在Python 3.8发布时,将有一个新的选项(计划于2019年10月20日),感谢PEP 572:分配表达式。 新的分配表达式操作员:= 允许您分配副本的结果,并且仍然使用它来呼叫更新,留下组合代码一个单一表达式,而不是两个陈述,改变:

newdict = dict1.copy()
newdict.update(dict2)

到:

(newdict := dict1.copy()).update(dict2)

(newdict := dict1.copy()).update(dict2) or newdict

newdict = {**dict1, **dict2}

脱包方法更清晰(对于任何人都知道一般的脱包,首先,你应该),不需要一个名字的结果(因此,它是更紧密的,当构建一个暂时的,即时转移到一个功能或包含在一个名单 / 双字或类似),并且几乎肯定更快,也是(在CPython上)相当于:

newdict = {}
newdict.update(dict1)
newdict.update(dict2)

它也更可扩展,因为结合三个法则是显而易见的:

 newdict = {**dict1, **dict2, **dict3}

 (newdict := dict1.copy()).update(dict2), newdict.update(dict3)

或沒有暫時的九個,但以真實的測試,每個九個結果:

 (newdict := dict1.copy()).update(dict2) or newdict.update(dict3)

2 词典

def union2(dict1, dict2):
    return dict(list(dict1.items()) + list(dict2.items()))

n 字典

def union(*dicts):
    return dict(itertools.chain.from_iterable(dct.items() for dct in dicts))

查看 https://mathieularose.com/how-not-to-flatten-a-list-of-lists-in-python/

>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> x, z = dict(x), x.update(y) or x
>>> x
{'a': 1, 'b': 2}
>>> y
{'c': 11, 'b': 10}
>>> z
{'a': 1, 'c': 11, 'b': 10}

深深的定律:

from typing import List, Dict
from copy import deepcopy

def merge_dicts(*from_dicts: List[Dict], no_copy: bool=False) -> Dict :
    """ no recursion deep merge of two dicts

    By default creates fresh Dict and merges all to it.

    no_copy = True, will merge all dicts to a fist one in a list without copy.
    Why? Sometime I need to combine one dictionary from "layers".
    The "layers" are not in use and dropped immediately after merging.
    """

    if no_copy:
        xerox = lambda x:x
    else:
        xerox = deepcopy

    result = xerox(from_dicts[0])

    for _from in from_dicts[1:]:
        merge_queue = [(result, _from)]
        for _to, _from in merge_queue:
            for k, v in _from.items():
                if k in _to and isinstance(_to[k], dict) and isinstance(v, dict):
                    # key collision add both are dicts.
                    # add to merging queue
                    merge_queue.append((_to[k], v))
                    continue
                _to[k] = xerox(v)

    return result

使用:

print("=============================")
print("merge all dicts to first one without copy.")
a0 = {"a":{"b":1}}
a1 = {"a":{"c":{"d":4}}}
a2 = {"a":{"c":{"f":5}, "d": 6}}
print(f"a0 id[{id(a0)}] value:{a0}")
print(f"a1 id[{id(a1)}] value:{a1}")
print(f"a2 id[{id(a2)}] value:{a2}")
r = merge_dicts(a0, a1, a2, no_copy=True)
print(f"r  id[{id(r)}] value:{r}")

print("=============================")
print("create fresh copy of all")
a0 = {"a":{"b":1}}
a1 = {"a":{"c":{"d":4}}}
a2 = {"a":{"c":{"f":5}, "d": 6}}
print(f"a0 id[{id(a0)}] value:{a0}")
print(f"a1 id[{id(a1)}] value:{a1}")
print(f"a2 id[{id(a2)}] value:{a2}")
r = merge_dicts(a0, a1, a2)
print(f"r  id[{id(r)}] value:{r}")

用一个细致的理解,你可以

x = {'a':1, 'b': 2}
y = {'b':10, 'c': 11}

dc = {xi:(x[xi] if xi not in list(y.keys()) 
           else y[xi]) for xi in list(x.keys())+(list(y.keys()))}

给予

>>> dc
{'a': 1, 'c': 11, 'b': 10}

注意合成,如果不明白

{ (some_key if condition else default_key):(something_if_true if condition 
          else something_if_false) for key, value in dict_.items() }

一个联合的OP的两个词典会是这样的东西:

{'a': 1, 'b': 2, 10, 'c': 11}

具体而言,两个实体(x 和 y)的联盟包含所有 x 和/或 y 的元素,不幸的是,OP 所要求的不是联盟,尽管职位的标题。

我的下面的代码既不优雅,也不是单线,但我认为它与联盟的意义一致。

从OP的例子:

x = {'a':1, 'b': 2}
y = {'b':10, 'c': 11}

z = {}
for k, v in x.items():
    if not k in z:
        z[k] = [(v)]
    else:
        z[k].append((v))
for k, v in y.items():
    if not k in z:
        z[k] = [(v)]
    else:
        z[k].append((v))

{'a': [1], 'b': [2, 10], 'c': [11]}

无论你想要的列表都可以改变,但上述将工作,如果一个词典包含列表(和列表)作为每个词典中的值。