7

I was wondering how to generate a random weibull distribution with 2-parameter (lambda, k) in python. I know that numpy has a numpy.random.weibull, but it only accepts the a parameter as the shape of the distribution.

Víctor Navarro
  • 812
  • 9
  • 12

2 Answers2

6

Well, if you sample a number from weibull distribution with scale parameter missing (which assumes scale is equal to 1), then to get it scale multiply by lambda.

x = numpy.random.weibull(a)
return lambda*x
Severin Pappadeux
  • 18,636
  • 3
  • 38
  • 64
5

Severin Pappadeux's answer is probably the simplest way to include the scale parameter. An alternative is to use scipy.stats.weibull_min. weibull_min has three parameters: shape, location and scale. You only want the shape and scale, so you would set the location to 0.

from scipy.stats import weibull_min

n = 100     # number of samples
k = 2.4     # shape
lam = 5.5   # scale

x = weibull_min.rvs(k, loc=0, scale=lam, size=n)
Warren Weckesser
  • 110,654
  • 19
  • 194
  • 214