python - 如何获取 OrderedDict 中的“下一个”项目？

Question

我正在使用 OrderedDict 随机访问列表，但现在想要next列表中的项目来自我拥有的项目：

foo = OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])
apple = foo['apple']

如何使用 justfoo和获取香蕉apple？

Answer 1

如果您可以访问有意保持私有的 OrderedDict 实现部分：

>>> class MyOrderedDict(OrderedDict):
...     def next_key(self, key):
...             next = self._OrderedDict__map[key][1]
...             if next is self._OrderedDict__root:
...                     raise ValueError("{!r} is the last key".format(key))
...             return next[2]
...     def first_key(self):
...             for key in self: return key
...             raise ValueError("OrderedDict() is empty")
... 
>>> od = MyOrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])
>>> od.next_key("apple")
'banana'
>>> od.next_key("banana")
'orange'
>>> od.next_key("orange")
'pear'
>>> od.next_key("pear")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 5, in next_key
ValueError: 'pear' is the last key
>>> od.first_key()
'apple'

Answer 2

I shudder to think how slow this will be on a list of size, but the only way I've come up with so far...

>>> foo.items()[foo.keys().index('apple') + 1]
('banana', 3)

Edit:

The example was slightly contrived; my actual collection is keyed by dates. If I need the entry after today; found a solution using dropwhile...

>>> foo = OrderedDict([(datetime.date(2000,1,1), 4), (datetime.date(2000,5,23), 3), datetime.date(2000,10,1), 2), (datetime.date(2000,12,31), 1)])
>>> today = datetime.date(2000,1,30)
>>> foo.items()[foo.keys().index((itertools.dropwhile(lambda d: d<today, foo)).next())]
(datetime.date(2000, 5, 23), 3)

Quite a mouthful.

Answer 3

Python 3.X

dict.items 将返回一个可迭代的 dict 视图对象而不是一个列表。我们需要将调用包装到一个列表中以使索引成为可能：

>>> from collections import OrderedDict
>>> 
>>> foo = OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])
>>> 
>>> def next_item(odic, key):
...     return list(odic)[list(odic.keys()).index(key) + 1]
... 
>>> next = next_item(foo, 'apple')
>>> print(next, foo[next])
banana 3

Answer 4

从您的代码中重新设计，我想这种方式会更好一些：

import collections as co
import datetime as dt
import itertools as it

foo = co.OrderedDict([
    (dt.date(2000,1,1), 4),
    (dt.date(2000,5,23), 3),
    (dt.date(2000,10,1), 2),
    (dt.date(2000,12,31), 1)
])
today = dt.date(2000,1,30)

fooiter = it.dropwhile(lambda d: d <= today, foo)
print next(fooiter)
print list(fooiter)

基本上在正确的地方有迭代器就足够了。

从任何位置开始迭代都会很酷，但不确定是否可能。需要一些思考。