将新项目添加到原始项目，排序，制作新字典：
>>> arr = {('a',1111),('b',2222),('f',3333)}
>>> arr = collections.OrderedDict(arr)
>>> new = ('c',4444)
>>> items = list(arr.items())
>>> items.append(new)
>>> items.sort()
>>> arr = collections.OrderedDict(items)
>>> arr
OrderedDict([('a', 1111), ('b', 2222), ('c', 4444), ('f', 3333)])
或者更多的选择：
- 子类 collections.OrderedDict
- 使用
move_to_end 方法作为指导，创建一个将遍历双向链表的新方法；找到插入的地方；然后插入新的密钥 - 也许可以在这里使用 bisect 或其他一些双向链表排序插入算法
- 覆盖
__setitem__ 方法并在中调用新方法 - 或者用算法替换add-new-key-to-the-end代码您在上一个项目符号中提出的。
保持键排序顺序的排序字典
我不知道如何使 OrderedDict 子类工作 - 它有许多名称被破坏的属性 - 只需要重写一两个方法，我不想花时间弄清楚名称修饰方面。
所以只需将 整个 OrderedDict 类从源 from here - to here 复制到一个单独的模块中，这样您就可以导入它，并包含这些导入。
from _weakref import proxy as _proxy
from collections import _Link, _OrderedDictKeysView
from collections import _OrderedDictItemsView, _OrderedDictValuesView
import _collections_abc
from _weakref import proxy as _proxy
from reprlib import recursive_repr as _recursive_repr
from operator import itemgetter as _itemgetter, eq as _eq
import bisect
然后在类中更改以下内容：
- 类名当然可以随便你。类名后面是一个文档字符串和一些描述类行为的 cmets - 这些应该更新。
class SortOrderedDict(dict):
- 覆盖
__setitem__ 方法。下面使用bisect 查找插入顺序。不知道是否真的有必要，它必须首先列出一个 dict 键视图，但这部分应该是快速 C 代码（？在这里猜测）
def __setitem__(self, key, value,
dict_setitem=dict.__setitem__, proxy=_proxy, Link=_Link):
'od.__setitem__(i, y) <==> od[i]=y'
# Setting a new item creates a new link in the linked list,
# inserted at its key sorted position - uses less than comparisons,
# and the inherited dictionary is updated with the new key/value pair.
if key not in self:
self.__map[key] = link = Link()
root = self.__root
last = root.prev
link.key = key
curr = root.next
if curr is root: # first item!
link.prev, link.next = last, root
last.next = link
root.prev = proxy(link)
elif link.key < root.next.key: # at the beginning?
#print(f'{link.key} before {root.next.key}')
soft_link = root.next
link.prev, link.next = root, soft_link
soft_link.prev = link
root.next = link
elif root.prev.key < link.key: # at the end?
#print(f'{link.key} at the end after {root.prev.key}')
soft_link = root.prev
link.prev, link.next = soft_link, root
soft_link.next = link
root.prev = proxy(link)
else: # in the middle somewhere - use bisect
keys = list(self.keys())
i = bisect.bisect_left(keys,key)
right = self.__map[keys[i]]
#print(f'{link.key} between {right.prev.key} and {right.key}')
soft_link = right.prev
link.prev,link.next = soft_link,right
right.prev = link
soft_link.next = link
dict_setitem(self, key, value)
-
添加
update 方法 - 此类是 dict 的子类，这会覆盖其更新方法，强制其使用 __setitem__。
def update(self,other):
try:
other = other.items()
except AttributeError:
pass
for k,v in other:
self[k] = v
- 将此行
update = __update = _collections_abc.MutableMapping.update 更改为
__update = update
- 在
__reduce__ 方法中，将 for k in vars(OrderedDict()): 中的类名更改为您为类命名的任何名称
for k in vars(SortOrderedDict()):
-
__eq__ 方法中的相同内容。将if isinstance(other, OrderedDict): 更改为
if isinstance(other, SortOrderedDict):
如果使用 bisect 似乎不值得，只需遍历链表直到找到插入点。 （上面列出的所有其他更改仍然适用）
def __setitem__(self, key, value,
dict_setitem=dict.__setitem__, proxy=_proxy, Link=_Link):
'od.__setitem__(i, y) <==> od[i]=y'
# Setting a new item creates a new link in the linked list,
# inserted at its key sorted position - uses less than comparisons,
# and the inherited dictionary is updated with the new key/value pair.
if key not in self:
self.__map[key] = link = Link()
root = self.__root
last = root.prev
link.key = key
curr = root.next
if curr is root: # first item!
link.prev, link.next = last, root
last.next = link
root.prev = proxy(link)
# traverse the linked list; find sorted insertion point; insert
while curr is not root:
if link.key < curr.key:
soft_link = curr.prev
soft_link.next = link
link.prev = soft_link
link.next = curr
curr.prev = link
break
elif curr.next is root:
link.prev, link.next = curr, root
curr.next = link
root.prev = proxy(link)
break
curr = curr.next
dict_setitem(self, key, value)
用法
>>> arr = {('a',1111),('f',3333),('b',2222)}
>>> arr = SortOrderedDict(arr)
>>> arr
SortOrderedDict([('a', 1111), ('b', 2222), ('f', 3333)])
>>> other = {k:v for k,v in zip('tvsnpqkl',range(8))}
>>> arr.update(other)
>>> arr
SortOrderedDict([('a', 1111), ('b', 2222), ('f', 3333), ('k', 6), ('l', 7), ('n', 3), ('p', 4), ('q', 5), ('s', 2), ('t', 0), ('v', 1)])
>>> b = SortOrderedDict((('a',1111),('f',3333),('b',2222)))
>>> b.update(other)
>>> arr == b
True
>>> b == arr
True
>>>