similar to: Integration in R

Hi R-users. I'm having difficulty with an integration in R via the package "cubature". I'm putting it with a simple example here. I wish to integrate a function like: f(x1,x2)=2/3*(x1+x2) in the interval 0<x1<x2<7. To be sure I tried it by hand and got 114.33, but the following R code is giving me 102.6667.

Hi, I am using adaptIntegrate from Cubature to do numerical integration on a double integral with a 1 x 2 vector x. Say the function is something simple to start like f(x)=x1*x2 and I wish to integrate x1 over (0,365-x2) and x2 over (0,365) f <- function(x) {(x[2])*(x[1])} # "x" is vector int1<-adaptIntegrate(f, lowerLimit = c(0, 0), upperLimit = c(365-x[2], 365)) I recieve

Dear list, I'm seeking some advice regarding a particular numerical integration I wish to perform. The integrand f takes two real arguments x and y and returns a vector of constant length N. The range of integration is [0, infty) for x and [a,b] (finite) for y. Since the integrand has values in R^N I did not find a built-in function to perform numerical quadrature, so I wrote my own after

Dear R-users, May I seek some suggestions from you. I have a long programme written in R with several 'for' loops inside. I just want to get them out by any elegant way (if there is!) to reduce the computational time of the main programme. For instance, is there any smart way for the following programme that will lessen time?

Dear All, May I seek your suggestion on a simple issue. I want to draw vertical lines at some positions in the following R plot. To be more specific, I wish to draw vertical lines at d=c(5.0,5.5,6) and they should go till p=c(0.12,0.60,0.20) . I haven't found any way out, though made several attempts. Please run the following commands first if you are interested in!

Hi all, it seems that there is problem with function "adaptIntegrate", when the integration limits is infinity. Please see the code below. The second integration does not seem to work. Can anyone familiar with this give some help? Thank you with much. Hanna library(mnormt) library(cubature) ff <- function(x, rho){ mu <- rep(0,3) Sigma

Dear all, I got an error message when running the following code. Can anyone give any suggestions on fixing this type of error? Thank you very much in advance. Hanna > integrand <- function(x, rho, a, b, z){ + x1 <- x[1] + x2 <- x[2] + Sigma <- matrix(c(1, rho, rho, 1), 2,2) + mu <- rep(0,2) + f <-

Hello Dear List members, as you can see (and guess) from the code below adaptIntegrate(f,lowerLimit=c(-1,-1),upperLimit=c(.9999,.9999)) $integral [1] 9.997e-09 $error [1] 1.665168e-16 $functionEvaluations [1] 17 $returnCode [1] 0 > adaptIntegrate(f,lowerLimit=c(-1,-1),upperLimit=c(1,1)) the last command runs for 45 mins now. -this one takes only less than sec:

I want to measure the error in the estimation of a 2 dimensional density function that is calculated using an integral but run into problems trying to integrate with adaptIntegrate because the integrand also calls the function adaptIntegrate. In particular I want \int \hat{f}(x,y) - f(x,y) dx dy where \hat{f}(x,y) = \int K(a,b, x, y) da db and in this simulation study I know what the true value

UseRs, I used the optim function valor.optim <- optim(c(1,1,1),logexp1,method ="BFGS",control=list(fnscale=-1),hessian=T); and I want to calculate the derivates, psi1<-valor.optim$par[1] psi2<-valor.optim$par[2] psi3<-valor.optim$par[3] a0=exp(psi1); a1=exp(psi2)/(20+exp(psi2)+exp(psi3)); a2=exp(psi3)/(20+exp(psi2)+exp(psi3))

Hello, I am trying to use adaptIntegrate function but I need to pass on a few additional parameters to the integrand. However, this function seems not to have the flexibility of passing on such additional parameters. Am I missing something or this is a known limitation. Is there a good alternative to such restrictions, if there at all are? Many thanks for your time. HC -- View this message in

Andr?, are you quite sure you have it working? In this thread someone from Open-Xchange stated that no, Dovecot doesn?t have SMTPUTF8 support implemented, and the same response was given by another Dovecot developer last September (it ?is being considered? was the answer then, see https://dovecot.org/pipermail/dovecot/2018-September/112887.html). I am using LMTP to deliver mail to Dovecot from

Hello r-help, I am trying to collapse or aggregate 'some' of a data frame. A very simplified version of my data frame looks like: > tester trip set num sex lfs1 lfs2 1 313 15 5 M 2 3 2 313 15 3 F 1 2 3 313 17 1 M 0 1 4 313 17 2 F 1 1 5 313 17 1 U 1 0 And I want to omit sex from the picture and just get an addition of num,

I wrote a simple log likelihood (for the ordinary least squares (OLS) model), in two ways. The first works out the likelihood. The second merely calls the first, but after transforming the variance parameter, so as to allow an unconstrained maximisation. So the second suffers a slight cost for one exp() and then it pays the cost of calling the first. I did performance measurement. One would

Dear R-user, May I seek your suggestion. I have a data file 'stop' to be scanned in R. But I want to ignore one specific number '21' there. Putting differently, I want to get all numbers in the file except 21. Is there any command to achieve it? -------------------------------------------------------------- b<-scan("F:\\stop.txt")

Dear list, [cross-posting from Stack Overflow where this question has remained unanswered for two weeks] I'd like to perform a numerical integration in one dimension, I = int_a^b f(x) dx where the integrand f: x in IR -> f(x) in IR^p is vector-valued. integrate() only allows scalar integrands, thus I would need to call it many (p=200 typically) times, which sounds suboptimal. The

Hello all, I have an issue where I am generating data and trying to confirm the estimates using a sem. I keep getting an error about the degree of freedom being negative "Error in sem.default(ram, S = S, N = N, raw = raw, data = data, pattern.number = pattern.number, : The model has negative degrees of freedom = -6" Can someone explain this error or tell me what is wrong with my

Hi All, I'm trying to use one of the (2D) numerical integration functions, which is not where the problem is. The function definition is as follows: adaptIntegrate(f, lowerLimit, upperLimit, ...) The problem is that I want to integrate a 3D function which has been parametrised such that it is a 2D function: eg. I want to integrate f(x_start+gradient_x*t1, y_start+gradient_y*t2) for t1 in

Dear R-users, I am trying to understand how the 'segmented'-package works to determine breakpoints and slopes of regression lines in broken-line regression models. However, I am not able to repeat the example on the "plant"-dataset, which was reported in the accompanying paper of the package. (V.M.R Muggeo, "Segmented: an R package to fit regression models with

My data: I have raw data points that form a logit style curve as if they were a time series. Which is to say they form 3 distinct lines with 3 distinct slopes in backwards z pattern. A certain class of my data looks essentially flat to the eye with marginal oscillation. What is important to me is the x value at which the state change is occurring, in other words, the break point Use of