[R] matrix with Loop

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

[R] simulation

Hello, i need to simulate 100 times, n=40 ,  the distribution has 90% from X~N(0,1) + 10% from X~N(20,10) Is my loop below correct? Thank you n=40 for(i in 1:100){ x-rnorm(40,0,1)  # 90% of n z-rnorm(40,20,10)  # 10% of n } x+z __

[R] R-HELP

please help; I want to know how to generate an interval-censored data of about 20% and a right censored data of about 30%  using the weibull distribution of say, x=rweibull(100,shape=1.2,scale=1.5) [[alternative HTML version deleted]] __

[R] R-Help

Can somebody 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. Thank you [[alternative HTML version deleted]]

[R] R-help mailing list submissions

R-help mailing list submissions [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide

[R] r-help; weibull parameter estimate

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])-

[R] r-help; weibull distribution

 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

[R] parameter estimate

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)

[R] r-help; parameter estimate

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)

[R] r-help; fisher information

Hello, i have used the code below to estimate the parameters of weibull distribution and i want to obtain the fisher information by providing the the next code but i receive errors anytime i try to, what do i do? by the way is my replication correct and is it placed at the right position for

[R] Fisher Imformation

Hello, i have used the code below to estimate the parameters of weibull distribution and i want to obtain the fisher information by providing the the next code but i receive errors anytime i try to, what do i do? by the way is my replication correct and is it placed at the right position for

[R] R- Fisher Information

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]

[R] standard error

Dear all,  I want to determine the standard error or the mean squared error for the parameter estimate for beta and eta base on the real data.  Any help on how to obtain these estimated errors. library(survival) d - data.frame(ob=c(149971, 70808, 133518, 145658, 175701, 50960, 126606, 82329),

[R] problem with ifelse

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

[R] Problem with Newton_Raphson

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

Re: [R] Problem with Newton_Raphson

Kelvin chris_kelvin2...@yahoo.com Cc: r-help@r-project.org r-help@r-project.org Sent: Thursday, September 20, 2012 8:52 PM Subject: Re: [R] Problem with Newton_Raphson On 20-09-2012, at 13:46, Christopher Kelvin wrote: Hello, I have being trying to estimate the parameters of the generalized

Re: [R] Problem with Newton_Raphson

Thank you very much for everything. Your suggestions were very helpful.  Chris - Original Message - From: Berend Hasselman b...@xs4all.nl To: Christopher Kelvin chris_kelvin2...@yahoo.com Cc: r-help@r-project.org r-help@r-project.org Sent: Thursday, September 20, 2012 10:06 PM Subject

[R] Optim Problem

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

[R] R-help; Censoring

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)

[R] R-help; generating censored data

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

[R] R-Help: Censoring data

Hello,  I want to estimate weibull parameters with 30% censored data. I have below the code for the censoring  but how it must be put into the likelihood equation to obtain the desire estimate is where i have a problem with,  can some body help?  My likelihood equation is for a random type-I

[R] R: Help; error in optim

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

[R] Interval censorin

Hello, May i know whether it is possible to generate data twice from Weibull distribution and use one as the start time and the  other as the end time, below is my code. Any suggestion on how to estimate the parameters of Weibull distribution with interval data will be highly appreciated. Thank

[R] Standard error

Hello, I have tried obtaining the value of standard error from the code below but i get different values when i compare it with the  standard error obtained from the hessian matrix. Can somebody help me out? Thank you n=100;rr=1000 p1=1.2;b=1.5 sq11=sq21=0 for (i in 1:rr){

[R] interval censoring

Hello, May i know whether it is possible to generate data twice from Weibull distribution and use one as the start time and the  other as the end time, below is my code. Any suggestion on how to estimate the parameters of Weibull distribution with interval data will be highly appreciated. Thank

[R] standard error

Hello, I have tried obtaining the value of standard error from the code below but i get different values when i compare it with the  standard error obtained from the hessian matrix. Can somebody help me out? Thank you n=100;rr=1000 p1=1.2;b=1.5 sq11=sq21=0 for (i in 1:rr){

[R] How to replace NA with zero (0)

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)

Re: [R] Clarification on Simulation and Iteration

On Saturday, August 1, 2015 3:32 AM, David Winsemius dwinsem...@comcast.net wrote: On Jul 31, 2015, at 6:36 PM, Christopher Kelvin via R-help wrote: Dear All, I am performing some simulations for a new model. I run about 10,000 iterations with a sample of 50 datasets and this returns one set

[R] Clarification on Simulation and Iteration

Dear All, I am performing some simulations for a new model. I run about 10,000 iterations with a sample of 50 datasets and this returns one set of 50 simulated data. Now, what I need to obtain is 10 sets of the 50 simulated data out of the 10,000 iterations and not just only 1 set. The model

[R] R2WinBUGS with Multivariate Logistic Regression

Dear R-User, I have written a simple code to analyze some data using Bayesian logistic regression via the R2WinBUGS package. The code when run in WinBUGS stops WinBUGS from running it and using the package returns no results also. I attach herewith, the code and a sample of the dataset. Any