如何将元素添加到 OrderedDict 的开头?

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【中文标题】如何将元素添加到 OrderedDict 的开头?【英文标题】:How to add an element to the beginning of an OrderedDict? 【发布时间】:2013-05-15 22:01:05 【问题描述】:

我有这个:

d1 = OrderedDict([('a', '1'), ('b', '2')])

如果我这样做:

d1.update('c':'3')

然后我明白了:

OrderedDict([('a', '1'), ('b', '2'), ('c', '3')])

但我想要这个:

[('c', '3'), ('a', '1'), ('b', '2')]

不创建新字典。

【问题讨论】:

我认为你应该重新设计你的程序 "OrderedDict 是一个 dict,它会记住第一次插入键的顺序。" @ZagoulkinDmitry 是对的 :) (docs.python.org/2/library/…) ***.com/questions/38987/… 【参考方案1】:

在 Python 2 中没有用于执行此操作的内置方法。如果需要,您需要编写一个 prepend() 方法/函数,该方法/函数在 OrderedDict 内部进行操作,复杂度为 O(1)。

对于 Python 3.2 及更高版本,您应该使用move_to_end 方法。该方法接受一个last 参数,该参数指示元素将被移动到OrderedDict 的底部(last=True)还是顶部(last=False)。

最后,如果您想要一个快速、肮脏和缓慢的解决方案,您可以从头开始创建一个新的OrderedDict

四种不同解决方案的详细信息:


扩展OrderedDict并添加一个新的实例方法

from collections import OrderedDict

class MyOrderedDict(OrderedDict):

    def prepend(self, key, value, dict_setitem=dict.__setitem__):

        root = self._OrderedDict__root
        first = root[1]

        if key in self:
            link = self._OrderedDict__map[key]
            link_prev, link_next, _ = link
            link_prev[1] = link_next
            link_next[0] = link_prev
            link[0] = root
            link[1] = first
            root[1] = first[0] = link
        else:
            root[1] = first[0] = self._OrderedDict__map[key] = [root, first, key]
            dict_setitem(self, key, value)

演示:

>>> d = MyOrderedDict([('a', '1'), ('b', '2')])
>>> d
MyOrderedDict([('a', '1'), ('b', '2')])
>>> d.prepend('c', 100)
>>> d
MyOrderedDict([('c', 100), ('a', '1'), ('b', '2')])
>>> d.prepend('a', d['a'])
>>> d
MyOrderedDict([('a', '1'), ('c', 100), ('b', '2')])
>>> d.prepend('d', 200)
>>> d
MyOrderedDict([('d', 200), ('a', '1'), ('c', 100), ('b', '2')])

操作OrderedDict对象的独立函数

这个函数通过接受dict对象、键和值来做同样的事情。我个人更喜欢这门课:

from collections import OrderedDict

def ordered_dict_prepend(dct, key, value, dict_setitem=dict.__setitem__):
    root = dct._OrderedDict__root
    first = root[1]

    if key in dct:
        link = dct._OrderedDict__map[key]
        link_prev, link_next, _ = link
        link_prev[1] = link_next
        link_next[0] = link_prev
        link[0] = root
        link[1] = first
        root[1] = first[0] = link
    else:
        root[1] = first[0] = dct._OrderedDict__map[key] = [root, first, key]
        dict_setitem(dct, key, value)

演示:

>>> d = OrderedDict([('a', '1'), ('b', '2')])
>>> ordered_dict_prepend(d, 'c', 100)
>>> d
OrderedDict([('c', 100), ('a', '1'), ('b', '2')])
>>> ordered_dict_prepend(d, 'a', d['a'])
>>> d
OrderedDict([('a', '1'), ('c', 100), ('b', '2')])
>>> ordered_dict_prepend(d, 'd', 500)
>>> d
OrderedDict([('d', 500), ('a', '1'), ('c', 100), ('b', '2')])

使用OrderedDict.move_to_end() (Python >= 3.2)

Python 3.2 introduced OrderedDict.move_to_end() 方法。使用它,我们可以在 O(1) 时间内将现有键移动到字典的任一端。

>>> d1 = OrderedDict([('a', '1'), ('b', '2')])
>>> d1.update('c':'3')
>>> d1.move_to_end('c', last=False)
>>> d1
OrderedDict([('c', '3'), ('a', '1'), ('b', '2')])

如果我们需要插入一个元素并将其移动到顶部,只需一步,我们可以直接使用它来创建一个prepend() 包装器(此处不提供)。


创建一个新的OrderedDict - 慢!!!

如果您不想这样做并且性能不是问题,那么最简单的方法是创建一个新的字典:

from itertools import chain, ifilterfalse
from collections import OrderedDict


def unique_everseen(iterable, key=None):
    "List unique elements, preserving order. Remember all elements ever seen."
    # unique_everseen('AAAABBBCCDAABBB') --> A B C D
    # unique_everseen('ABBCcAD', str.lower) --> A B C D
    seen = set()
    seen_add = seen.add
    if key is None:
        for element in ifilterfalse(seen.__contains__, iterable):
            seen_add(element)
            yield element
    else:
        for element in iterable:
            k = key(element)
            if k not in seen:
                seen_add(k)
                yield element

d1 = OrderedDict([('a', '1'), ('b', '2'),('c', 4)])
d2 = OrderedDict([('c', 3), ('e', 5)])   #dict containing items to be added at the front
new_dic = OrderedDict((k, d2.get(k, d1.get(k))) for k in \
                                           unique_everseen(chain(d2, d1)))
print new_dic

输出:

OrderedDict([('c', 3), ('e', 5), ('a', '1'), ('b', '2')])

【讨论】:

注意如果c已经存在,这不会更新旧值 @IARI 如果您指的是move_to_end,那么问题上没有 Python 3 标签,move_to_end 仅适用于 Python 3.2+。我将更新我的答案以包含基于 Python 3 的解决方案。非常感谢您的更新! @AshwiniChaudhary 既然 Python 3 已经添加了move_to_front,那么实现move_to_front 方法而不是单独的prepend 方法可能更好?如果您需要从同一个代码库同时支持 Python 2 和 Python 3,这将使您的代码更具可移植性。 dict_setitem=dict.__setitem__ 作为prepend 的参数背后的原因是什么?为什么会/应该通过不同的二传手? 上面的ordered_dict_prepend中一定有bug。调用ordered_dict_prepend(d, 'c', 100) 两次并尝试打印结果字典(只需在Python 的控制台中输入d)会导致Python 进程不断占用内存。使用 Python 2.7.10 测试【参考方案2】:

如果您需要不存在的功能,只需使用您想要的任何内容扩展类即可:

from collections import OrderedDict

class OrderedDictWithPrepend(OrderedDict):
    def prepend(self, other):
        ins = []
        if hasattr(other, 'viewitems'):
            other = other.viewitems()
        for key, val in other:
            if key in self:
                self[key] = val
            else:
                ins.append((key, val))
        if ins:
            items = self.items()
            self.clear()
            self.update(ins)
            self.update(items)

效率不高,但有效:

o = OrderedDictWithPrepend()

o['a'] = 1
o['b'] = 2
print o
# OrderedDictWithPrepend([('a', 1), ('b', 2)])

o.prepend('c': 3)
print o
# OrderedDictWithPrepend([('c', 3), ('a', 1), ('b', 2)])

o.prepend([('a',11),('d',55),('e',66)])
print o
# OrderedDictWithPrepend([('d', 55), ('e', 66), ('c', 3), ('a', 11), ('b', 2)])

【讨论】:

【参考方案3】:

我建议向这个纯 Python ActiveState recipe 添加一个 prepend() 方法或从中派生一个子类。考虑到用于排序的底层数据结构是链表,这样做的代码可能相当有效。

更新

为了证明这种方法是可行的,下面是执行建议的代码。作为奖励,我还做了一些额外的小改动,以便在 Python 2.7.15 和 3.7.1 中工作。

prepend() 方法已添加到配方中的类中,并已根据已添加的另一个名为 move_to_end() 的方法实现,该方法已添加到 Python 3.2 中的 OrderedDict

prepend() 也可以直接实现,几乎与@Ashwini Chaudhary 的answer 开头所示的完全一样——这样做可能会稍微快一些,但这留给有动力的读者作为练习...

# Ordered Dictionary for Py2.4 from https://code.activestate.com/recipes/576693

# Backport of OrderedDict() class that runs on Python 2.4, 2.5, 2.6, 2.7 and pypy.
# Passes Python2.7's test suite and incorporates all the latest updates.

try:
    from thread import get_ident as _get_ident
except ImportError:  # Python 3
#    from dummy_thread import get_ident as _get_ident
    from _thread import get_ident as _get_ident  # Changed - martineau

try:
    from _abcoll import KeysView, ValuesView, ItemsView
except ImportError:
    pass

class MyOrderedDict(dict):
    'Dictionary that remembers insertion order'
    # An inherited dict maps keys to values.
    # The inherited dict provides __getitem__, __len__, __contains__, and get.
    # The remaining methods are order-aware.
    # Big-O running times for all methods are the same as for regular dictionaries.

    # The internal self.__map dictionary 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].

    def __init__(self, *args, **kwds):
        '''Initialize an ordered dictionary.  Signature is the same as for
        regular dictionaries, but keyword arguments are not recommended
        because their insertion order is arbitrary.

        '''
        if len(args) > 1:
            raise TypeError('expected at most 1 arguments, got %d' % len(args))
        try:
            self.__root
        except AttributeError:
            self.__root = root = []  # sentinel node
            root[:] = [root, root, None]
            self.__map = 
        self.__update(*args, **kwds)

    def prepend(self, key, value):  # Added to recipe.
        self.update(key: value)
        self.move_to_end(key, last=False)

    #### Derived from cpython 3.2 source code.
    def move_to_end(self, key, last=True):  # Added to recipe.
        '''Move an existing element to the end (or beginning if last==False).

        Raises KeyError if the element does not exist.
        When last=True, acts like a fast version of self[key]=self.pop(key).
        '''
        PREV, NEXT, KEY = 0, 1, 2

        link = self.__map[key]
        link_prev = link[PREV]
        link_next = link[NEXT]
        link_prev[NEXT] = link_next
        link_next[PREV] = link_prev
        root = self.__root

        if last:
            last = root[PREV]
            link[PREV] = last
            link[NEXT] = root
            last[NEXT] = root[PREV] = link
        else:
            first = root[NEXT]
            link[PREV] = root
            link[NEXT] = first
            root[NEXT] = first[PREV] = link
    ####

    def __setitem__(self, key, value, dict_setitem=dict.__setitem__):
        'od.__setitem__(i, y) <==> od[i]=y'
        # Setting a new item creates a new link which goes at the end of the linked
        # list, and the inherited dictionary is updated with the new key/value pair.
        if key not in self:
            root = self.__root
            last = root[0]
            last[1] = root[0] = self.__map[key] = [last, root, key]
        dict_setitem(self, key, value)

    def __delitem__(self, key, dict_delitem=dict.__delitem__):
        'od.__delitem__(y) <==> del od[y]'
        # Deleting an existing item uses self.__map to find the link which is
        # then removed by updating the links in the predecessor and successor nodes.
        dict_delitem(self, key)
        link_prev, link_next, key = self.__map.pop(key)
        link_prev[1] = link_next
        link_next[0] = link_prev

    def __iter__(self):
        'od.__iter__() <==> iter(od)'
        root = self.__root
        curr = root[1]
        while curr is not root:
            yield curr[2]
            curr = curr[1]

    def __reversed__(self):
        'od.__reversed__() <==> reversed(od)'
        root = self.__root
        curr = root[0]
        while curr is not root:
            yield curr[2]
            curr = curr[0]

    def clear(self):
        'od.clear() -> None.  Remove all items from od.'
        try:
            for node in self.__map.itervalues():
                del node[:]
            root = self.__root
            root[:] = [root, root, None]
            self.__map.clear()
        except AttributeError:
            pass
        dict.clear(self)

    def popitem(self, last=True):
        '''od.popitem() -> (k, v), return and remove a (key, value) pair.
        Pairs are returned in LIFO order if last is true or FIFO order if false.

        '''
        if not self:
            raise KeyError('dictionary is empty')
        root = self.__root
        if last:
            link = root[0]
            link_prev = link[0]
            link_prev[1] = root
            root[0] = link_prev
        else:
            link = root[1]
            link_next = link[1]
            root[1] = link_next
            link_next[0] = root
        key = link[2]
        del self.__map[key]
        value = dict.pop(self, key)
        return key, value

    # -- the following methods do not depend on the internal structure --

    def keys(self):
        'od.keys() -> list of keys in od'
        return list(self)

    def values(self):
        'od.values() -> list of values in od'
        return [self[key] for key in self]

    def items(self):
        'od.items() -> list of (key, value) pairs in od'
        return [(key, self[key]) for key in self]

    def iterkeys(self):
        'od.iterkeys() -> an iterator over the keys in od'
        return iter(self)

    def itervalues(self):
        'od.itervalues -> an iterator over the values in od'
        for k in self:
            yield self[k]

    def iteritems(self):
        'od.iteritems -> an iterator over the (key, value) items in od'
        for k in self:
            yield (k, self[k])

    def update(*args, **kwds):
        '''od.update(E, **F) -> None.  Update od from dict/iterable E and F.

        If E is a dict instance, does:           for k in E: od[k] = E[k]
        If E has a .keys() method, does:         for k in E.keys(): od[k] = E[k]
        Or if E is an iterable of items, does:   for k, v in E: od[k] = v
        In either case, this is followed by:     for k, v in F.items(): od[k] = v

        '''
        if len(args) > 2:
            raise TypeError('update() takes at most 2 positional '
                            'arguments (%d given)' % (len(args),))
        elif not args:
            raise TypeError('update() takes at least 1 argument (0 given)')
        self = args[0]
        # Make progressively weaker assumptions about "other"
        other = ()
        if len(args) == 2:
            other = args[1]
        if isinstance(other, dict):
            for key in other:
                self[key] = other[key]
        elif hasattr(other, 'keys'):
            for key in other.keys():
                self[key] = other[key]
        else:
            for key, value in other:
                self[key] = value
        for key, value in kwds.items():
            self[key] = value

    __update = update  # let subclasses override update without breaking __init__

    __marker = object()

    def pop(self, key, default=__marker):
        '''od.pop(k[,d]) -> v, remove specified key and return the corresponding value.
        If key is not found, d is returned if given, otherwise KeyError is raised.

        '''
        if key in self:
            result = self[key]
            del self[key]
            return result
        if default is self.__marker:
            raise KeyError(key)
        return default

    def setdefault(self, key, default=None):
        'od.setdefault(k[,d]) -> od.get(k,d), also set od[k]=d if k not in od'
        if key in self:
            return self[key]
        self[key] = default
        return default

    def __repr__(self, _repr_running=):
        'od.__repr__() <==> repr(od)'
        call_key = id(self), _get_ident()
        if call_key in _repr_running:
            return '...'
        _repr_running[call_key] = 1
        try:
            if not self:
                return '%s()' % (self.__class__.__name__,)
            return '%s(%r)' % (self.__class__.__name__, self.items())
        finally:
            del _repr_running[call_key]

    def __reduce__(self):
        'Return state information for pickling'
        items = [[k, self[k]] for k in self]
        inst_dict = vars(self).copy()
        for k in vars(MyOrderedDict()):
            inst_dict.pop(k, None)
        if inst_dict:
            return (self.__class__, (items,), inst_dict)
        return self.__class__, (items,)

    def copy(self):
        'od.copy() -> a shallow copy of od'
        return self.__class__(self)

    @classmethod
    def fromkeys(cls, iterable, value=None):
        '''OD.fromkeys(S[, v]) -> New ordered dictionary with keys from S
        and values equal to v (which defaults to None).

        '''
        d = cls()
        for key in iterable:
            d[key] = value
        return d

    def __eq__(self, other):
        '''od.__eq__(y) <==> od==y.  Comparison to another OD is order-sensitive
        while comparison to a regular mapping is order-insensitive.

        '''
        if isinstance(other, MyOrderedDict):
            return len(self)==len(other) and self.items() == other.items()
        return dict.__eq__(self, other)

    def __ne__(self, other):
        return not self == other

    # -- the following methods are only used in Python 2.7 --

    def viewkeys(self):
        "od.viewkeys() -> a set-like object providing a view on od's keys"
        return KeysView(self)

    def viewvalues(self):
        "od.viewvalues() -> an object providing a view on od's values"
        return ValuesView(self)

    def viewitems(self):
        "od.viewitems() -> a set-like object providing a view on od's items"
        return ItemsView(self)

if __name__ == '__main__':

    d1 = MyOrderedDict([('a', '1'), ('b', '2')])
    d1.update('c':'3')
    print(d1)  # -> MyOrderedDict([('a', '1'), ('b', '2'), ('c', '3')])

    d2 = MyOrderedDict([('a', '1'), ('b', '2')])
    d2.prepend('c', 100)
    print(d2)  # -> MyOrderedDict([('c', 100), ('a', '1'), ('b', '2')])

【讨论】:

@LazyLeopard:它不是线程安全的。您可能会发现 How to make built-in containers (sets, dicts, lists) thread safe? 感兴趣。【参考方案4】:

您必须创建OrderedDict 的新实例。如果您的密钥是唯一的:

d1=OrderedDict([("a",1),("b",2)])
d2=OrderedDict([("c",3),("d",99)])
both=OrderedDict(list(d2.items()) + list(d1.items()))
print(both)

#OrderedDict([('c', 3), ('d', 99), ('a', 1), ('b', 2)])

但如果不是,请注意,您可能不希望这种行为:

d1=OrderedDict([("a",1),("b",2)])
d2=OrderedDict([("c",3),("b",99)])
both=OrderedDict(list(d2.items()) + list(d1.items()))
print(both)

#OrderedDict([('c', 3), ('b', 2), ('a', 1)])

【讨论】:

在python3中,items方法不再返回一个列表,而是一个视图,它的作用就像一个集合。在这种情况下,您需要使用集合并集,因为与 + 连接不起作用: dict(x.items() | y.items()) @TheDemz 我认为 set union 不会保留顺序,从而使生成的 OrderedDict 中的项目的最终顺序不稳定? @max 是的,它不稳定。从 dict.keys()、dict.values() 和 dict.items() 返回的对象称为字典视图。它们是惰性序列,可以看到底层字典的变化。要强制字典视图成为完整列表,请使用 list(dictview)。请参阅字典视图对象。 docs.python.org/3.4/library/…【参考方案5】:

编辑(2019-02-03) 请注意,以下答案仅适用于旧版本的 Python。最近,OrderedDict 已用 C 重写。此外,这确实触及了令人不悦的双下划线属性。

为了类似的目的,我刚刚在我的一个项目中编写了OrderedDict 的一个子类。 Here's the gist.

与大多数这些解决方案不同,插入操作也是常数时间O(1)(它们不需要您重建数据结构)。

>>> d1 = ListDict([('a', '1'), ('b', '2')])
>>> d1.insert_before('a', ('c', 3))
>>> d1
ListDict([('c', 3), ('a', '1'), ('b', '2')])

【讨论】:

我在 Python 3.4 上使用它时得到TypeError: '_Link' object does not support indexing 它也不适用于 python 3.5: AttributeError: 'ListDict' object has no attribute '_OrderedDict__map' 这不再有效,因为从 Python 3.5 开始,OrderedDict 一直是 rewritten in C,并且这个子类犯了与内部人员混为一谈的禁忌(实际上是颠倒名称修改以访问 __ 属性)。跨度> 【参考方案6】:

现在可以使用 move_to_end(key, last=True)

>>> d = OrderedDict.fromkeys('abcde')
>>> d.move_to_end('b')
>>> ''.join(d.keys())
'acdeb'
>>> d.move_to_end('b', last=False)
>>> ''.join(d.keys())
'bacde'

https://docs.python.org/3/library/collections.html#collections.OrderedDict.move_to_end

【讨论】:

【参考方案7】:

如果您知道需要一个“c”键,但不知道值,请在创建 dict 时插入带有虚拟值的“c”。

d1 = OrderedDict([('c', None), ('a', '1'), ('b', '2')])

稍后再更改值。

d1['c'] = 3

【讨论】:

有没有办法插入一个有序的字典,这样元素仍然是排序的(增加/减少)。 有序字典不按项目的任何属性排序。它是按插入顺序排列的,与物品本身无关。在 CPthon 3.6 和 Python 3.7 中,所有 dicts 都是如此有序的,几乎没有理由使用 OrderedDict。【参考方案8】:

FWIW 这是我为插入任意索引位置而编写的快速脏代码。不一定有效,但它可以就地工作。

class OrderedDictInsert(OrderedDict):
    def insert(self, index, key, value):
        self[key] = value
        for ii, k in enumerate(list(self.keys())):
            if ii >= index and k != key:
                self.move_to_end(k)

【讨论】:

【参考方案9】:

您可能希望完全使用不同的结构,但在 python 2.7 中有一些方法。

d1 = OrderedDict([('a', '1'), ('b', '2')])
d2 = OrderedDict(c='3')
d2.update(d1)

d2 将包含

>>> d2
OrderedDict([('c', '3'), ('a', '1'), ('b', '2')])

正如其他人所说,在 python 3.2 中,您可以使用OrderedDict.move_to_end('c', last=False) 在插入后移动给定的键。

注意:请注意,由于创建新的 OrderedDict 和复制旧值,第一个选项对于大型数据集的速度较慢。

【讨论】:

【参考方案10】:

我在尝试使用@Ashwini Chaudhary 打印或保存字典时遇到了一个无限循环,使用 Python 2.7 回答。但我设法减少了他的代码,让它在这里工作:

def move_to_dict_beginning(dictionary, key):
    """
        Move a OrderedDict item to its beginning, or add it to its beginning.
        Compatible with Python 2.7
    """

    if sys.version_info[0] < 3:
        value = dictionary[key]
        del dictionary[key]
        root = dictionary._OrderedDict__root

        first = root[1]
        root[1] = first[0] = dictionary._OrderedDict__map[key] = [root, first, key]
        dict.__setitem__(dictionary, key, value)

    else:
        dictionary.move_to_end( key, last=False )

【讨论】:

这太棒了!为我工作!【参考方案11】:

这是一个默认的有序字典,允许在任何位置插入项目并使用 .运算符创建键:

from collections import OrderedDict

class defdict(OrderedDict):

    _protected = ["_OrderedDict__root", "_OrderedDict__map", "_cb"]    
    _cb = None

    def __init__(self, cb=None):
        super(defdict, self).__init__()
        self._cb = cb

    def __setattr__(self, name, value):
        # if the attr is not in self._protected set a key
        if name in self._protected:
            OrderedDict.__setattr__(self, name, value)
        else:
            OrderedDict.__setitem__(self, name, value)

    def __getattr__(self, name):
        if name in self._protected:
            return OrderedDict.__getattr__(self, name)
        else:
            # implements missing keys
            # if there is a callable _cb, create a key with its value
            try:
                return OrderedDict.__getitem__(self, name)
            except KeyError as e:
                if callable(self._cb):
                    value = self[name] = self._cb()
                    return value
                raise e

    def insert(self, index, name, value):
        items = [(k, v) for k, v in self.items()]
        items.insert(index, (name, value))
        self.clear()
        for k, v in items:
            self[k] = v


asd = defdict(lambda: 10)
asd.k1 = "Hey"
asd.k3 = "Bye"
asd.k4 = "Hello"
asd.insert(1, "k2", "New item")
print asd.k5 # access a missing key will create one when there is a callback
# 10
asd.k6 += 5  # adding to a missing key
print asd.k6
# 15
print asd.keys()
# ['k1', 'k2', 'k3', 'k4', 'k5', 'k6']
print asd.values()
# ['Hey', 'New item', 'Bye', 'Hello', 10, 15]

【讨论】:

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