Python:使用一个或多个关键字参数扩展方法的优雅习惯是什么？

Question

我正在尝试使用单个关键字参数扩展方法,同时保持对方法的其余部分的公正性; 我只是想传递它.尝试0:

class SomeSuperclass(object):
    pass # in reality: some implementation for some_method

class SomeClass(SomeSuperclass):
    def some_method(self, my_kwarg=42, *args, **kwargs):
        super(SomeClass, self).some_method(*args, **kwargs)
        do_something_interesting_with(my_kwarg)

SomeClass().some_method('arg 0', 'arg 1', some_kwargs=5, my_kwarg=8)

这不起作用:

Traceback (most recent call last):
  File "test.py", line 9, in <module>
    SomeClass().some_method('arg 0', 'arg 1', some_kwargs=5, my_kwarg=8)
TypeError: some_method() got multiple values for keyword argument 'my_kwarg'

我完全理解为什么以上不起作用,但我正在寻找一种很好的方法来使它工作.这是我目前(丑陋)的解决方法:

class SomeClass(SomeSuperclass):
    def some_method(self, *args, **kwargs):
        my_kwarg = kwargs.get('my_kwarg', 42) 
        if 'my_kwarg' in kwargs:
            del kwargs['my_kwarg']
        super(SomeClass, self).some_method(*args, **kwargs)
        do_something_interesting_with(my_kwarg)

这是每个kwarg 3条线路......

Answer 1

我会这样使用dict.pop():

class SomeSuperclass(object):
  def some_method(self, *args, **kwargs):
    print 'SomeSuperclass: args=%s kwargs=%s' % (args, kwargs)

class SomeClass(SomeSuperclass):
  def some_method(self, *args, **kwargs):
    my_kwarg = kwargs.pop('my_kwarg', 42) # 42 is the default
    print 'SomeClass: my_kwarg=%s' % my_kwarg
    super(SomeClass, self).some_method(*args, **kwargs)

SomeClass().some_method('arg 0', 'arg 1', some_kwargs=5, my_kwarg=8)
SomeClass().some_method('arg 0', 'arg 1', some_kwargs=5)

运行时打印出来:

SomeClass: my_kwarg=8
SomeSuperclass: args=('arg 0', 'arg 1') kwargs={'some_kwargs': 5}
SomeClass: my_kwarg=42
SomeSuperclass: args=('arg 0', 'arg 1') kwargs={'some_kwargs': 5}