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')]
无需创建新字典。
问问题
144352 次
11 回答
100
在 Python 2 中没有内置的方法可以做到这一点。如果你需要这个，你需要编写一个在O(1) 复杂度的内部prepend()操作的方法/函数。OrderedDict
对于 Python 3.2 及更高版本，您应该使用该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底部 ( last=True) 还是顶部 ( last=False) OrderedDict。
最后，如果你想要一个快速、肮脏和缓慢的解决方案，你可以OrderedDict从头开始创建一个新的。
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')])
OrderedDict操作对象的独立函数
这个函数通过接受 dict 对象、键和值来做同样的事情。我个人更喜欢这个类：
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，只需一步完成，我们可以直接使用它来创建一个prepend()包装器（此处未展示）。
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')])
于 2013-05-21T07:59:45.597 回答
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已用 C 重写。此外，这确实触及了令人不悦的双下划线属性。
OrderedDict我只是为了类似的目的在我的一个项目中编写了一个子类。这是要点。
与大多数这些解决方案不同，插入操作也是恒定时间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')])
于 2013-08-20T04:20:20.280 回答
13
你必须创建一个新的实例OrderedDict。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，您可能需要也可能不需要这种行为：
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)])
于 2013-08-06T12:44:27.890 回答
6
如果您知道您需要一个“c”键，但不知道该值，请在创建 dict 时插入带有虚拟值的“c”。
d1 = OrderedDict([('c', None), ('a', '1'), ('b', '2')])
并稍后更改该值。
d1['c'] = 3
于 2015-03-25T16:45:13.213 回答
5
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
于 2015-01-22T11:28:42.327 回答
4
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)
于 2018-05-18T16:39:31.200 回答
3
您可能希望完全使用不同的结构，但在python 2.7中有一些方法可以做到这一点。
d1 = OrderedDict([('a', '1'), ('b', '2')])
d2 = OrderedDict(c='3')
d2.update(d1)
然后 d2 将包含
>>> 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)在插入后移动给定的键。
注意：考虑到由于创建新的 OrderedDict 和复制旧值，第一个选项对于大型数据集较慢。
于 2018-06-08T13:19:49.433 回答
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该类：
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)])
于 2013-05-21T08:55:06.820 回答
1
我建议向prepend()这个纯 Python ActiveState 配方添加一个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。
一个prepend()方法已添加到配方中的类中，并已根据已添加的另一种方法实现实现，该方法已move_to_end()添加到OrderedDictPython 3.2 中。
prepend()也可以直接实现，几乎与@Ashwini Chaudhary 的答案开头所显示的完全一样——这样做可能会导致它稍微快一点，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')])
于 2013-05-21T09:39:53.817 回答
1
尝试使用@Ashwini Chaudhary回答 Python打印或保存字典时出现无限循环2.7。但我设法减少了他的代码，并让它在这里工作：
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 )
于 2019-01-11T11:56:13.023 回答
0
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]
于 2019-07-04T18:31:39.303 回答