similar to: Fitting weibull and exponential distributions to left censoring data

Hi, Is it normal to get intercept in the list of covariates in the output of survreg function with standard error, z, p.value etc? Does it mean that intercept was fitted with the covariates? Does Value column represent coefficients or some thing else? Regards, ------------------------------------------------- tmp = survreg(Surv(futime, fustat) ~ ecog.ps + rx, ovarian,

Dear R-users I would like to fit weibull parameters using "Method of moments" in order to provide the inital values of the parameter to de function 'fitdistr' . I don`t have much experience with maths and I don't know how to do it. Can anyone please put me in the rigth direction? Borja [[alternative HTML version deleted]]

Dear All, I have two questions regarding distribution fitting. I have several datasets, all left-truncated at x=1, that I am attempting to fit distributions to (lognormal, weibull and exponential). I had been using fitdistr in the MASS package as follows: fitdistr<-(x,"weibull") However, this does not take into consideration the truncation at x=1. I read another posting in this

Dear R community, I'm trying to remove a loop from my code but I'm stock and I can't find a good way to do it. Hopefully one of you will have something clever to propose. Here is a simplified example: I have a squared matrix: > nom.plac2 <- c("102", "103", "301", "303","304", "403") > poids2 <- matrix(NA, 6,6,

Hi there, can somebody give me a guide as to how to generate data from weibull distribution with censoring for example, the code below generates only failure data, what do i add to get the censored data, either right or interval censoring q<-rweibull(100,2,10). Thank you Grace Kam student, University of Ghana [[alternative HTML version deleted]]

Hi all, I'm trying to generate a Weibull distribution including four covariates in the model. Here is the code I used: T = rweibull(200, shape=1.3, scale=0.004*exp(-(-2.5*b1+2.5*b2+0.9*x1-1.3*x2)/1.3)) C = rweibull(n, shape=1.5, scale=0.008) #censoring time time = pmin(T,C) #observed time is min of censored and true event = time==T # set to 1 if event is observed

Dear list, I'm looking for a function to generate (simulate) a random Weibull point process. Can anyone help? Cheers, Torbj?rn Ergon, University of Oslo

Dear All; I tried to use fitdistr() in the MASS library to fit a mixture distribution of the 3-parameter Weibull, but the optimization failed. Looking at the source code, it seems to indicate the error occurs at if (res$convergence > 0) stop("optimization failed"). The procedures I tested are as following: >w3den <- function(x, a,b,c)

I am trying to generate predictions from a weibull survival curve but it seems that the predictions assume that the shape(scale for survfit) parameter is one(Exponential but with a strange rate estimate?). Here is an examle of the problem, the smaller the shape is the worse the discrepancy. ### Set Parameters scale<-10 shape<-.85 ### Find Mean scale*gamma(1 + 1/shape) ### Simulate Data

Hello, I wish to?censor 10% of my sample units of 50 from a Weibull distribution. Below is the code for it. I will need to know whether what i have done is correct and if not, can i have any suggestion to improve it? Thank you ?p=2;b=120 n=50 r=45 t<-rweibull(r,shape=p,scale=b) meantrue<-gamma(1+(1/p))*b meantrue cen<- runif(n-r,min=0,max=meantrue) cen Chris Guure Researcher,

Hello, If i write a function as below using log of weibull distribution i do not get the required results in estimating the parameters what do i do, please a/b * (t/b)^a-1 * exp(-t/b)^a n=500 x<-rweibull(n,2,2) z<-function(p) {(-n*log(p[1])+n*log(p[2])- (p[1]-1)*sum(log(x))+(p[1]-1)*log(p[2])+(sum(x/p[2])^(p[1]))  )} zz<-optim(c(0.5,0.5),z) zz [[alternative HTML version deleted]]

Hello, ?can i implement this as 10% censored data where t gives me failure and x censored. Thank you p=2;b=120 n=50 set.seed(132); r<-sample(1:50,45) t<-rweibull(r,shape=p,scale=b) t set.seed(123);? cens <- sample(1:50, 5)? x<-runif(cens,shape=p,scale=b)? x Chris Guure Researcher, Institute for Mathematical Research UPM

Hi I am using R to fit a survival function to my data (with a weibull distribution). Data: Survival of individuals in relation to 4 treatments ('a','b','c','g') syntax: ---- > survreg(Surv(date2)~males2, dist='weibull') But I have some problems interpreting the outcome and getting the parameters for each curve. --------- Value Std.

Hi, I am trying to fit a non-linear model for a parasite dataset. Initially, I tried log-transforming the data and conducting a 2-way ANCOVA, and found that the equal variance of populations and normality assumptions were violated. Gaba et al. (2005) suggests that the Weibull Distribution is best for highly aggregated parasite distributions, and performs better (lower type 1 and 2 error rates)

Hello, is there a quick way of estimating Weibull parameters for some data points that are assumed to be Weibull-distributed? I guess I'm just too lazy to set up a Maximum-Likelihood estimation... ...but maybe there is a simpler way? Thanks for any hint (and yes, I've read help(Weibull) ;) Kaspar Pflugshaupt -- Kaspar Pflugshaupt Geobotanical Institute ETH Zurich, Switzerland

Buenas días me gustaria saber cual es la libreria mas actual que contiene al algoritmo de maxima entropia que permite modelar nichos ecologicos; a su vez que libreria presenta mejor temática en la representacion de mapas. Saludos¡ [[alternative HTML version deleted]]

In splus rexp() and rweibull() are related: > set.seed(153) > rexp(1) [1] 0.0493267 > set.seed(153) > rweibull(1, shape=1) [1] 0.0493267 (you can also try shape =2, then rweibull = sqrt(rexp) ) However, in rw0.64.1 (on Win NT) they are different > .Random.seed <- 1:4 > rexp(1) [1] 1.412030 > .Random.seed <- 1:4 > rweibull(1, shape=1) [1] 2.054032 May be rweibull

hie when i use plot.survfit to plot more than one graph why I only see the last graph how do i see the other graphs.for example n=20 n1=n/2 n2=n/4 a11=4;a12=4 ;a21=4 ;a22=4 t1<-array(1,c(n1)) t2<-array(2,c(n1)) treatgrp=matrix(c(t1,t2))

hi could you help me to solve this issue Question: Using command rweibull(100,8,15), simulate n = 100 realizations from Weibull(8; 15) distribution. Using the simulated sample, compute the sample mean, variance and standard deviation of these observations. I am trying like this sim<-rweibull(100,8,15) # simulated sample SM<-mean(sim) # simulated sample mean var(sim) # variance

I generate random vector from Weibull distribution sampWB <-urweibull(sampleSize, shape=shape.true, scale=scale.true, lb=0, ub=Inf) how can I create subvector containing 30% of samplesize of sampWB which should be assigned as Censored data? The probability for each value in sampWB can be uniform to be included in the subvector.