Python:如何实现 __getattr__()？

Question

我的类(class)有一个字典，例如:

class MyClass(object):
    def __init__(self):
        self.data = {'a': 'v1', 'b': 'v2'}

那我想用 dict 的 key 和 MyClass 实例来访问 dict，例如:

ob = MyClass()
v = ob.a   # Here I expect ob.a returns 'v1'

我知道这应该由 __getattr__ 实现，但我是 Python 新手，我不完全知道如何实现它。

Answer 1

class MyClass(object):

    def __init__(self):
        self.data = {'a': 'v1', 'b': 'v2'}

    def __getattr__(self, attr):
        return self.data[attr]

---

>>> ob = MyClass()
>>> v = ob.a
>>> v
'v1'

在实现 __setattr__ 时要小心，您需要进行一些修改:

class MyClass(object):

    def __init__(self):
        # prevents infinite recursion from self.data = {'a': 'v1', 'b': 'v2'}
        # as now we have __setattr__, which will call __getattr__ when the line
        # self.data[k] tries to access self.data, won't find it in the instance 
        # dictionary and return self.data[k] will in turn call __getattr__
        # for the same reason and so on.... so we manually set data initially
        super(MyClass, self).__setattr__('data', {'a': 'v1', 'b': 'v2'})

    def __setattr__(self, k, v):
        self.data[k] = v

    def __getattr__(self, k):
        # we don't need a special call to super here because getattr is only 
        # called when an attribute is NOT found in the instance's dictionary
        try:
            return self.data[k]
        except KeyError:
            raise AttributeError

---

>>> ob = MyClass()
>>> ob.c = 1
>>> ob.c
1

如果您不需要设置属性，只需使用命名元组例如。

>>> from collections import namedtuple
>>> MyClass = namedtuple("MyClass", ["a", "b"])
>>> ob = MyClass(a=1, b=2)
>>> ob.a
1

如果你想要默认参数，你可以围绕它编写一个包装类:

class MyClass(namedtuple("MyClass", ["a", "b"])):

    def __new__(cls, a="v1", b="v2"):
        return super(MyClass, cls).__new__(cls, a, b)

或者它作为一个函数看起来更好:

def MyClass(a="v1", b="v2", cls=namedtuple("MyClass", ["a", "b"])):
    return cls(a, b)

---

>>> ob = MyClass()
>>> ob.a
'v1'