我可以提供一个变体，检查是否输入了第三个值

>>> def example(x, y, z=None):
...   if z is None:
...     z = y
...     y = 0 # то самое значение по умолчанию
...   return x, y, z
...
>>> example(1, 2, 3)
(1, 2, 3)
>>> example(1, 2)
(1, 0, 2)
>>>

如果值为 2，则表示第二个被写入y，然后手动传输到z

None只有当- 是第三个参数的有效值时才会出现问题，并且它可以以自然的方式到达那里：

example(1, 2, None)

UPD

我将用解决方案更新我的答案*args：

>>> def example(*xyz):
...   if len(xyz) == 2:
...     x, z = xyz
...     y = 0
...   elif len(xyz) == 3:
...     x, y, z = xyz
...   else:
...     raise TypeError('example() takes 2 or 3 positional arguments but {} were given'.format(len(xyz)))
...
...   return x, y, z
...
>>> example(1)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 8, in example
TypeError: example() takes 2 or 3 positional arguments but 1 were given
>>> example(1, 2)
(1, 0, 2)
>>> example(1, 2, 3)
(1, 2, 3)
>>> example(1, 2, 3, 4)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 8, in example
TypeError: example() takes 2 or 3 positional arguments but 4 were given

该解决方案对参数类型具有抵抗力，并且参数的存在None不会以任何方式损害它。

如何将带有参数名称的参数传递给函数

def foo(x, y=0, z=0):
    pass

foo(x=5, z=3)

def example(default=('x', 'y', 'z'), n=0, **kw):
    return [kw.get(k, n) for k in default]

print(example())  # [0, 0, 0]
print(example(x=1, z=2))  # [1, 0, 2]