def dict_merge(a, b):
  c = a.copy()
  c.update(b)
  return c

new = dict_merge(old, extras)

在如此阴影和可疑的答案中,这个闪发光的例子是在Python中合并独裁的唯一和唯一好的方式,由独裁者为生活Guido van Rossum自己支持! 另一个人提出了一半,但没有把它放在一个功能上。

print dict_merge(
      {'color':'red', 'model':'Mini'},
      {'model':'Ferrari', 'owner':'Carl'})

给了:

{'color': 'red', 'owner': 'Carl', 'model': 'Ferrari'}

在Python3中,项目方法不再返回一个列表,而是一个视图,它像一个集一样作用。

dict(x.items() | y.items())

dict(x.viewitems() | y.viewitems())

编辑:

首先,请注意,在 Python 3 中, dic(x、 **y) 技巧不会工作,除非 y 中的键是线条。

此外,Raymond Hettinger的链路图答案是相当优雅的,因为它可以作为论点采取任意数量的论点,但从论点看起来它顺序地通过每个搜索的所有论点的列表:

In [1]: from collections import ChainMap
In [2]: from string import ascii_uppercase as up, ascii_lowercase as lo; x = dict(zip(lo, up)); y = dict(zip(up, lo))
In [3]: chainmap_dict = ChainMap(y, x)
In [4]: union_dict = dict(x.items() | y.items())
In [5]: timeit for k in union_dict: union_dict[k]
100000 loops, best of 3: 2.15 µs per loop
In [6]: timeit for k in chainmap_dict: chainmap_dict[k]
10000 loops, best of 3: 27.1 µs per loop

我是 Chainmap 的粉丝,但看起来不太实用,在那里可能有很多搜索。

我想要一些类似的东西,但有能力说明双键中的值是如何合并的,所以我打破了这个(但没有重量测试)。

def merge(d1, d2, merge_fn=lambda x,y:y):
    """
    Merges two dictionaries, non-destructively, combining 
    values on duplicate keys as defined by the optional merge
    function.  The default behavior replaces the values in d1
    with corresponding values in d2.  (There is no other generally
    applicable merge strategy, but often you'll have homogeneous 
    types in your dicts, so specifying a merge technique can be 
    valuable.)

    Examples:

    >>> d1
    {'a': 1, 'c': 3, 'b': 2}
    >>> merge(d1, d1)
    {'a': 1, 'c': 3, 'b': 2}
    >>> merge(d1, d1, lambda x,y: x+y)
    {'a': 2, 'c': 6, 'b': 4}

    """
    result = dict(d1)
    for k,v in d2.iteritems():
        if k in result:
            result[k] = merge_fn(result[k], v)
        else:
            result[k] = v
    return result

此外,当您使用.items() (Python 3.0 之前),您正在创建一个新的列表,包含从字典中的项目. 如果您的字典是大,那么它是相当多的顶部(两个大列表将被扔掉,一旦合并的字典创建)。更新() 可以更有效地工作,因为它可以通过第二个字典项目为项目。

在时间方面:

>>> timeit.Timer("dict(x, **y)", "x = dict(zip(range(1000), range(1000)))\ny=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
15.52571702003479
>>> timeit.Timer("temp = x.copy()\ntemp.update(y)", "x = dict(zip(range(1000), range(1000)))\ny=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
15.694622993469238
>>> timeit.Timer("dict(x.items() + y.items())", "x = dict(zip(range(1000), range(1000)))\ny=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
41.484580039978027

此外,字典创建的关键词论点仅在Python 2.3中添加,而复制()和更新()将在较旧版本中工作。

x = {'a': 1, 'b': 2}
y = {'b': 3, 'c': 4}

>>> z
{'a': 1, 'b': 3, 'c': 4}

z = {**x, **y}

z = {**x, 'foo': 1, 'bar': 2, **y}

>>> z
{'a': 1, 'b': 3, 'foo': 1, 'bar': 2, 'c': 4}

z = x.copy()
z.update(y) # which returns None since it mutates z

def merge_two_dicts(x, y):
    """Given two dictionaries, merge them into a new dict as a shallow copy."""
    z = x.copy()
    z.update(y)
    return z

z = merge_two_dicts(x, y)

def merge_dicts(*dict_args):
    """
    Given any number of dictionaries, shallow copy and merge into a new dict,
    precedence goes to key-value pairs in latter dictionaries.
    """
    result = {}
    for dictionary in dict_args:
        result.update(dictionary)
    return result

z = merge_dicts(a, b, c, d, e, f, g) 

和 g 的关键值对将先行于字典 a 到 f 等。

z = dict(x.items() + y.items())

>>> c = dict(a.items() + b.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: 'dict_items' and 'dict_items'

同样,在 Python 3 (viewitems() 在 Python 2.7) 中采取元素的合并也会失败,当值是不可破坏的对象(如列表,例如)。即使您的值是可破坏的,因为套件是无形的,行为与先例无定义。

>>> c = dict(a.items() | b.items())

>>> x = {'a': []}
>>> y = {'b': []}
>>> dict(x.items() | y.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: 'list'

>>> x = {'a': 2}
>>> y = {'a': 1}
>>> dict(x.items() | y.items())
{'a': 2}

另一个你不应该使用的黑客:

z = dict(x, **y)

字典的目的是采取可触摸的密钥(例如,frozensets或tuples),但这种方法在Python 3中失败,当密钥不是线条时。

>>> c = dict(a, **b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: keyword arguments must be strings

和

dict(a=1, b=10, c=11)

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

>>> foo(**{('a', 'b'): None})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() keywords must be strings
>>> dict(**{('a', 'b'): None})
{('a', 'b'): None}

我的答案: merge_two_dicts(x,y)实际上对我来说看起来更清楚,如果我们实际上对可读性感兴趣。

from copy import deepcopy

def dict_of_dicts_merge(x, y):
    z = {}
    overlapping_keys = x.keys() & y.keys()
    for key in overlapping_keys:
        z[key] = dict_of_dicts_merge(x[key], y[key])
    for key in x.keys() - overlapping_keys:
        z[key] = deepcopy(x[key])
    for key in y.keys() - overlapping_keys:
        z[key] = deepcopy(y[key])
    return z

>>> x = {'a':{1:{}}, 'b': {2:{}}}
>>> y = {'b':{10:{}}, 'c': {11:{}}}
>>> dict_of_dicts_merge(x, y)
{'b': {2: {}, 10: {}}, 'a': {1: {}}, 'c': {11: {}}}

{k: v for d in dicts for k, v in d.items()} # iteritems in Python 2.7

dict((k, v) for d in dicts for k, v in d.items()) # iteritems in Python 2

from itertools import chain
z = dict(chain(x.items(), y.items())) # iteritems in Python 2

from timeit import repeat
from itertools import chain

x = dict.fromkeys('abcdefg')
y = dict.fromkeys('efghijk')

def merge_two_dicts(x, y):
    z = x.copy()
    z.update(y)
    return z

min(repeat(lambda: {**x, **y}))
min(repeat(lambda: merge_two_dicts(x, y)))
min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
min(repeat(lambda: dict(chain(x.items(), y.items()))))
min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))

>>> min(repeat(lambda: {**x, **y}))
1.0804965235292912
>>> min(repeat(lambda: merge_two_dicts(x, y)))
1.636518670246005
>>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
3.1779992282390594
>>> min(repeat(lambda: dict(chain(x.items(), y.items()))))
2.740647904574871
>>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
4.266070580109954

$ uname -a
Linux nixos 4.19.113 #1-NixOS SMP Wed Mar 25 07:06:15 UTC 2020 x86_64 GNU/Linux

词典中的资源

您可以使用 toolz.merge([x, y]) 为此。