similar to: Weibull parameter estimation

Dear R-users I have some datasets, all left-censoring, and I would like to fit distributions to (weibull,exponential, etc..). I read one solution using the function survreg in the survival package. i.e survreg(Surv(...)~1, dist="weibull") but it returns only the scale parameter. Does anyone know how to successfully fit the exponential, weibull etc... distributions to left-censoring

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,

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

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

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.

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

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)

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 list, I would like to simulate individual survival times from a model that has been fitted using the survreg procedure (library survival). Output shown below. My plan is to extract the shape and scale arguments for use with rweibull() since my error terms are assumed to be Weibull, but it does not make any sense. The mean survival time is easy to predict, but I would like to simulate

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)

Hola, ¿qué tal? Tengo una pregunta de esta

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

Hey all, I am trying to use the survreg function in R to estimate the mean and standard deviation to come up with the MLE of alpha and lambda for the weibull distribution. I am doing the following: times<-c(10,13,18,19,23,30,36,38,54,56,59,75,93,97,104,107,107,107) censor<-c(1,0,0,1,0,1,1,0,0,0,1,1,1,1,0,1,0,0) survreg(Surv(times,censor),dist='weibull') and I get the following

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]]

I am trying to extract the shape and scale parameters of a wind speed distribution for different sites. I can do this in a clunky way, but I was hoping to find a way using data.table or plyr. However, when I try I am met with the following: set.seed(144) weib.dist<-rweibull(10000,shape=3,scale=8) weib.test<-data.table(cbind(1:10,weib.dist))

I was wondering if someone familiar with survival analysis can help me with the following. I would like to fit a Weibull curve, that may be dependent on a covariate, my dataframe "labdata" that has the fields "cov", "time", and "censor". Do I do the following? wieb<-survreg(Surv(labdata$time, labadata$censor)~labdata$cov,

Dear R-Users I have a list with two vectors of doubles tha have different lengths. I want to export it to a file and I also want to print it in two columns. I try with write.table but it need vectors of the same length. Does anyone know how to do it? Thanks Borja [[alternative HTML version deleted]]

Hi All, I have conducted the following survival analysis which appears to be OK (thanks BRipley for solving my earlier problem). > surv.mod1 <- survreg( Surv(timep1, relall6)~randgrpc, data=Dataset, dist="weibull", scale = 1) > summary(surv.mod1) Call: survreg(formula = Surv(timep1, relall6) ~ randgrpc, data = Dataset, dist = "weibull", scale = 1)

Hi I have been asked to provide Weibull parameters from a paper using Kaplan Meir survival analysis. This is something I am not familiar with. The survival analysis in R works nicely and is the same as commercial software (only the graphs are superior in R). The Weibull does not and produces an error (see below). Any ideas why this error should occur? My approach may be spurious.

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