Python 3.5 TypeError: got multiple values for argument [duplicate]

Question

def f(a, b, *args):
    return (a, b, args)

f(a=3, b=5)  
(3, 5, ())

whereas:

f(a=3, b=5, *[1,2,3])  
TypeError: got multiple values for argument 'b'

Why it behaves like this?
Any particular reason?

Answer 1

In the documentation for calls:

If the syntax *expression appears in the function call, expression must evaluate to an iterable. Elements from these iterables are treated as if they were additional positional arguments. For the call f(x1, x2, *y, x3, x4), if y evaluates to a sequence y1, ..., yM, this is equivalent to a call with M+4 positional arguments x1, x2, y1, ..., yM, x3, x4.

And, this is followed by:

A consequence of this is that although the *expression syntax may appear after explicit keyword arguments, it is processed before the keyword arguments (and any **expression arguments – see below).

(emphasis mine)

So Python will first process the *args as positional arguments, assign a value to b and re-assign it with b=5 resulting in an error for the keyword argument having multiple values.

Answer 2

The problem is the keywords. You are not allowed positional arguments after keyword arguments.

f(3, 5, *[1,2,3])

works fine, in that it passes a tuple with the values 1,2,3. The same as:

f(3, 5, *(1,2,3))

Keywords must always come after positional arguments, so it is your function declaration which is incorrect:

def f(*args, a, b):
    return (a, b, args)

print(f(a=3, b=5))
print(f(*[1,2,3], a=3, b=5))

Gives:

(3, 5, ())
(3, 5, (1, 2, 3))