similar to: weibull model

Dear everyone in the Mailing list:   It is easy to get the unique elements in a column. But I would like to get rid of those rows that the elements of this column are redundant. Or sometimes, to have a look at the rows that the elements of this column are redundant is also important. I guess it boils down to throw out the index of the redundant elements.   With millions of rows, how can I

Dear Helpers:   I spend more than half a day to solve this problem in R:   Let x be a vector of a string of 0s and 1s, such as x<-c(0,0,1,1,0,0,0,0,1,1,0,0,0,0). It can be a very long vector. How to sub vectors of 0s? In the above example, I would like get the vectors (0,0), (0,0,0,0), (0,0,0,0). I can use which(x==0) to get the index of the 0 elements, but I don't know how to get the

Hi, everyone I am trying to estimate 3 parameters for my survival function. It's very complicated. The negative loglikelihood function is: l<- function(m1,m2,b) -sum( d*( log(m1) + log(m2) + log(1- exp(-(b + m2)*t)) ) + (m1/b - d)*log(m2 + b*exp(-(b + m2)*t) ) + m1*t - m1/b*log(b+m2) ) here d and t are given, "sum" means sum over these two vairables. the parameters

Dear All: I have a question on using coxph for multiple genes: I have written code to loop through all 22283 genes in the Hgu-133A and apply coxph on survival data. However, I don't know how to work with the result for each gene: survtest<-coxph(Surv(pcc.primary.stg.3.cox[,'fup_interval'],pcc.primary.stg.

I have a test directory for the survival suite, and dyn.load has ceased to work in it. Below shows the log: tmt1075% R --vanilla R version 2.12.2 (2011-02-25) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-unknown-linux-gnu (64-bit) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain

Dear Everyone in list: I am writing some codes to automate the process of fitting linear models where the names of variables of models are produced and stored in character vectors. But I have problems to pass the vectors to the lm( ) because I don't know how to strip the quotation marks automatically. Here are the codes of the example of lm( ): ## Annette Dobson (1990) "An Introduction

Hi, I wrote a function that worked well in R, but not in S-Plus, can anyone suggest a solution? > f.coxph.zph<-function(x) { cox.fit <- coxph(Surv(time.cox, status.cox) ~ x, na.action = na.exclude, method = "breslow") fit.zph<-cox.zph(cox.fit,transform='log') fit.zph$table[,3] } yyy is my data frame that contains survial time, censor status and predictor

The following works, you need to include x=TRUE in the call to coxph. Passing the time and status variables as additional arguments is a matter of personal preference. f.coxph.zph<-function(x, timeVar, statusVar) { cox.fit <- coxph(Surv(timeVar, statusVar) ~ x, na.action = na.exclude, method = "breslow", x=TRUE) fit.zph<-cox.zph(cox.fit) fit.zph$table[,3] } time.cox <-

Dear All: I have a question on using coxph for multiple genes: I have written code to loop through all 22283 genes in the Hgu-133A and apply coxph on survival data. However, I don't know how to work with the result for each gene: survtest<-coxph(Surv(pcc.primary.stg.3.cox[,'fup_interval'],pcc.primary.stg.

Dear Knowledgeable R Community Members, Please excuse my ignorance, I apologize in advance if this is an easy question, but I am a bit stumped and could use a little guidance. I have a finite mixture modeling problem -- for example, a 2-component Weibull mixture -- where the components have a large overlap, and I am trying to adapt the "mclust" package which concern to normal

Dear R-users, Using Maximum-likelihood Fitting (fitdistr function) I've got the next error: > fitdistr(datos,"weibull",lower=0) Error in optim(x = c(1.4625e-06, 0.257854, 0.0001217545, 0.11421005, 0.028721576, : L-BFGS-B *needs finite values of 'fn' * where "datos" is a vector of length=1000 between 1.4625e-06 and 0.8867114 I add the lower argument in

Hi, I have a data set which is assumed to follow weibull distr'. How can I find of cdf for this data. For example, for normal data I used (package - lmomco) >cdfnor(15,parnor(lmom.ub(c(df$V1)))) Also, lmomco package does not have functions for finding cdf for some of the distributions like lognormal. Is there any other package, which can handle these distributions?

I need to be able to run a generalized linear model with a log() link and a Weibull family, or something similar to deal with an extreme value distribution. I actually have a large dataset where this is apparently necessary. It has to do with recovery of forensic samples from surfaces, where as much powder as possible is collected. This apparently causes the results to conform to some type

 Please, Help me, How do I generate data from the weibull distribution if the data contain both failure and interval censored, For example, I want to generate n=100, shape=2 and scale =4 with 30% interval censored.  What about right censoring Thank you  [[alternative HTML version deleted]]

Dear list, dear Frank, I try to fit a Weibull survival regression model with package Design: sclear <- psm(sobj~V1+V2,dist="weibull") sobj is a one-dimensional survival object (no event indicators), V1 and V2 are factors. I get the following result: Parametric Survival Model: Weibull Distribution psm(formula = sobj ~ V1 + V2, dist = "weibull") Obs Events

Hello everyone, This is Kangmin. I am trying to produce shape and scale of my wind data. My data is based on wind speed frequency with 1km/hr increment. data is described below. Windspeed (km/h) Frequency 1 351 2 147 3 317 4 378 5 527 6 667 7 865 8 970 9 987 10 907 11 905 12 642 13 1000 14 983 15 847 16 842 17 757 18 698 19 632 20 626 21 599 22 529 23 325 24 391

Dear all, I get the following error message. And I cannot quite work out what is wrong. I think the optim gets infinite values. Certainly my data do not have any infinite values. How can I solve this? fitdistr(A1, "weibull") Error in optim(start, mylogfn, x = x, hessian = TRUE, ...) : non-finite value supplied by optim I am using R version 1.9.1 on RedHat Linux, Kernel 2.6.8.

How do you fit a Weibull distribution in R?

HI Every one Could some one provide me definitions of following bivariate distributions gamma, exponencial, Weibull, half-normal , Rayleigh, Erlang,chi-square thanks A.S. Qureshi

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.