According to (for instance)
http://en.wikipedia.org/wiki/Weibull_distribution the Weibull
distribution has two parameters: lambda 0 is the scale parameter
(real) and k 0 is the shape parameter (real). However, the
numpy.random.weibull function has only a single 'a' parameter (except
for the
D.Hendriks (Dennis) wrote:
According to (for instance)
http://en.wikipedia.org/wiki/Weibull_distribution the Weibull
distribution has two parameters: lambda 0 is the scale parameter
(real) and k 0 is the shape parameter (real). However, the
numpy.random.weibull function has only a
Robert Kern wrote:
D.Hendriks (Dennis) wrote:
According to (for instance)
http://en.wikipedia.org/wiki/Weibull_distribution the Weibull
distribution has two parameters: lambda 0 is the scale parameter
(real) and k 0 is the shape parameter (real). However, the
numpy.random.weibull
On Mon, 12 Nov 2007, D.Hendriks (Dennis) apparently wrote:
All of this makes me doubt the correctness of the formula
you proposed.
It is always a good idea to hesitate before doubting Robert.
Alan G Isaac wrote:
On Mon, 12 Nov 2007, D.Hendriks (Dennis) apparently wrote:
All of this makes me doubt the correctness of the formula
you proposed.
It is always a good idea to hesitate before doubting Robert.
D.Hendriks (Dennis) wrote:
Alan G Isaac wrote:
On Mon, 12 Nov 2007, D.Hendriks (Dennis) apparently wrote:
All of this makes me doubt the correctness of the formula
you proposed.
It is always a good idea to hesitate before doubting Robert.
D.Hendriks (Dennis) wrote:
Alan G Isaac wrote:
On Mon, 12 Nov 2007, D.Hendriks (Dennis) apparently wrote:
All of this makes me doubt the correctness of the formula
you proposed.
It is always a good idea to hesitate before doubting Robert.
