在 python 继承层次结构中管理 args 和 kwargs [重复]

Question

有没有像我在这段代码中尝试过的那样在继承层次结构中管理 args 和 kwargs 的任何“好”方法。我的意思是不使用 kwargs 中的指定键或类似的东西获取值...

应该显示1 2 3 4：

class Parent(object):
    def __init__(self, motherArg1, motherArg2=100):
        self.motherArg1 = motherArg1
        self.motherArg2 = motherArg2
    def printParent(self):
        print self.motherArg1
        print self.motherArg2

class Child(Parent):
    def __init__(self, childArg1, *args, childArg2=100, **kwargs): # Doesn't work here
        super(Child, self).__init__(*args, **kwargs)
        self.childArg1 = childArg1
        self.childArg2 = childArg2
    def printChild(self):
        print self.childArg1
        print self.childArg2

child = Child(1, 3, childArg2=2, motherArg2=4)
child.printChild()
child.printParent()

语法不好：预期为“）；”在 *args 之后。

而def __init__(self, childArg1, childArg2=100, *args, **kwargs) 是正确的语法但不起作用。

1. 当我尝试此语法和 child = Child(1, childArg2=2, 3, motherArg2=4) 时，我得到 SyntaxError: non-keyword arg after keyword arg
2. 当我尝试 child = Child(1, 3, childArg2=2, motherArg2=4) 时，我得到 TypeError: __init__() got multiple values for keyword argument 'childArg2'

Answer 1

在 Python 2 中，您必须将 *args 参数放在 任何显式关键字参数之后：

def __init__(self, childArg1, childArg2=100, *args, **kwargs):

但是，您不能再使用其他位置参数无论如何，因为它们被childArg2 参数捕获：

>>> child = Child(1, 3, childArg2=2, motherArg2=4)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: __init__() got multiple values for keyword argument 'childArg2'

您唯一的选择是从 **kwargs 字典中获取关键字参数：

def __init__(self, childArg1, *args, **kwargs):
    childArg2 = kwargs.pop('childArg2', 2)

这使得childArg2 仅作为显式关键字参数起作用，而*args 捕获所有额外的位置参数：

>>> class Parent(object):
...     def __init__(self, motherArg1, motherArg2=100):
...         self.motherArg1 = motherArg1
...         self.motherArg2 = motherArg2
...     def printParent(self):
...         print self.motherArg1
...         print self.motherArg2
...
>>> class Child(Parent):
...     def __init__(self, childArg1, *args, **kwargs):
...         childArg2 = kwargs.pop('childArg2', 2)
...         super(Child, self).__init__(*args, **kwargs)
...         self.childArg1 = childArg1
...         self.childArg2 = childArg2
...     def printChild(self):
...         print self.childArg1
...         print self.childArg2
...
>>> child = Child(1, 3, childArg2=2, motherArg2=4)
>>> child.printChild()
1
2
>>> child.printParent()
3
4

Answer 2

当您将父级的 printChild 重命名为 printParent（并修复 prints）时，它已经在 python 3 中工作，如建议的那样：

1
2
3
4

但是你也可以让它在 python2 上工作。您可以通过删除 kwargs 中的条目来做到这一点，这些条目应该与子级相关，然后再传递给父级。

代码（适用于 python3）：

class Parent(object):
    def __init__(self, motherArg1, motherArg2=100):
        self.motherArg1 = motherArg1
        self.motherArg2 = motherArg2
    def printParent(self):
        print(self.motherArg1)
        print(self.motherArg2)

class Child(Parent):
    def __init__(self, childArg1, *args, childArg2=100, **kwargs):
        super(Child, self).__init__(*args, **kwargs)
        self.childArg1 = childArg1
        self.childArg2 = childArg2
    def printChild(self):
        print(self.childArg1)
        print(self.childArg2)

child = Child(1, 3, childArg2=2, motherArg2=4)
child.printChild()
child.printParent()

python2的代码

class Parent(object):
    def __init__(self, motherArg1, motherArg2=100):
        self.motherArg1 = motherArg1
        self.motherArg2 = motherArg2
    def printParent(self):
        print(self.motherArg1)
        print(self.motherArg2)

class Child(Parent):
    def __init__(self, childArg1, *args, **kwargs): 
        # this shows the concept, it can be formulated more elegantly
        # with @Martijn Pieters answer's 'pop':
        if 'childArg2' in kwargs:
            childArg2 = kwargs['childArg2']
            del kwargs['childArg2']
        else:
            childArg2 = 2
        super(Child, self).__init__(*args, **kwargs)
        self.childArg1 = childArg1
        self.childArg2 = childArg2
    def printChild(self):
        print(self.childArg1)
        print(self.childArg2)

child = Child(1, 3, childArg2=2, motherArg2=4)
child.printChild()
child.printParent()