我有这个多维词典：
a = {'a' : 'b', 'c' : {'d' : 'e'}}
并编写了简单的函数来压缩dict：
def __flatten(self, dictionary, level = []):
tmp_dict = {}
for key, val in dictionary.items():
if type(val) == dict:
tmp_dict.update(self.__flatten(val, level + [key]))
else:
tmp_dict['.'.join(level + [key])] = val
return tmp_dict
用dict a调用此函数后，我得到结果：
{'a' : 'b', 'c.d' : 'e'}
现在，在对这个扁平的字典做出一些指示之后，我需要从那个扁平的字母中构建新的多维字典。例如：
>> unflatten({'a' : 0, 'c.d' : 1))
{'a' : 0, 'c' : {'d' : 1}}
我唯一的问题是我没有功能unflatten :)
有人能帮忙吗？我不知道该怎么做。
修改
另一个例子：
{'a' : 'b', 'c.d.e.f.g.h.i.j.k.l.m.n.o.p.r.s.t.u.w' : 'z'}
应该在不平坦之后：
{'a': 'b', 'c': {'d': {'e': {'f': {'g': {'h': {'i': {'j': {'k': {'l': {'m': {'n': {'o': {'p': {'r': {'s': {'t': {'u': {'w': 'z'}}}}}}}}}}}}}}}}}}}
另一个：
{'a' : 'b', 'c.d' : 'z', 'c.e' : 1}
要：
{'a' : 'b', 'c' : {'d' : 'z', 'e' : 1}}
我知道，这大大增加了任务的难度。这就是为什么我有这个问题，并在几小时内找不到解决方案..
答案 0 :(得分：11)
def unflatten(dictionary):
resultDict = dict()
for key, value in dictionary.iteritems():
parts = key.split(".")
d = resultDict
for part in parts[:-1]:
if part not in d:
d[part] = dict()
d = d[part]
d[parts[-1]] = value
return resultDict
答案 1 :(得分：2)
from collections import defaultdict
def unflatten(d):
ret = defaultdict(dict)
for k,v in d.items():
k1,delim,k2 = k.partition('.')
if delim:
ret[k1].update({k2:v})
else:
ret[k1] = v
return ret
答案 2 :(得分：1)
作为粗略草稿（可以在变量名称选择中使用稍微改进，也许可以使用稳健性，但它适用于给出的示例）：
def unflatten(d):
result = {}
for k,v in d.iteritems():
if '.' in k:
k1, k2 = k.split('.', 1)
v = {k2: v}
k = k1
result[k] = v
return result
答案 3 :(得分：0)
这是一个利用Python 3.5+功能的功能，例如键入和解构分配。 Try the tests out on repl.it。
from typing import Any, Dict
def unflatten(
d: Dict[str, Any],
base: Dict[str, Any] = None,
) -> Dict[str, Any]:
"""Convert any keys containing dotted paths to nested dicts
>>> unflatten({'a': 12, 'b': 13, 'c': 14}) # no expansion
{'a': 12, 'b': 13, 'c': 14}
>>> unflatten({'a.b.c': 12}) # dotted path expansion
{'a': {'b': {'c': 12}}}
>>> unflatten({'a.b.c': 12, 'a': {'b.d': 13}}) # merging
{'a': {'b': {'c': 12, 'd': 13}}}
>>> unflatten({'a.b': 12, 'a': {'b': 13}}) # insertion-order overwrites
{'a': {'b': 13}}
>>> unflatten({'a': {}}) # insertion-order overwrites
{'a': {}}
"""
if base is None:
base = {}
for key, value in d.items():
root = base
###
# If a dotted path is encountered, create nested dicts for all but
# the last level, then change root to that last level, and key to
# the final key in the path.
#
# This allows one final setitem at the bottom of the loop.
#
if '.' in key:
*parts, key = key.split('.')
for part in parts:
root.setdefault(part, {})
root = root[part]
if isinstance(value, dict):
value = unflatten(value, root.get(key, {}))
root[key] = value
return base
答案 4 :(得分：0)
我一年前用 Python 2 和 3 编写，我在下面进行了改编。这是为了更容易检查给定的字典是否是更大字典的子集，而不管是以扁平形式还是脚手架形式提供的。
一个额外的功能：如果有连续的整数索引（如 0、1、2、3、4 等），这也会将它们转换回列表。
def unflatten_dictionary(field_dict):
field_dict = dict(field_dict)
new_field_dict = dict()
field_keys = list(field_dict)
field_keys.sort()
for each_key in field_keys:
field_value = field_dict[each_key]
processed_key = str(each_key)
current_key = None
current_subkey = None
for i in range(len(processed_key)):
if processed_key[i] == "[":
current_key = processed_key[:i]
start_subscript_index = i + 1
end_subscript_index = processed_key.index("]")
current_subkey = int(processed_key[start_subscript_index : end_subscript_index])
# reserve the remainder descendant keys to be processed later in a recursive call
if len(processed_key[end_subscript_index:]) > 1:
current_subkey = "{}.{}".format(current_subkey, processed_key[end_subscript_index + 2:])
break
# next child key is a dictionary
elif processed_key[i] == ".":
split_work = processed_key.split(".", 1)
if len(split_work) > 1:
current_key, current_subkey = split_work
else:
current_key = split_work[0]
break
if current_subkey is not None:
if current_key.isdigit():
current_key = int(current_key)
if current_key not in new_field_dict:
new_field_dict[current_key] = dict()
new_field_dict[current_key][current_subkey] = field_value
else:
new_field_dict[each_key] = field_value
# Recursively unflatten each dictionary on each depth before returning back to the caller.
all_digits = True
highest_digit = -1
for each_key, each_item in new_field_dict.items():
if isinstance(each_item, dict):
new_field_dict[each_key] = unflatten_dictionary(each_item)
# validate the keys can safely converted to a sequential list.
all_digits &= str(each_key).isdigit()
if all_digits:
next_digit = int(each_key)
if next_digit > highest_digit:
highest_digit = next_digit
# If all digits and can be sequential order, convert to list.
if all_digits and highest_digit == (len(new_field_dict) - 1):
digit_keys = list(new_field_dict)
digit_keys.sort()
new_list = []
for k in digit_keys:
i = int(k)
if len(new_list) <= i:
# Pre-populate missing list elements if the array index keys are out of order
# and the current element is ahead of the current length boundary.
while len(new_list) <= i:
new_list.append(None)
new_list[i] = new_field_dict[k]
new_field_dict = new_list
return new_field_dict
# Test
if __name__ == '__main__':
input_dict = {'a[0]': 1,
'a[1]': 10,
'a[2]': 5,
'b': 10,
'c.test.0': "hi",
'c.test.1': "bye",
"c.head.shoulders": "richard",
"c.head.knees": 'toes',
"z.trick.or[0]": "treat",
"z.trick.or[1]": "halloween",
"z.trick.and.then[0]": "he",
"z.trick.and.then[1]": "it",
"some[0].nested.field[0]": 42,
"some[0].nested.field[1]": 43,
"some[2].nested.field[0]": 44,
"mixed": {
"statement": "test",
"break[0]": True,
"break[1]": False,
}}
expected_dict = {'a': [1, 10, 5],
'b': 10,
'c': {
'test': ['hi', 'bye'],
'head': {
'shoulders': 'richard',
'knees' : 'toes'
}
},
'z': {
'trick': {
'or': ["treat", "halloween"],
'and': {
'then': ["he", "it"]
}
}
},
'some': {
0: {
'nested': {
'field': [42, 43]
}
},
2: {
'nested': {
'field': [44]
}
}
},
"mixed": {
"statement": "test",
"break": [True, False]
}}
# test
print("Input:")
print(input_dict)
print("====================================")
print("Output:")
actual_dict = unflatten_dictionary(input_dict)
print(actual_dict)
print("====================================")
print(f"Test passed? {expected_dict==actual_dict}")