默认参数的python函数

Question

如何在方法内部访问 qux？我真的必须在 foo() 的主体中再次定义它，还是有办法从 self 导入它？

class Baz(object):

    qux = lambda x : x + '_quux'

    def foo(self, bar=qux('fred')):
        print bar

        print qux('waldo')
        # NameError: global name 'qux' is not defined

        print Baz.qux('waldo')
        # TypeError: unbound method <lambda>() must be called with Baz instance as first argument (got str instance instead)

        print Baz.qux(self, 'waldo')
        # TypeError: <lambda>() takes exactly 1 argument (2 given)

        print self.qux('waldo')
        # TypeError: <lambda>() takes exactly 1 argument (2 given)                

Answer 1

如果您希望能够访问类外的方法，那么您必须使用 classmethod或 staticmethod .

class Baz(object):
  qux = staticmethod(lambda x : x + '_quux')

但是don't do that .