python 迭代地和优雅地合并字典？

Question

我有 2 个字典，例如：

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

我想要的合并结果应该是：

z = {3: {'a':1}, 2: {'a':3, 'b':1, 'c':2}}

我试过dict(x.items() + y.items())和Counter，但只得到

{2: {'a': 2, 'c': 2}, 3: {'a': 1}}

结果。

如何合并其值也是 dict 本身的 dict？

---

我已经解释了How to merge two Python dictionaries in a single expression?之间的区别 是不是我的字典的值也是字典。

Answer 1

这个呢：

对于 Python2.x：

from collections import Counter

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

t=x.copy()
print(dict({k: (dict(Counter(t.pop(k, None)) + Counter(v))) for k, v in y.items()},**t))

结果：

{1: {'a': 1, 'b': 1}, 2: {'a': 3, 'c': 2, 'b': 1}, 3: {'a': 1}}

对于 Python3.5+：

{**{k: (dict(Counter(t.pop(k, None)) + Counter(v))) for k, v in y.items()},**t}

Answer 2

这里有一个可能的解决方案。虽然它不使用任何库，但它相当复杂。

它还可以与任何两个字典一起使用，独立于哪个字典最大。

{key: {ikey: x.get(key, {}).get(ikey, 0) + y.get(key, {}).get(ikey, 0)  
    for ikey in x.get(key, {}).keys()+y.get(key, {}).keys()} 
 for key in x.keys()+y.keys()}

输出：

{2: {'a': 3, 'b': 1, 'c': 2}, 3: {'a': 1}}

Answer 3

假设对于某个键的任何相同深度，在两个字典中，值的类型相同（两个字典或两个数字，我不知道如何定义将数字合并到字典中）。

def merge(d1, d2):
    for i in d2:
        if i not in d1:
            continue
       if isinstance(d1[i], dict) and isinstance(d2[i], dict):
            merge(d1[i], d2[i])
        else:
            d2[i] += d1[i]
        d1.pop(i)
    for j in d1:
        d2[j] = d1[j]

这很简单，但可能并不优雅。

Answer 4

使用collections.Counter“合并”您要合并的项目（仅公共键），并保持其他项目不变，在字典理解中使用双三元表达式，迭代键的联合：

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

# pre-compute common keys, one-liners are not the ultimate goal in life!
common_keys = set(x) & set(y)

# now generate the merged dict
result = {k:collections.Counter(x[k])+collections.Counter(y[k]) if k in common_keys else x[k] if k in x else y[k] for k in set(x)|set(y)}
print(result)

结果：

{2: Counter({'a': 3, 'c': 2, 'b': 1}), 3: {'a': 1}}

注意：我们可以通过替换来避免预先计算公共键：

if k in common_keys

通过

if k in x and k in y

Answer 5

这行得通：

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


def key_merge (dict1, dict2):   
## function which checks for any matching keys, and merges them   
if len(dict1.keys()) == 0 or len(dict2.keys()) == 0:
    return {}
else:    
    for key in dict1.keys():
        if key in dict2.keys():
          return {key:{ k: dict1[key].get(k, 0) + dict2[key].get(k, 0) for k in set(dict1[key])|set(dict2[key]) }}

z = {**x, **y, **key_merge(x, y)}

再次，是否优雅由您决定！

Answer 6

我不知道你对“优雅”的定义是什么，但假设你的意思是可读，那么也许这种方式适合你。

from collections import Counter


def counter_dict(in_dict):
    """
    Create a dict of Counters from a dict of dicts.
    """
    return dict((k, Counter(v)) for (k, v) in in_dict.items())


def merge_counter_dicts(a, b):
    """
    Create a dict of Counters from two dicts of Counters.
    Where keys exist in both counts are summed.
    """
    out_dict = a.copy()
    for k in b.keys():
        out_dict[k] = out_dict.setdefault(k, Counter()) + b[k]
    return out_dict

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

xc = counter_dict(x)
xy = counter_dict(y)

print merge_counter_dicts(xc, xy)

Answer 7

# What about something like this:

# merge_nested_dicts.py

# tested under Python3.6
# assuming dict_02 overwrites dict_01
# one liner functional style
def deep_merge(dict_01, dict_02):
    return {k: {**dict_01.get(k), **dict_02.get(k)} if k in dict_01 and
                isinstance(dict_01.get(k), dict) and
                isinstance(dict_02.get(k), dict) else v 
                for k, v in {**dict_01, **dict_02}.items()}              

if __name__ == '__main__':
    y = {2: {'a': 1, 'b': 1}, 1: {'a': 1, 'b': 1}}
    x = {3: {'a': 1}, 2: {'a': 2, 'c': 2}}

    print(x)
    print(y)
    print(deep_merge(x, y))

'''
{3: {'a': 1}, 2: {'a': 2, 'c': 2}}
{2: {'a': 1, 'b': 1}, 1: {'a': 1, 'b': 1}}
{3: {'a': 1}, 2: {'a': 1, 'c': 2, 'b': 1}, 1: {'a': 1, 'b': 1}}
'''