Question

I am attempting to extract Weibull distribution parameters (shape ‘k’ and scale ‘lambda’) that satisfy a certain mean and variance. In this example, the mean is 4 and the variance is 8. It is a 2-unknowns and 2-equations type of problem.

Since this algorithm works with Excel 2010’s GRG Solver, I am certain it is about the way I am framing the problem, or potentially, the libraries I am using. I am not overly familiar with optimization libraries, so please let me know where the error is.

Below is the script:

from scipy.optimize import fmin_cg
import math

def weibull_mu(k, lmda):                  #Formula can be found on wikipedia
    return lmda*math.gamma(1+1/k)
def weibull_var(k, lmda):                 #Formula can be found on wikipedia
    return lmda**2*math.gamma(1+2/k)-weibull_mu(k, lmda)**2

def min_function(arggs):
    actual_mean = 4                          # specific to this example
    actual_var = 8                           # specific to this example
    k = arggs[0]
    lmda = arggs[1]
    output = [weibull_mu(k, lmda)-(var_wei)]
    output.append(weibull_var(k, lmda)-(actual_var)**2-(actual_mean)**2)
    return output

print fmin(min_function, [1,1])

This script gives me the following error:

[...]
  File "C:\Program Files\Python27\lib\site-packages\scipy\optimize\optimize.py", line 278, in fmin
    fsim[0] = func(x0)
ValueError: setting an array element with a sequence.

Answer 1

I managed to get it to work thanks to Anders Gustafsson’s comment (thank you). This script now works if one returns only a scalar (in this case I used something along the lines of least-squares). Also, bounds were added by changing the optimization function to “fmin_l_bfgs_b” (again, thanks to Anders Gustafsson).

I only changed the min_function definition relative to the question.

from scipy.optimize import fmin_l_bfgs_b
import math

def weibull_mu(k, lmda):
    return lmda*math.gamma(1+1/k)
def weibull_var(k, lmda):
    return lmda**2*math.gamma(1+2/k)-weibull_mu(k, lmda)**2

def min_function(arggs):
    actual_mean = 4.                    # specific to this example
    actual_var = 8.                     # specific to this example
    k = arggs[0]
    lmda = arggs[1]
    extracted_var = weibull_var(k, lmda)
    extracted_mean = weibull_mu(k, lmda)
    output = (extracted_var - actual_var)**2 + (extracted_mean - actual_mean)**2
    return output

print fmin_l_bfgs_b(min_function, best_guess, approx_grad = True, bounds = [(.0000001,None),(.0000001,None)], disp = False)

Note: Please feel free to use this script for your own or professional use.