Re: [R] Simulating from a Weibull distribution

2014-09-02 Thread David Winsemius
On Sep 2, 2014, at 6:52 PM, Lucy Leigh wrote: > 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

Re: [R] Simulating from a Weibull distribution

2014-09-02 Thread Lucy Leigh
Awesome, thankyou! From: Joshua Wiley [mailto:jwiley.ps...@gmail.com] Sent: Wednesday, 3 September 2014 12:04 PM To: Lucy Leigh Cc: r-help@R-project.org Subject: Re: [R] Simulating from a Weibull distribution Hi Lucy, Try the gamlss.dist package, specifically the rWEI2() function. Cheers

Re: [R] Simulating from a Weibull distribution

2014-09-02 Thread Duncan Murdoch
On 02/09/2014, 9:52 PM, Lucy Leigh wrote: > 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 w

Re: [R] Simulating from a Weibull distribution

2014-09-02 Thread Joshua Wiley
Hi Lucy, Try the gamlss.dist package, specifically the rWEI2() function. Cheers, Josh On Wed, Sep 3, 2014 at 11:52 AM, Lucy Leigh wrote: > Hi, > I wish to simulate some data from a Weibull distribution. The rweibull > function in R uses the parameterisation > > 'with shape parameter a and sc

[R] Simulating from a Weibull distribution

2014-09-02 Thread Lucy Leigh
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 d