如何將元素添加到OrderedDict的開頭？

Question

我有這個：

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

如果我這樣做：

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

然后我明白了：

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

但我想要這個：

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

沒有創建新詞典。

Answer 1

在Python 2中沒有內置的方法。如果需要這個，你需要編寫一個prepend()方法/函數，它在OrderedDict內部操作，復雜度為O（1）。

對於Python 3.2及更高版本，您應該使用move_to_end方法。 該方法接受last參數，該參數指示元素是否將移動到OrderedDict的底部（ last=True ）或top（ last=False ）。

最后，如果你想要一個快速，骯臟和緩慢的解決方案，你可以從頭開始創建一個新的OrderedDict 。

四種不同解決方案的詳細信息：

---

擴展OrderedDict並添加新的實例方法

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對象的獨立函數

這個函數通過接受dict對象，鍵和值來做同樣的事情。 我個人更喜歡上課：

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OrderedDict.move_to_end()方法。 使用它，我們可以在O（1）時間內將現有密鑰移動到字典的任一端。

>>> 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')])

如果我們需要插入一個元素並將其移到頂部，一步一步，我們可以直接使用它來創建一個prepend()包裝器（此處未提供）。

---

創建一個新的OrderedDict - 慢!!!

如果您不想這樣做並且性能不是問題，那么最簡單的方法是創建一個新的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

輸出：

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

---

Answer 2

編輯（2019-02-03） 請注意，以下答案僅適用於舊版本的Python。 最近， OrderedDict已經在C中重寫。此外，這確實觸及了雙下划線屬性，這是不贊成的。

我剛剛在我的一個項目中寫了一個OrderedDict的子類，用於類似的目的。 這是要點 。

與大多數這些解決方案不同，插入操作也是恆定時間O(1) （它們不需要您重建數據結構）。

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

Answer 3

你必須創建一個OrderedDict的新實例。 如果您的鑰匙是唯一的：

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)])

但如果沒有，請注意，您可能需要或不需要此行為：

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)])

Answer 4

如果您知道需要'c'鍵但不知道該值，則在創建dict時插入帶有虛擬值的'c'。

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

並在以后更改該值。

d1['c'] = 3

Answer 5

現在可以使用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

Answer 6

如果您需要不存在的功能，只需使用您想要的任何內容擴展類：

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)

不是非常有效，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)])

Answer 7

FWIW這是我為插入任意索引位置而編寫的快速臟代碼。 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)

Answer 8

您可能希望完全使用不同的結構，但有一些方法可以在python 2.7中實現 。

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

然后d2將包含

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

正如其他人所提到的，在python 3.2中，您可以使用OrderedDict.move_to_end('c', last=False)在插入后移動給定的鍵。

注意：考慮到由於創建新的OrderedDict和復制舊值，大型數據集的第一個選項較慢。

Answer 9

我建議在這個純Python ActiveState配方中添加一個prepend()方法，或者從中派生一個子類。 考慮到訂購的基礎數據結構是鏈表，這樣做的代碼可能是相當有效的。

更新

為了證明這種方法是可行的，下面是執行建議的代碼。 作為獎勵，我還進行了一些額外的微小改動，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類中，並已根據另一個名為move_to_end() ，該方法已添加到Python 3.2中的OrderedDict中。

prepend()也可以直接實現，幾乎完全如@Ashwini Chaudhary的答案開頭所示 - 這樣做可能會導致它稍微快一些，但這仍然留給了有動力的讀者練習......

# 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')])

Answer 10

在嘗試使用Python 2.7 @Ashwini Chaudhary答案打印或保存字典時，我得到了一個無限循環。 但我設法減少了他的代碼，並讓它在這里工作：

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 )

Answer 11

這是一個默認的有序dict，允許在任何位置插入項目並使用。 運算符來創建鍵：

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]