为什么这不可能在Python中？

Question

def myfunc(a,b=2):
    print("Called with", a, b)
    return
p1 = functools.partial(myfunc, b=4)
p1("foobar", 4)

Why do I get a syntax error when I run that last line? It works if i do: myfunc("foobar", 4)

Answer 1

'partial' already sets 'b' to 4; if you want another value you should explicitely set parameter 'b':

>>> p1("foobar")
('Called with', 'foobar', 4)

>>> p1("foobar", b=5)
('Called with', 'foobar', 5)

Answer 2

Because you are supplying it with a b in functools.partial call. Your call should look like:

p1("foobar")

or you could just get rid of b=4 in p1 = functools.partial(myfunc, b=4) and make it something like:

p1 = functools.partial(myfunc)

Answer 3

I think this is because python allows you to use optional parameters in any order. When you work with more statically typed languages, there is usually a constraint on using optional parameters in the order defined in the function/method. So for instance, in python this is legal:

def myfunc(a=1,b=2,c=3):
    print a,b,c

myfunc(c=99, b=13, a=12)

Because you can specify optional parameters in any order, I think python explicitly needs to know which parameters are assigned to which local function variables.