Question

我想组合两个嵌套词典

d1 = {"admin": {"key1": "v2"}}
d2 = {"admin": {"key2": "v3"},
      "user": {"something": "else"}}

这应该结合到：

d = {"admin": {"key1": "v2",
               "key2": "v3"},
     "user": {"something": "else"}}

除了迭代第一个键之外，还有一种简单的方法吗？

Answer 1

您可以使用defaultdict将第一级的每个值都设为dict。从dict中使用update()之后。

>>> from collections import defaultdict
>>> d = defaultdict(dict)
>>> d1 = {"admin": {"key1": "v2"}}
>>> d2 = {"admin": {"key2": "v3"},
...       "user": {"something": "else"}}
>>> for dd in [d1,d2]:
...     for k,v in dd.items():
...         d[k].update(v)
... 
>>> d
defaultdict(<type 'dict'>, {'admin': {'key2': 'v3', 'key1': 'v2'}, 'user': {'something': 'else'}})

或者以更紧凑的形式，您可以使用map() ...我真的不喜欢它但是很有机会

>>> for dd in [d1,d2]:
...     map(lambda x:d[x[0]].update(x[1]),dd.items())

Answer 2

如果只有一个嵌套级别：

>>> d1 = {"admin": {"key1": "v2"}}
>>> d2 = {"admin": {"key2": "v3"},
...       "user": {"something": "else"}}
>>> keys = list(d1) + list(d2)
>>> d = {k: dict(d1.get(k, {}).items() + d2.get(k, {}).items()) for k in keys}
>>> d
{'admin': {'key1': 'v2', 'key2': 'v3'}, 'user': {'something': 'else'}}

更深的嵌套需要递归。

Answer 3

from itertools import chain
from collections import defaultdict
new_d = defaultdict(dict)

for k,v in chain(d1.iteritems(),d2.iteritems()):
    new_d[k].update(v)
print(new_d)
defaultdict(<type 'dict'>, {'admin': {'key2': 'v3', 'key1': 'v2'}, 'user': {'something': 'else'}})

Answer 4

from copy import deepcopy

d1 = {"admin": {"key1": "v2"}}
d2 = {"admin": {"key2": "v3"},
      "user": {"something": "else"}}

def merge(one, two):
    if isinstance(two, dict):
        result = deepcopy(one)
        for key, value in two.iteritems():
            if key in result and isinstance(result[key], dict):
                result[key] = merge(result[key], value)
            else:
                result[key] = deepcopy(value)
        return result
    return two

print merge(d1, d2)

输出

{'admin': {'key2': 'v3', 'key1': 'v2'}, 'user': {'something': 'else'}}

python：将两个嵌套字典与字典组合为顶级键的值

4 个答案: