method to calculate Weibull parameters `c` and `scale` from mean and standard deviation

I’ve been struggling with the problem of how to calculate the Weibull parameters c and scale (required by scipy.stats.weibull_min) from mean and standard. From a few searches, it’s clear that others have wrestled with the same problem. It would be helpful if a robust solution could be incorporated into scipy.stats. My code, which is not currently working, follows.

def weibull_c_and_scale(mn, sd):
   """
   Overview
   --------

   Various definitions of the Weibull distribution are in use.  This function
   uses an iterative technique to convert from one parameterization to another.
   Inputs are the mean and standard deviation; outputs are the parameters `c`
   and `scale`, which are required by `scipy.stats.weibull_min`.


   Note
   ----

   (1) I'm following the definitions of `c` and `scale` from ref 1 below.

   (2) The last statement of the objective function uses

       log(sqrt(var_) / sd)

   rather than the simpler

       sqrt(var_) - sd

   because the latter is strongly nonlinear, which tends to inflate the number
   of optimization steps.


   References
   ----------

   1. https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.weibull_min.html
   2. https://en.wikipedia.org/wiki/Weibull_distribution
   """

   def f(x):
      c, scale= x
      mn_, var_= stats.weibull_min.stats(c, moments='mv')
      return (mn_ - mn) **2 + log(sqrt(var_) / sd) **2

   result= optimize.minimize(f, x0=[1.0, 1.0], method='L-BFGS-B',
     bounds=[(0, None), (0, None)])

   return result.x


def weibull_c_and_scale_test(mn=1):
   """
   This function validates `weibull_c_from_sd`, testing the behavior for ten
   values of standard deviation spanning 0.1 to 100.
   """

   print("     mn     mn_actual      sd      sd_actual     c       scale")
   format= 6 * '{:10.6f} '

   for sd in [0.1, 0.2, 0.5, 1, 2, 5, 10, 20, 50, 100]:
      c, scale= weibull_c_and_scale(mn=mn, sd=sd)
      mn_actual= stats.weibull_min(c=c, scale=scale).mean()
      sd_actual= stats.weibull_min(c=c, scale=scale).std()
      print(format.format(mn, mn_actual, sd, sd_actual, c, scale))

[mdhaber]

@tupui just a note: I think fit solves a different problem. This is about converting from one set of known parameters to a different parameterization, (whereas fit gets unknown parameters from data). The difficulty is that the equations relating the two parameterizations don’t seem to be solvable in closed form.

[Phillip-M-Feldman]

This is very nice! My thanks to both of you.

What are the chances of getting this implemented in SciPy?

On Mon, May 18, 2020 at 5:49 PM Warren Weckesser [e-mail] wrote:

@siddhantwahal https://github.com/siddhantwahal, nice!

I wouldn’t be surprised if there are other formulations of the problem that are more robust than what I came up with; I stopped when I got results that seemed “pretty good”. And further investigation might show that different formulations work best in different parameter ranges. Getting something like this to work as a black box solver can end up being a lot of work. Allowing the user to provide an initial guess is a reasonable way to give the user some control for those cases where the current implementation’s default value fails.

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