Integration with "quad" and "quadrature" from Python/SciPy

After reading this and that, it occurs to me that both "quad" and "quadrature" should be interchangeable*, atleast syntax-wise. Strangely it does seem they are not:

from scipy.integrate import quad as q
#from scipy.integrate import quadrature as q

def myfunc(x):
    return x

def integr():
    return q(myfunc, 0, 1)[0]

print integr()


def myfunc2(x, y):
    return x + y

def integr2(y):
    return q(myfunc2, 0, 1, args=(y))[0]
    #return q(myfunc2, 0, 1, args=[y])[0] 

print integr2(10)

... the example runs fine for "quad", but not for "quadrature" - I end up with:

Traceback (most recent call last):
  File "./test.py", line 38, in <module>
    print integr2(10)
  File "./test.py", line 36, in integr2
    return q(myfunc2, 0, 1, args=(y))[0]
  File "/usr/lib/python2.6/dist-packages/scipy/integrate/quadrature.py", line 136, in quadrature
    newval = fixed_quad(vfunc, a, b, (), n)[0]
  File "/usr/lib/python2.6/dist-packages/scipy/integrate/quadrature.py", line 48, in fixed_quad
    return (b-a)/2.0*sum(w*func(y,*args),0), None
  File "/usr/lib/python2.6/dist-packages/scipy/integrate/quadrature.py", line 77, in vfunc
    return func(x, *args)
TypeError: myfunc2() argument after * must be a sequence, not int

I have to switch the args tuple to a list (cf. commented line in integr2) even though the documentation says it should be a tuple. It seemed this is what the interpreter complains about ... (right?)

Is this intended? Or am I doing something wrong? In the end I'd like to be able to choose integration methods afterwards without having to change too much of the rest of the code.

*Actually开发者_如何转开发 I don't really get how to choose between the two. I do understand the difference between Gaussian quadrature and adaptive quadrature, but I don't know what "adaptive Gaussian quadrature" is supposed to mean - is the number of nodes adapted, if so how!?

---

The problem is in the line return q(myfunc2, 0, 1, args=(y))[0], specifically in the args=(y) part. What you want is args=(y,) (notice the comma after y) or args=[y].

The issue is that in Python tuples are created with commas, not with parentheses. Look:

>>> a = (1,)
>>> b = (1)
>>> print a, type(a)
(1,) <type 'tuple'>
>>> print b, type(b)
1 <type 'int'>