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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