__init __（）为关键字参数“ bar”获得了多个值

Question

What seems like a simple pattern for allowing a variable number of positional arguments along with a defaulted named arg doesn't work. I think I can overcome this by using **kwargs , but is there something more straightforward I am doing wrong?

Why doesn't this work?

class Foo(obect):
    """Baffling"""
    def __init__(self, bar=None, *args):
        pass

foo = Foo(1) #works
foo = Foo(1, bar=2) # explodes

----> 1 foo = Foo(1, bar='baz')
TypeError: __init__() got multiple values for keyword argument 'bar'

How is the above code passing a duplicate bar keyword argument?

Answer 1

In Python, named attributes are used after non-named (like *args ). So you should change your __init__ as:

def __init__(self, *args, bar=None):

Here is the example:

class Foo(object):
    """Baffling"""
    def __init__(self, *args, bar=None):
        self.bar = bar

foo = Foo(1, bar=2) # explodes
foo.bar

Will return:

Out[...]: 2

---

PS Anyway it is not a good idea to mix *args , **kwargs and named attributes. I recommend you to avoid it.

Answer 2

Named arguments can be passed both by position and by name:

>>> def foo(a):
...     print(a)
...
>>> foo(1)
1
>>> foo(a=2)
2

If you pass positional and keyword arguments, they get assigned in order. In your case, 1 gets assigned to the first positional argument ( bar ), then bar=2 assigns the the argument named bar . Thus, both assign to the name bar , creating a conflict.

---

You can pass additional arguments after your named one:

>>> def foo(bar=None, *args):
...    print('args=%r, bar=%r' % (args, bar))
...
... foo(2, 1)  # bar=2, *args=(1,)
args=(1,), bar=2

In Python3, you can also make your parameter keyword only:

>>> def foo(*args, bar=None):
...    print('args=%r, bar=%r' % (args, bar))
...
... foo(1, bar=2)
args=(1,), bar=2

This also works when you do not take any variadic arguments:

>>> def foo(*, bar=None):
...    print('args=%r, bar=%r' % ('undefined', bar))
...
... foo(bar=2)
args='undefined', bar=2
>>> foo(1, bar=2)
TypeError: foo() takes 0 positional arguments but 1 positional argument (and 1 keyword-only argument) were given

In Python2, only **kwargs is allowed after *args . You can emulate named parameters by popping them from kwargs :

>>> def foo(*args, **kwargs):
...    bar = kwargs.pop("bar")
...    print('args=%r, bar=%r' % (args, bar))
...
... foo(1, bar=2)
args=(1,), bar=2

If you want named-only parameters without variadic positional and named parameters, you must raise errors yourself:

>>> def foo(*args, **kwargs):
...    bar = kwargs.pop('bar')
...    if args or kwargs:
...        raise TypeError
...    print('args=%r, bar=%r' % ('undefined', bar))
...
... foo(bar=2)
args='undefined', bar=2
>>> foo(1, bar=2)
TypeError