippled. +> path.expand("a ~ b") [1] "a /home/user b" +> path.expand("a ~ b ~") [1] "a /home/user b /home/user" I want to open a file regardless whether its name contains any character unless 0. The unix filesystem allow the creation of such files, it

When `length( skewed.probs ) > 200' uniform samples are generated in R-devel. R-3.5.1 behaves as expected. `epsilon` can be a lot bigger than illustrated and still the uniform distribution is produced. Chuck > set.seed(123) > > epsilon <- 1e-10 > > ## uniform to 200 then small > p200 <- prop.table( rep( c(1, epsilon), c(200, 999-200))) > ## uniform to 201

I have a matrix of frequency counts from 0-160. x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)) I would like to apply a function creating a new column (x[,2])containing values equal to: a) log(x[m,1]) if x[m,1] > 0; and b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero values +1) for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472 whereas x[3,2] should

Hi, How can I check if a certain ... argument has been passed on to my user-defined function or not? foo <- function(data, ...) { ### here I want to check whether xlab was passed with the ... arguments ### or if the ... arguments did not contain an xlab argument } I tried missing(xlab) , exists(xlab) and several other things but did not find a solution. TIA, Mark