similar to: apply() to a function that has more than 2 arguments

Dear all, I would like to use the apply or a similar function belonging to this family, but applying for each column (or row) but let say for each q columns. For example I would like to apply a function FUN for the first q columns of matrix X then for q+1:2*q and so on. If I do apply (X, 2, FUN) it applies for each column and not for every q columns. Is that possible with any similar function?

Dear all, I am trying to use the integrate function in R but it seems that it does not work in my example and I cannot figure out why. I create a function Mu1 (which works fine) and try to integrate by the code: n <- 100 Ctrl <- as.matrix(cbind(runif(n, -30, 30))) W <- Ctrl + as.matrix(rnorm(n)) Rsp <- (W>as.matrix(sample(10, n, T)))*1 Mu1 <- function(x, Y=Rsp, Xc=Ctrl){ x

Hello, I?m new at programming and I will greatly appreciate if you can help me with this. I have a very large matrix (hundreds of rows and columns), with the first raw filled with different numbers (between 0 and 1). The rest of the matrix is filled with values 0, 1, 2. What I need is to replace the values in the matrix (except the first row will has to remain intact). In each column I have to

Hello R people, How can I compute the mean of the "Pulse_rate" column of the data frame or matrix from the following character object called "str_got". It has 14 entries and each entry has 8 values, separated by commas. Please go thru the following R commands to know how I tried to unstring and unlist the values to form a data frame. > str_got [1]

Dear all, I am using glm function in order to estimate a logit model i.e. glm(Y ~ data[,2] + data[,3], family = binomial(link = "logit")). I also created a function that estimates logit model and I would like it to compare it with the glm function. So, does anyone know what optimizer or optimization method glm uses in order to derive the result? Thank you Dimitris -- View this

Dear all, I would like to create a code for semiparametric Klein and Spady's estimator. For that I created a function that provides the log-likelihood function for each observation (so it is a function of betas and i, where i denotes the observation). Now, in order to maximize the log-likelihood function, I have to sum these log-likelihood functions for each i and so to get another function

I'm reading in some data from a csv file, and it's reading in some of the columns as character variables instead of numeric. I know I can fix this by doing as.numeric for each of the columns, but the problem is I have a LOT of different quantitative variables that I would have to do this for. I've been trying something along the following but it hasn't really been working

Hello, everybody Say I have nm1 <- c(rep(1,10), rep(0,10)) then I can do: diff(nm1) to see where I have shift in value but what if I have nm2 <- c(rep("SPZ8", 10), rep("SPX9", 10)) how can I produce the same ouput as diff(nm1) does, that is zeros everywhere except for one place where SPZ8 changes to SPX9 (there should be 1 there)? What if I have a matrix of characters

Dear all, I am trying to make some matrix operations (whose size I think is smaller than what R allows) but the operations are not feasible when they run in one session but it is feasible if they run separately while each operation is totally independent of the other. I run the code in one session the error that appears is: Error: cannot allocate vector of size 915.5 Mb R(16467,0xa0421540)

I thought the "apply" functions are faster than for loops, but my most recent test shows that apply actually takes a significantly longer than a for loop. Am I missing something? It doesn't matter much if I do column wise calculations rather than row wise ## Example of how apply is SLOWER than for loop: #rm(list=ls()) ## DEFINE VARIABLES mu=0.05 ; sigma=0.20 ; dt=.25 ; T=50 ;

Dear list, Following a recent enquiry, I've been playing with the idea of creating a colour gradient for a polygon, using the Grid package. The idea is to draw a number of stripes of different colours, using the grid.clip function. Below is my current attempt at this, library(grid) rotate.polygon <- function(g, angle=0){ # utility function, works fine matR <- matrix(c(cos(angle),

Hi everyone, My goal is to sample from a two-dimensional grid. Consider the following example of code: n.grid <- 500 muA.grid <- seq(-4,4, length=n.grid) muB.grid <- seq(-4,4, length=n.grid) mu.p <- matrix(NA, nrow=n.grid, ncol=n.grid) for(i in 1:n.grid){ for(j in 1:n.grid){ mu.p[i,j] <- dnorm(muA.grid[i], 0, 1)*dnorm(muB.grid[j], 0, 0.5) } } mu.p <-

Hi R fans I am trying to make a program to ask the user to choose data file: print("Choose data file please !") matr=read.table(file.choose(),dec=".",header=TRUE) The problem is that the prompt Choose data file please ! comes after I have chosen the file. What am doing wrong? Thanks in advance Ralf Finne

Hi all, myint=function(mu,sigma){ integrate(function(x) dnorm(x,mu,sigma)/(1+exp(-x)),-Inf,Inf)$value } mymu=seq(-3,3,length(1000)) mysigma=seq(0,1,length(500))[-1] k=1 v=c() for (j in 1:length(mymu)) { for (i in 1:length(mysigma)) { v[k]=myint(mymu[j],mysigma[i]) k=k+1 } } Basically, I want to investigate for what values of mu and sigma, the integral is divergent. Is there another way

Hello, One way of preventing that is to use ?force. Just put force(l) right after the commented out print and before you change 'u'. Hope this helps, Rui Barradas Citando Matthias Gondan <matthias-gondan at gmx.de>: > Dear R developers, > > sessionInfo() below > > Please have a look at the following two versions of the same function: > > 1. Intended

Mathias, If it's any comfort, I appreciated the example; 'expected' behaviour maybe, but a very nice example for staff/student training! S Ellison > -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Matthias > Gondan > Sent: 02 September 2017 18:22 > To: r-help at r-project.org > Subject: [R] Strange lazy evaluation of

Dear Bill, All makes perfect sense (including the late evaluation). I actually discovered the problem by looking at old code which used your proposed solution. Still I find it strange (and, hnestly, I don?t like R?s behavior in this respect), and I am wondering why u is not being copied to L just before u is assigned a new value. Of course, this would require the R interpreter to track all these

Another way to avoid the problem is to not redefine variables that are arguments. E.g., > Su3 <- function(u=100, l=u, mu=0.53, sigma2=4.3^2, verbose) { if (verbose) { print(c(u, l, mu)) } uNormalized <- u/sqrt(sigma2) lNormalized <- l/sqrt(sigma2) muNormalized <- mu/sqrt(sigma2) c(uNormalized, lNormalized, muNormalized) } > Su3(verbose=TRUE)

Hi, I have a problem with Kolmogorov-Smirnov test fit. I try fit distribution to my data. Actualy I create two test: - # First Kolmogorov-Smirnov Tests fit - # Second Kolmogorov-Smirnov Tests fit see below. This two test return difrent result and i don't know which is properly. Which result is properly? The first test return lower D = 0.0234 and lower p-value = 0.00304. The lower 'D'

I am trying to reset the default arguments in the rnbinom function with the following example code: params <- c("size" = 1, "mu" = 1) formals(rnbinom)[names(params)] <- params rnbinom(n = 10) It returns the following: Error in rnbinom(n = 10) : argument "prob" is missing, with no default If I set the defaults with this code: params <- c("size"