search for: chris_kelvin2001

Dear all, ?The code below is used to generate interval censored data but unfortunately there is an error with the ifelse which i am not able to rectify. ?Can somebody help correct it for me. Thank you t<-rexp(20,0.2)? v<-c(0,m,999)? y<-function(t,v){ ? z<-numeric(length(t (( ? ? s<-numeric(length(t (( ? ? ? for(i in 1:length(t)){ ? ? ? ? for(j in 1:length(v-1))? ? ? ? ? { ifelse

Hello! I got something to ask..whether you can help me with the R program...i got this for example 5x4 matrix..and i want to find: ?i) mean for each row of the matrix ii) median for each column of the matrix and i need to do this using a loop function...below is my program..u try to check it for me as the output that i got is not what i desired...thanks.. data<-rnorm(20,0,1)

Hello, ?When i generate data with the code below there appear NA as part of the generated data, i prefer to have zero (0) instead of NA on my data. Is there a command i can issue to replace the NA with zero (0) even if it is after generating the data?? Thank you library(survival) p1<-0.8;b<-1.5;rr<-1000 for(i in 1:rr){ r<-runif(45,min=0,max=1) t<-rweibull(45,p1,b)

...lt methods. When I tried "all.methods", one (uobyqa) seemed to lock up. This is a fairly ill-conditioned problem. Best, JN On 04/14/2012 06:00 AM, r-help-request at r-project.org wrote: > Message: 13 > Date: Fri, 13 Apr 2012 03:54:43 -0700 (PDT) > From: Christopher Kelvin <chris_kelvin2001 at yahoo.com> > To: "r-help at r-project.org" <r-help at r-project.org> > Subject: [R] R-Help: Censoring data > Message-ID: > <1334314483.27693.YahooMailNeo at web65408.mail.ac4.yahoo.com> > Content-Type: text/plain; charset=iso-8859-1 > > Hello, >...

Hello, I have being trying to estimate the parameters of the?generalized?exponential distribution. The random number generation for the GE distribution is?x<-(-log(1-U^(1/p1))/b), where U stands for uniform dist. The data i have generated to estimate the parameters is right censored and the code is given below; The problem is that, the newton-Raphson approach isnt working and i do not know what

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]]

 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]]

I need help, the codes below estimates the weibull parameters with complete failure, my question is how do i change the state to include some censoring (may be right, type-I or type-II) to generate and estimate the parameters. thank you x=rweibull(10,2,2) library(survival) d<-data.frame(ob=c(x),state=1) s <- Surv(d$ob,d$state) sr <- survreg(s~1,dist="weibull")

Dear All, Can you help me, with the code below how do I obtain the fisher information from it. Is my q<-replicate(1000,x) the right way to do simulation. thank you. x<-rweibull(100,0.8,1.5) q<-replicate(1000,x) z<-function(p){ beta<-p[1] eta<-p[2] log1<-(n*log(beta)-n*beta*log(eta)+(beta-1)*sum(log(x))-sum((x/eta)^beta)) return(-log1) } zz<-optim(c(0.5,0.5),z) zz Chris

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, When i run the code below from Weibull distribution with 30% censoring by using optim i get an error form R, which states that Error in optim(start, fn = z, data = q, hessian = T) :? ? objective function in optim evaluates to length 25 not 1 can somebody?help me remove this error. Is my censoring approach correct. n=25;rr=1000 p=1.5;b=1.2 for (i in 1:rr){ q<-c(t,cen)

Hello, I want to estimate the exponential parameter by using?optim?with the following input, where t contains 40% of the data and q contains 60% of the data within an interval. In implementing the code command for optim i want it to contain both the t and q data so i can obtain the correct estimate. Is there any suggestion as to how this can be done. I have tried h<-c(t,q) but it is not working

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