Displaying 20 results from an estimated 33 matches for "rweibul".

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rweibull

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
> .Rando...

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
return(cbind(time,event,T,C))
And then, I used this da...

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

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,

2003 Jul 28

1

Optimization failed in fitting mixture 3-parameter Weibull distri bution using fitdistr()

...The procedures I tested are as following:
>w3den <- function(x, a,b,c) {c/b*((x -a)/b)^(c-1)*exp(-((x-a)/b)^c)} #
define 3-parameter Weibull density
>w3den <- function(x, a,b,c) {c/b*((x -a)/b)^(c-1)*exp(-((x-a)/b)^c)}
>set.seed(123)
> x3 <- rweibull(100, shape = 4, scale = 100) #
Distribution 1
> fitdistr(x3, w3den, start= list(a = 8.8, b = 90.77, c = 3.678)) #
Fitting 3-parameter
a b c
8.9487415 90.6667712 3.6722124
(15.1462445) (16.0657103) ( 0.7582376)
Warning...

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 individual survival times.
I am probably missing something completely obvious. Any hints or advice are appreciated.
Thanks
Sixten
> summar...

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
[[alternative HTML version deleted]]

2009 Apr 29

12

Una pregunta de estadística (marginalmente relacionada con R)

Hola, ¿qué tal?
Tengo una pregunta de esta

1997 Jul 09

1

R-beta: Problem with `rpois'

There is a problem with `rpois'. It does seem to take care about the
order of the arguments. This is an example:
> rpois(n=1,lambda=2)
[1] 3
> rpois(lambda=2,n=1)
[1] 2 0
It obviously uses the first argument as the number of samples to be
drawn, which is wrong.
I used Version 0.49 Beta (April 23, 1997).
Fredrik

1997 Jul 09

1

R-beta: Problem with `rpois'

There is a problem with `rpois'. It does seem to take care about the
order of the arguments. This is an example:
> rpois(n=1,lambda=2)
[1] 3
> rpois(lambda=2,n=1)
[1] 2 0
It obviously uses the first argument as the number of samples to be
drawn, which is wrong.
I used Version 0.49 Beta (April 23, 1997).
Fredrik

2010 Feb 09

3

Goodness

..."ad.test()", donde al pasarle la muestra te devuelve un AD(anderson darling) y p-value de la distribución, entonces comparo el p-value con la normal, y si cuadra, pues teoricamente deberia de probar con las otras distribuciones. Es decir, deberia de calcular los parametros para la "rweibull" por ejemplo, y aplicarla por ejemplo a 100 iteraciones, y volver a pasarle el ad.test(), y ver si el p-value obtenido coincide. Aunque esta forma de hacerlo no me convence, pues tendria que ir probando una a una cada distribucion, ir calculandole los parametros y tal, seria una forma muy man...

2018 Mar 06

4

Capturing warning within user-defined function

...erflow but with no response. So, if
anyone can help, I'd be appreciative.
(sidenote: I used rgamma to create my sampling weights because that's what
most resembles the distribution of my weights and it's close enough to
reproduce the convergence issue. If I used rnorm or even rlnorm or rweibull
I couldn't reproduce it. Just FYI.)
Best,
Jen
[[alternative HTML version deleted]]

2007 Mar 08

2

Using logarithmic y-axis (density) in a histogram

Hi,
I am searching for a possibility to display a logarithimic y-axis in a histogram. With plot that's easy (e.g.
plot(1:10, log="y")
but for histograms this does not work the same way: I tried
hist(rnorm(1000), freq=FALSE, seq(-4, 4, .5), ylim=c(0.001, 0.5), log="y")
Which gives the expected histogram but also warnings for my log="y" command

2014 Sep 03

3

Simulating from a Weibull distribution

Hi,
I wish to simulate some data from a Weibull distribution. The rweibull function in R uses the parameterisation
'with shape parameter a and scale parameter b has density given by f(x) = (a/b) (x/b)^(a-1) exp(- (x/b)^a)'.
However, it would be much more useful for me to simulate data using a different parameterisation of the
Weibull, with shape a1...

2018 Mar 06

0

Capturing warning within user-defined function

...nse. So, if
> anyone can help, I'd be appreciative.
>
> (sidenote: I used rgamma to create my sampling weights because that's what
> most resembles the distribution of my weights and it's close enough to
> reproduce the convergence issue. If I used rnorm or even rlnorm or rweibull
> I couldn't reproduce it. Just FYI.)
>
> Best,
>
> Jen
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/...

2018 Mar 06

0

Capturing warning within user-defined function

...nse. So, if
> anyone can help, I'd be appreciative.
>
> (sidenote: I used rgamma to create my sampling weights because that's what
> most resembles the distribution of my weights and it's close enough to
> reproduce the convergence issue. If I used rnorm or even rlnorm or rweibull
> I couldn't reproduce it. Just FYI.)
>
> Best,
>
> Jen
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/...

2006 Dec 04

0

How to calculate area between ECDF and CDF?

...nce as noted above I'll be
doing the area calculation 1e5 times or so per test, a computationally
frugal solution would be much appreciated!
Here's some code that I've been toying with:
#set up some true parameters
shape=2
scale=.5
shift=.3
n=10
#generate some observed data
obs=obs=rweibull(10,shape,scale)+shift
#lets say that the following are the estimated parameters from whatever
estimation process I'm using
est.shape=1.9
est.scale=.6
est.shift=.35
#Calculate area between ECDF and CDF of the function defined by the
#estimated parameters
# ???
#The following would work if th...

2003 Aug 06

1

probability plot correlation coefficient

As a newbie to R, I'm still rather at a loss for finding information
(the commands names can be rather arcane)so I'm just posting my question:
I would like to estimate the shape coefficient of diverse
distributions (Weibull, gamma and Tukey-Lambda specifically, but other
could be of interest)
- Does R have a PPCC utility to estimate such parameter?(maximum value
of correlation coef)

2010 Jan 28

4

Problems with fitdistr

Hi,
I want to estimate parameters of weibull distribution. For this, I am using
fitdistr() function in MASS package.But when I give fitdistr(c,"weibull") I
get a Error as follows:-
Error in optim(x = c(4L, 41L, 20L, 6L, 12L, 6L, 7L, 13L, 2L, 8L, 22L,
:
non-finite value supplied by optim
Any help or suggestions are most welcomed
--
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1997 Dec 13

1

R-beta: Compile error; R-0.60.1, Solaris 2.6, gcc 2.7.2.1

...c -o punif.o
gcc -g -I../include -c qunif.c -o qunif.o
gcc -g -I../include -c runif.c -o runif.o
gcc -g -I../include -c dweibull.c -o dweibull.o
gcc -g -I../include -c pweibull.c -o pweibull.o
gcc -g -I../include -c qweibull.c -o qweibull.o
gcc -g -I../include -c rweibull.c -o rweibull.o
ar cr ../lib/libmath.a fsquare.o fcube.o fint.o fmax.o fmin2.o imin.o
imax.o fsign.o round.o prec.o sign.o i1mach.o d1mach.o beta.o gamma.o
lbeta.o lgamma.o choose.o dpsifn.o digamma.o sexp.o snorm.o sunif.o
dbeta.o pbeta.o qbeta.o rbeta.o dbinom.o pbinom.o qbinom.o rbinom....