Issue
I was doing a numerical integration where instead of using
scipy.integrate.dblquad
I tried to build the double integral directly using
scipy.integrate.quad
From camz's answer: https://stackoverflow.com/a/30786163/11141816
a=1
b=2
def integrand(theta, t):
return sin(t)*sin(theta);
def theta_integral(t):
# Note scipy will pass args as the second argument
return integrate.quad(integrand, b-a, t , args=(t))[0]
integrate.quad(theta_integral, b-a,b+a )
However, the part where it was confusing was how the args was passed through the function. For example, in the post args=(t) was pass through to the second argument t of the function integrand automatically, not the first argument theta, where in the scipy document https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.quad.html the only reference was
args: tuple, optional
Extra arguments to pass to func.
and in fact their arguments had a comma , to be args=(1,), instead of args=(1) directly.
What if I decide to pass several variables, i.e.
def integrand(theta, t,a,b):
return sin(t)*sin(theta)*a*b;
how would I know which args(t,a,b) or args(t,a,b,) were the correct args() for the function?
How to pass args() to integrate.quad in scipy?
Solution
In [214]: from scipy.integrate import quad
A simple example. I'm not using the args, just showing them:
In [215]: def foo(x,a,b,c):
...: print(x,a,b,c)
...: return x**2
...:
In [216]: quad(foo, 0, 1, (1,(1,2,3),np.arange(2)))
0.5 1 (1, 2, 3) [0 1]
0.013046735741414128 1 (1, 2, 3) [0 1]
0.9869532642585859 1 (1, 2, 3) [0 1]
0.06746831665550773 1 (1, 2, 3) [0 1]
0.9325316833444923 1 (1, 2, 3) [0 1]
0.16029521585048778 1 (1, 2, 3) [0 1]
0.8397047841495122 1 (1, 2, 3) [0 1]
...
0.35280356864926987 1 (1, 2, 3) [0 1]
0.6471964313507301 1 (1, 2, 3) [0 1]
Out[216]: (0.33333333333333337, 3.700743415417189e-15)
That's all there is to args.
Looking at the code for quad, the first line is
if not isinstance(args, tuple):
args = (args,)
So if you use args=(t), that will be changed to args=(t,). Remember, in python (2) is the same as 2; the () just group operations in complex cases, otherwise they are ignored. But (1,) is a tuple, as (1,2). It's really the comma that defines the tuple.
With your inner function, and an added print:
In [232]: def integrand(theta, t):
...: print(theta, t)
...: return np.sin(t)*np.sin(theta)
...:
In [233]: integrand(1,2) # pass 2 numbers
1 2
Out[233]: 0.7651474012342926
In [234]: integrand(1,np.arange(3)) # is still works if the 2nd is an array:
1 [0 1 2]
Out[234]: array([0. , 0.70807342, 0.7651474 ])
t is an array, and so is np.sin(t), and also the return. But you aren't passing arrays. Calling it via quad, the theta varies, the t does not:
In [235]: quad(integrand, 0, 1, args=(1,))
0.5 1
0.013046735741414128 1
0.9869532642585859 1
...
0.6471964313507301 1
Out[235]: (0.38682227139505565, 4.2945899214059184e-15)
In [236]: quad(integrand, 0, 1, args=(1,2))
---------------------------------------------------------------------------
...
TypeError: integrand() takes 2 positional arguments but 3 were given
The actual error is raised in compiled code, so isn't quite as transparent, but the effect the same as if I tried to pass 3 arguments:
In [237]: integrand(1,1,2)
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
Input In [237], in <cell line: 1>()
----> 1 integrand(1,1,2)
TypeError: integrand() takes 2 positional arguments but 3 were given
Answered By - hpaulj Answer Checked By - Katrina (PHPFixing Volunteer)
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