这个解决方案很容易使用，它被用作一个普通的字典，但你可以使用求和函数。

class SumDict(dict):
    def __add__(self, y):
        return {x: self.get(x, 0) + y.get(x, 0) for x in set(self).union(y)}

A = SumDict({'a': 1, 'c': 2})
B = SumDict({'b': 3, 'c': 4})  # Also works: B = {'b': 3, 'c': 4}
print(A + B)  # OUTPUT {'a': 1, 'b': 3, 'c': 6}

这是一个很一般的解。你可以处理任意数量的dict +键，只在一些dict +容易使用任何聚合函数你想要:

def aggregate_dicts(dicts, operation=sum):
    """Aggregate a sequence of dictionaries using `operation`."""
    all_keys = set().union(*[el.keys() for el in dicts])
    return {k: operation([dic.get(k, None) for dic in dicts]) for k in all_keys}

例子:

dicts_same_keys = [{'x': 0, 'y': 1}, {'x': 1, 'y': 2}, {'x': 2, 'y': 3}]
aggregate_dicts(dicts_same_keys, operation=sum)
#{'x': 3, 'y': 6}

不相同的键和泛型聚合示例:

dicts_diff_keys = [{'x': 0, 'y': 1}, {'x': 1, 'y': 2}, {'x': 2, 'y': 3, 'c': 4}]

def mean_no_none(l):
    l_no_none = [el for el in l if el is not None]
    return sum(l_no_none) / len(l_no_none)

aggregate_dicts(dicts_diff_keys, operation=mean_no_none)
# {'x': 1.0, 'c': 4.0, 'y': 2.0}

import itertools
import collections

dictA = {'a':1, 'b':2, 'c':3}
dictB = {'b':3, 'c':4, 'd':5}

new_dict = collections.defaultdict(int)
# use dict.items() instead of dict.iteritems() for Python3
for k, v in itertools.chain(dictA.iteritems(), dictB.iteritems()):
    new_dict[k] += v

print dict(new_dict)

# OUTPUT
{'a': 1, 'c': 7, 'b': 5, 'd': 5}

OR

你也可以像@Martijn上面提到的那样使用Counter。

另外，请注意a.update(b)比a + b快2倍

from collections import Counter
a = Counter({'menu': 20, 'good': 15, 'happy': 10, 'bar': 5})
b = Counter({'menu': 1, 'good': 1, 'bar': 3})

%timeit a + b;
## 100000 loops, best of 3: 8.62 µs per loop
## The slowest run took 4.04 times longer than the fastest. This could mean that an intermediate result is being cached.

%timeit a.update(b)
## 100000 loops, best of 3: 4.51 µs per loop

没有额外进口的那个!

它们是一种python标准，叫做EAFP(请求原谅比请求许可更容易)。下面的代码基于该python标准。

# The A and B dictionaries
A = {'a': 1, 'b': 2, 'c': 3}
B = {'b': 3, 'c': 4, 'd': 5}

# The final dictionary. Will contain the final outputs.
newdict = {}

# Make sure every key of A and B get into the final dictionary 'newdict'.
newdict.update(A)
newdict.update(B)

# Iterate through each key of A.
for i in A.keys():

    # If same key exist on B, its values from A and B will add together and
    # get included in the final dictionary 'newdict'.
    try:
        addition = A[i] + B[i]
        newdict[i] = addition

    # If current key does not exist in dictionary B, it will give a KeyError,
    # catch it and continue looping.
    except KeyError:
        continue

编辑:感谢jerzyk提出的改进建议。

将三个字典a,b,c合并到一行中，没有任何其他模块或库

如果我们有三个字典

a = {"a":9}
b = {"b":7}
c = {'b': 2, 'd': 90}

将所有内容合并为一行，并使用返回一个dict对象

c = dict(a.items() + b.items() + c.items())

返回

{'a': 9, 'b': 2, 'd': 90}