有序字典的深度复制是否期望保留其顺序？

Question

我编写了一个小程序并测试它确实保持了顺序。然而，我仍然想确保deepcopy能够做到这一点。

import copy
import collections

a_dict = collections.OrderedDict()
a_dict['m'] = 10
a_dict['u'] = 15
a_dict['c'] = 5
a_dict['h'] = 25
a_dict['a'] = 55
a_dict['s'] = 30

print(a_dict)

other_dict = copy.deepcopy(a_dict)

other_dict['g'] = 75
other_dict['r'] = 35

print(other_dict)

该程序的输出是

OrderedDict([('m', 10), ('u', 15), ('c', 5), ('h', 25), ('a', 55), ('s', 30)])
OrderedDict([('m', 10), ('u', 15), ('c', 5), ('h', 25), ('a', 55), ('s', 30), ('g', 75), ('r', 35)])

Answer 1

在 CPython 中，顺序似乎被保留了。我通过检查 的实施情况得出了这个结论deepcopy。在这种情况下，它将在您的对象上找到用于酸洗的__reduce_ex__or方法：__reduce__OrderedDict

（https://github.com/python/cpython/blob/master/Lib/copy.py#L159-L161）

def deepcopy(x, memo=None, _nil=[]):
...
                    reductor = getattr(x, "__reduce_ex__", None)
                    if reductor is not None:
                        rv = reductor(4)
                    else:
                        reductor = getattr(x, "__reduce__", None)
                        if reductor:
                            rv = reductor()

这些返回odict_iterator用于构造的对象，因此将保留顺序：

>>> a = {}
>>> b = collections.OrderedDict()
>>> a['a'] = 1
>>> b['a'] = 1
>>> a.__reduce_ex__(4)
(<function __newobj__ at 0x10471a158>, (<class 'dict'>,), None, None, <dict_itemiterator object at 0x104b5d958>)
>>> a.__reduce__()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/copyreg.py", line 65, in _reduce_ex
    raise TypeError("can't pickle %s objects" % base.__name__)
TypeError: can't pickle dict objects
>>> b.__reduce_ex__(4)
(<class 'collections.OrderedDict'>, (), None, None, <odict_iterator object at 0x104c02d58>)
>>> b.__reduce__()
(<class 'collections.OrderedDict'>, (), None, None, <odict_iterator object at 0x104c5c780>)
>>>