如何在OrderedDict的开头添加元素?


93

我有这个:

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

如果我这样做:

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

然后我得到这个:

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

但我想要这个:

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

而不创建新字典。


3
我认为您应该重新设计程序
Dmitry Zagorulkin

2
“ OrderedDict是一种可以记住首次插入键的顺序的字典。” @ZagorulkinDmitry是正确的:)(docs.python.org/2/library/...
MARKON

Answers:


86

在Python 2中没有执行此操作的内置方法。如果需要此操作,则需要编写一种以O(1)复杂度prepend()OrderedDict内部操作的方法/函数。

对于Python 3.2和更高版本,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的底部(last=True)还是顶部(last=FalseOrderedDict

最后,如果您想要一种快速,肮脏且缓慢的解决方案,则可以OrderedDict从头开始创建一个新的解决方案。

四种不同解决方案的详细信息:


扩展OrderedDict并添加新的实例方法

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')])

操纵OrderedDict对象的独立功能

该函数通过接受dict对象,键和值来做同样的事情。我个人更喜欢上课:

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引入了该OrderedDict.move_to_end()方法。使用它,我们可以在O(1)时间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')])

如果只需一步就可以插入一个元素并将其移到顶部,则可以直接使用它来创建prepend()包装器(此处未显示)。


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dict:

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')])


请注意,如果c已经存在,则不会更新旧值
jamylak

1
@IARI如果您指的是move_to_end这个问题,则没有Python 3标记,move_to_end仅适用于Python 3.2+。我将更新答案以包括基于Python 3的解决方案。不过非常感谢您的更新!
Ashwini Chaudhary

1
@AshwiniChaudhary既然已经添加了Python 3 move_to_front,也许最好实现一个move_to_front方法而不是一个单独的prepend方法?如果您需要同时支持同一代码库中的Python 2和Python 3,这将使您的代码更易于移植。
m000

1
dict_setitem=dict.__setitem__作为参数的背后原因是什么prepend?为什么/应该通过一个不同的二传手?
2016年

2
必须有一个错误ordered_dict_prepend上面。调用ordered_dict_prepend(d, 'c', 100)两次并尝试打印结果字典(只需d在Python的控制台中输入)就会导致Python进程不断获取内存。与Python 2.7.10进行测试
彼得·Dobrogost

15

编辑(2019-02-03) 请注意,以下答案仅适用于旧版本的Python。最近,OrderedDict已经用C进行了重写。此外,这确实触及了不赞成使用的双下划线属性。

我只是OrderedDict出于类似的目的在我的项目中编写了的子类。这是要点

O(1)与大多数这些解决方案不同,插入操作也是固定时间的(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')])

TypeError: '_Link' object does not support indexing在Python 3.4上使用它时得到。
RickardSjogren

1
它也不适用于python 3.5:AttributeError:'ListDict'对象没有属性'_OrderedDict__map'–
maggie

2
这不再起作用,因为从Python 3.5OrderedDict开始已经用C重写了该子类,并且该子类犯了忌讳内部的禁忌(实际上是反转名称修饰以访问__属性)。
两位炼金术士

13

您必须创建一个新的实例OrderedDict。如果您的键是唯一的:

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)])

但是,如果不是这样,请提防,因为您可能会或可能不会希望这种行为:

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)])

3
在python3中,items方法不再返回列表,而是返回一个像set一样的视图。在这种情况下,您需要采用集合并集,因为与+串联将不起作用:dict(x.items()| y.items())
Demz,2013年

1
我以为@TheDemz我认为set union不会保留顺序,从而使生成的项目的最终顺序OrderedDict不稳定?
最多

@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,可以看到基础词典中的变化。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/...
该DEMZ

6

如果知道要使用“ c”键,但不知道该值,则在创建字典时将“ c”与虚拟值一起插入。

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

稍后再更改值。

d1['c'] = 3

有没有办法插入orderdict中,使元素仍然排序(增加/减少)。
Eswar

排序的字典不按项目的任何属性排序。它是按插入顺序排序的,与项目本身无关。在CPthon 3.6和Python 3.7中,所有字典都是如此排序的,几乎没有理由使用OrderedDict。
特里·扬·里迪


3

FWIW这是我编写的用于插入任意索引位置的快速n脏代码。不一定有效,但可以就地工作。

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)

3

您可能希望完全使用其他结构,但是在python 2.7中有多种方法可以使用它。

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')])

就像其他人提到的那样,在python 3.2中,您可以OrderedDict.move_to_end('c', last=False)在插入后移动给定键。

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


2

如果您需要不具备的功能,只需使用所需的任何类扩展类:

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,但是可以工作:

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)])

1

我建议向prepend()此纯Python ActiveState配方添加方法或从中获取一个子类。鉴于用于排序的基础数据结构是一个链表,因此这样做的代码可能相当高效。

更新资料

为了证明这种方法是可行的,下面是执行建议操作的代码。另外,我还做了一些其他小的更改,以在Python 2.7.15和3.7.1中都可以使用。

prepend()在配方中的类中已添加了一个方法,并且已根据添加的另一个方法实施了该方法move_to_end(),该方法是OrderedDict在Python 3.2中添加的。

prepend()也可以直接实现,几乎完全与@Ashwini Chaudhary回答开头所示的一样—这样做可能会导致其速度稍快,但这留给有动机的读者练习……

# 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')])

@LazyLeopard:它不是线程安全的。您可能会发现如何使内置容器(集合,字典,列表)具有线程安全性?出于兴趣。
martineau

1

在使用@Ashwini Chaudhary答案和Python尝试打印或保存字典时,出现了无限循环2.7。但是我设法减少了他的代码,并使其在这里工作:

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 )

这太棒了!为我工作!
Souvik Ray

0

这是默认的,有序的字典,它允许在任何位置插入项目并使用。操作员创建密钥:

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]
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