CODEWITHSUNDEEP
pythonscipy
Monica
Monica

Reputation: 1070

Integration using the quad method python

I'm trying to reproduce the error function using scipy, not using the erf method.

This is my code:

#!/usr/bin/env python
import numpy as np
import math
from scipy.integrate import quad

def error_func(y):
    return 2/math.sqrt(np.pi)*quad(np.exp(-y**2), 1, np.inf, args=(y))[0]

g = [error_func(x) for x in np.arange(-1,1,0.2)]

print g

This code returns the following error message:

File "./test.py", line 9, in <module>
    g = [error_func(x) for x in np.arange(-1,1,0.2)]
  File "./test.py", line 7, in error_func
    return 2/math.sqrt(np.pi)*quad(np.exp(-y**2), 1, np.inf, args=(y))[0]
  File "/usr/local/lib/python2.7/site-packages/scipy/integrate/quadpack.py", line 316, in quad
    points)
  File "/usr/local/lib/python2.7/site-packages/scipy/integrate/quadpack.py", line 383, in _quad
    return _quadpack._qagie(func,bound,infbounds,args,full_output,epsabs,epsrel,limit)
quadpack.error: quad: first argument is not callable

If I understand correctly, the first argument of quad must be a function. I thing that I pass it right. What is going wrong in my code?

Upvotes: 0

Views: 2106

Answers (1)

user2357112
user2357112

Reputation: 280572

Your first problem is that

np.exp(-y**2)

isn't a function. It's a number; specifically, it's the value of e^(-y^2). If you want to define a function that maps y to np.exp(-y**2), the easiest way is with the lambda syntax:

lambda y: np.exp(-y**2)

You have other problems, though:

(y) isn't a tuple in args=(y); it's just y in grouping parentheses. A 1-element tuple would be (y,).

Also, you shouldn't be passing the args argument to quad anyway, since your function doesn't need any additional arguments beyond the one we're integrating over.

Finally, your integration bounds are wrong; you should be integrating from 0 to the argument of the error function:

def error_func(x):
    return 2 / math.sqrt(np.pi) * quad(lambda y: np.exp(-y**2), 0, x)[0]

Upvotes: 2

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