Displaying 20 results from an estimated 4000 matches similar to: "sample mean, variance and SD"
2011 Apr 27
3
MASS fitdistr with plyr or data.table?
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))
2012 Mar 06
1
Scale parameter in Weibull distribution
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
2006 Sep 11
2
Translating R code + library into Fortran?
Hi all,
I'm running a monte carlo test of a neural network tool I've developed,
and it looks like it's going to take a very long time if I run it in R
so I'm interested in translating my code (included below) into something
faster like Fortran (which I'll have to learn from scratch). However, as
you'll see my code loads the nnet library and uses it quite a bit, and I
2003 Jul 28
1
Optimization failed in fitting mixture 3-parameter Weibull distri bution using fitdistr()
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)
2012 Apr 11
1
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 Mathematical Research
UPM
2004 Jul 28
2
Simulation from a model fitted by survreg.
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
2012 Dec 13
2
simulate time data
Hi,
Does anyone know how to write a command to generate time-to-event data for Cox's regression?
Scomet
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2012 Feb 05
2
R-Censoring
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
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1999 Aug 30
1
rexp and rweibull
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
2009 Jul 16
2
Weibull Prediction?
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
2012 Jan 29
1
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])-
(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
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2007 Oct 18
1
programming question
hie
i'm tryimg to generate two survival data using the following code (I know its ugly ) but it seems to repeat two of the variables can any one tell me whats the porblem.
n=20
n1=n/2
n2=n/4
a11=1 ;a12=1.4 ;a21=16 ;a22=a12 * a21
t1<-array(1,c(n1))
t2<-array(2,c(n1))
treatgrp=matrix(c(t1,t2))
2007 Oct 29
3
using survfit
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))
2012 Apr 11
1
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)
meantrue<-gamma(1+(1/p))*b
meantrue
cen<- runif(n-r,min=0,max=meantrue)
cen
Chris Guure
Researcher,
2012 Apr 16
1
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 censoring approach correct.
n=25;rr=1000
p=1.5;b=1.2
for (i in 1:rr){
q<-c(t,cen)
2008 Oct 28
2
Fitting weibull and exponential distributions to left censoring data
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
2004 Nov 08
1
coxph models with frailty
Dear R users:
I'm generating the following survival data:
set.seed(123)
n=200 #sample size
x=rbinom(n,size=1,prob=.5) #binomial treatment
v=rgamma(n,shape=1,scale=1) #gamma frailty
w=rweibull(n,shape=1,scale=1) #Weibull deviates
b=-log(2) #treatment's slope
t=exp( -x*b -log(v) + log(w) ) #failure times
c=rep(1,n) #uncensored indicator
id=seq(1:n) #individual frailty indicator
2009 Nov 13
2
survreg function in survival package
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,
2007 Jun 18
1
psm/survreg coefficient values ?
I am using psm to model some parametric survival data, the data is for
length of stay in an emergency department. There are several ways a
patient's stay in the emergency department can end (discharge, admit, etc..)
so I am looking at modeling the effects of several covariates on the various
outcomes. Initially I am trying to fit a survival model for each type of
outcome using the psm
2008 Oct 22
2
Weibull parameter estimation
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
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