[Numpy-discussion] numpy.vectorize fails, howto avoid hardcoding parameters?

[josef.pktd at gmail.com]

On Mon, Aug 2, 2010 at 3:35 PM, Alex Kraus <alex_work at live.de> wrote:
>
> Hi,
>
> I am trying to create a function that calculates the integral of another function. The integral function should later be used in scipy.optimize.leastsq(f, ...), so ideally it should have the format:
>
> def f(x, *param)
>
> so that it works for a variable number of parameters. While my code works for a fixed number of parameters I cannot get it to work with a variable number of parameters. It seems that numpy.vectorize fails here.
>
> Is there a different/better way to do this?
>
> from scipy.integrate import quad
> import numpy as np
>
> def integrand_function(x, a, b, c):
>     result = a*x**2 + np.exp(b*x) + np.cos(a*c)
>     return result
>
> def define_integral(f, lower, upper):
>     assert(lower < upper)
>
>     def function(a, b, c):
>         result = quad(f, lower, upper, args=(a, b, c))[0]
>         return result
>     return np.vectorize(function)
>
> def integrand_function_param(x, *param):
>     a, b, c = param
>     result = a*x**2 + np.exp(b*x) + np.cos(a*c)
>     return result
>
> def define_integral_param(f, lower, upper):
>     assert(lower < upper)
>
>     def function(a, *param):
>         print(param)
>         result = quad(f, lower, upper, args=(a, param))[0]
>         return result
>     return np.vectorize(function)
>
> a = np.array([1,2,3,4])
>
> print(integrand_function(1,2,3,4))
> # 21.9400368894
>
> f = define_integral(integrand_function, 0.0, 2.0)
> print(f(a, 1,2))
> # [  8.22342909  10.41510219  16.30939667  16.7647227 ]
>
> print(integrand_function_param(1,2,3,4))
> # 21.9400368894
>
> fp = define_integral_param(integrand_function_param, 0.0, 2.0)
> print(fp(a, 1,2))
> # ValueError: mismatch between python function inputs and received arguments

vectorize cannot infer correctly the number of arguments with *params

see http://projects.scipy.org/scipy/ticket/422

the solution that I used in scipy.stats.distributions was to
explicitly specify the number of input arguments.

nin = ...

(I don't remember all the details)

Josef


>
> # fp should later be used in scipy.optimize.leastsq(fp, ...
>
> Any help is very appreciated!
> Alexander
>
>
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