如何在python OrderedDict上使用字符串鍵而不是整數進行切片？

Question

由於OrderedDict具有列表（帶有序元素）和字典（帶鍵而不是索引）的功能，因此您可以使用鍵進行切片似乎很自然。

>>> from collections import OrderedDict
>>> cities = OrderedDict((('san francisco', 650), ('new york', 212), ('shanghai', 8621), ('barcelona', 42423)))
>>> test['shanghai':]  # I want all the cities from shanghai to the end of the list
TypeError: unhashable type

有趣的是，由於OrderedDictionary.__getslice__沒有實現，這不是你看到的錯誤。 我嘗試將自己的__getslice__方法添加到OrderedDict ，但我一直__getslice__這個TypeError問題。 似乎Python正在進行某種類型檢查以強制切片鍵只是整數，然后才傳遞給__getslice__函數，unpythonic如何！

>>> class BetterOrderedDict(OrderedDict):
        def __getslice__(self, start=None, end=None, step=1):
            return 'potato'

>>> test = BetterOrderedDict((('one', 1), ('two', 2), ('three', 3), ('four', 4)))
>>> print test[1:4]
'potato'                           # ok this makes sense so far

>>> test['one':'four']
TypeError: unhashable type         # WTF, strings are hashable!

所以我的問題是，為什么我__getslice__非int切片，什么樣的類型檢查阻止切片鍵甚至到達我的__getslice__函數，我可以通過在C中使用綁定實現我的BetterOrderedDict來覆蓋它嗎？

Answer 1

__getslice__是不推薦使用切片的方式。 相反，您應該使用__getitem__處理slice對象：

from collections import OrderedDict

class SlicableDict(OrderedDict):
    def __getitem__(self, key):
        if isinstance(key, slice):
            return 'potato({},{},{})'.format(key.start, key.stop, key.step)
        return super(SlicableDict, self).__getitem__(key)

>>> s = SlicableDict(a=1, b=2, c=3)
>>> s
SlicableDict([('a', 1), ('c', 3), ('b', 2)])
>>> s['a']
1
>>> s['a':'c']
'potato(a,c,None)'

如果您需要的不僅僅是土豆，那么您可以按照以下方式實施所有三種切片操作：

def _key_slice_to_index_slice(items, key_slice):
    try:
        if key_slice.start is None:
            start = None
        else:
            start = next(idx for idx, (key, value) in enumerate(items)
                         if key == key_slice.start)
        if key_slice.stop is None:
            stop = None
        else:
            stop = next(idx for idx, (key, value) in enumerate(items)
                        if key == key_slice.stop)
    except StopIteration:
        raise KeyError
    return slice(start, stop, key_slice.step)

class SlicableDict(OrderedDict):
    def __getitem__(self, key):
        if isinstance(key, slice):
            items = self.items()
            index_slice = _key_slice_to_index_slice(items, key)
            return SlicableDict(items[index_slice])
        return super(SlicableDict, self).__getitem__(key)

    def __setitem__(self, key, value):
        if isinstance(key, slice):
            items = self.items()
            index_slice = _key_slice_to_index_slice(items, key)
            items[index_slice] = value.items()
            self.clear()
            self.update(items)
            return
        return super(SlicableDict, self).__setitem__(key, value)

    def __delitem__(self, key):
        if isinstance(key, slice):
            items = self.items()
            index_slice = _key_slice_to_index_slice(items, key)
            del items[index_slice]
            self.clear()
            self.update(items)
            return
        return super(SlicableDict, self).__delitem__(key)

Answer 2

這是您期望的切片功能的實際實現。

OrderedDict內部以雙向鏈表的形式維護鍵的順序。 引用Python 2.7.9中的實際注釋 ，

# The internal self.__map dict maps keys to links in a doubly linked list.
# The circular doubly linked list starts and ends with a sentinel element.
# The sentinel element never gets deleted (this simplifies the algorithm).
# Each link is stored as a list of length three:  [PREV, NEXT, KEY].

現在，為了對字典進行切片，我們需要迭代雙向鏈表__root ，它實際上是一個私有變量，受名稱修改機制的保護。

注意：這涉及使用OrderedDict的內部數據結構的hacky名稱unmangling。

from collections import OrderedDict

class SlicableDict(OrderedDict):
    def __getitem__(self, key):
        if isinstance(key, slice):
            # Unmangle `__root` to access the doubly linked list
            root = getattr(self, "_OrderedDict__root")
            # By default, make `start` as the first element, `end` as the last
            start, end = root[1][2], root[0][2]
            start = key.start or start
            end = key.stop or end
            step = key.step or 1
            curr, result, begun, counter = root[1], [], False, 0

            # Begin iterating
            curr, result, begun = root[1], [], False
            while curr is not root:
                # If the end value is reached, `break` and `return`
                if curr[2] == end:
                    break
                # If starting value is matched, start appending to `result`
                if curr[2] == start:
                    begun = True
                if begun:
                    if counter % step == 0:
                        result.append((curr[2], self[curr[2]]))
                    counter += 1

                # Make the `curr` point to the next element
                curr = curr[1]

            return result

        return super(SlicableDict, self).__getitem__(key)

幾個樣本運行：

>>> s = SlicableDict(a=1, b=2, c=3, d=4)
>>> s
SlicableDict([('a', 1), ('c', 3), ('b', 2), ('e', 5), ('d', 4), ('f', 6)])
>>> s['a':'c']
[('a', 1)]
>>> s['a':]
[('a', 1), ('c', 3), ('b', 2), ('e', 5), ('d', 4)]
>>> s[:'a']
[]
>>> s['a':'f':2]
[('a', 1), ('b', 2), ('d', 4)]

Answer 3

嘗試這個（非常難看）實現

class SliceOrdered(OrderedDict):

    def __getitem__(self, key):
        if isinstance(key, slice):
            tmp = OrderedDict()
            i_self = iter(self)
            for k in i_self:
                if key.start <= k <= key.stop:
                    tmp[k] = self[k]
                    if key.step is not None and key.step > 1:
                        for _ in range(key.step-1):
                            try:
                                next(i_self)
                            except StopIteration:
                                break
            return tmp
        else:
            return super(SliceOrdered, self).__getitem__(key)

演示（Python3.4）

>>> s = SliceOrdered([('a',2), ('b',2), ('c',3), ('d',4)])
>>> s['a':'c']
OrderedDict([('a', 2), ('b', 2), ('c', 3)])
>>> s['a':'d':2]
OrderedDict([('a', 2), ('c', 3)])

NB這可能只有作用，因為在這個例子中， OrderedDict不僅被排序，而且還被排序。 在未排序的字典中，切片'a':'c'不一定包含'b' ，所以我的if key.start <= k <= key.stop邏輯可能會失敗。 以下代碼應該尊重：

class SliceOrdered(OrderedDict):
    def __getitem__(self, key):
        if not isinstance(key, slice):
            return super(SliceOrdered,self).__getitem__(key)
        tmp = OrderedDict()
        step = key.step or 1
        accumulating = False
        i_self = iter(self)
        for k in i_self:
            if k == key.start:
                accumulating = True
            if accumulating:
                tmp[k] = self[k]
                for _ in range(step-1):
                    next(i_self)
            if k == key.stop:
                accumulating = False
                break
        return tmp