from collections import defaultdict
from itertools import chain

class DictHelper:

@staticmethod
def merge_dictionaries(*dictionaries, override=True):
    merged_dict = defaultdict(set)
    all_unique_keys = set(chain(*[list(dictionary.keys()) for dictionary in dictionaries]))  # Build a set using all dict keys
    for key in all_unique_keys:
        keys_value_type = list(set(filter(lambda obj_type: obj_type != type(None), [type(dictionary.get(key, None)) for dictionary in dictionaries])))
        # Establish the object type for each key, return None if key is not present in dict and remove None from final result
        if len(keys_value_type) != 1:
            raise Exception("Different objects type for same key: {keys_value_type}".format(keys_value_type=keys_value_type))

        if keys_value_type[0] == list:
            values = list(chain(*[dictionary.get(key, []) for dictionary in dictionaries]))  # Extract the value for each key
            merged_dict[key].update(values)

        elif keys_value_type[0] == dict:
            # Extract all dictionaries by key and enter in recursion
            dicts_to_merge = list(filter(lambda obj: obj != None, [dictionary.get(key, None) for dictionary in dictionaries]))
            merged_dict[key] = DictHelper.merge_dictionaries(*dicts_to_merge)

        else:
            # if override => get value from last dictionary else make a list of all values
            values = list(filter(lambda obj: obj != None, [dictionary.get(key, None) for dictionary in dictionaries]))
            merged_dict[key] = values[-1] if override else values

    return dict(merged_dict)



if __name__ == '__main__':
  d1 = {'aaaaaaaaa': ['to short', 'to long'], 'bbbbb': ['to short', 'to long'], "cccccc": ["the is a test"]}
  d2 = {'aaaaaaaaa': ['field is not a bool'], 'bbbbb': ['field is not a bool']}
  d3 = {'aaaaaaaaa': ['filed is not a string', "to short"], 'bbbbb': ['field is not an integer']}
  print(DictHelper.merge_dictionaries(d1, d2, d3))

  d4 = {"a": {"x": 1, "y": 2, "z": 3, "d": {"x1": 10}}}
  d5 = {"a": {"x": 10, "y": 20, "d": {"x2": 20}}}
  print(DictHelper.merge_dictionaries(d4, d5))

输出:

{'bbbbb': {'to long', 'field is not an integer', 'to short', 'field is not a bool'}, 
'aaaaaaaaa': {'to long', 'to short', 'filed is not a string', 'field is not a bool'}, 
'cccccc': {'the is a test'}}

{'a': {'y': 20, 'd': {'x1': 10, 'x2': 20}, 'z': 3, 'x': 10}}

from collections import defaultdict
from itertools import chain

class DictHelper:

@staticmethod
def merge_dictionaries(*dictionaries, override=True):
    merged_dict = defaultdict(set)
    all_unique_keys = set(chain(*[list(dictionary.keys()) for dictionary in dictionaries]))  # Build a set using all dict keys
    for key in all_unique_keys:
        keys_value_type = list(set(filter(lambda obj_type: obj_type != type(None), [type(dictionary.get(key, None)) for dictionary in dictionaries])))
        # Establish the object type for each key, return None if key is not present in dict and remove None from final result
        if len(keys_value_type) != 1:
            raise Exception("Different objects type for same key: {keys_value_type}".format(keys_value_type=keys_value_type))

        if keys_value_type[0] == list:
            values = list(chain(*[dictionary.get(key, []) for dictionary in dictionaries]))  # Extract the value for each key
            merged_dict[key].update(values)

        elif keys_value_type[0] == dict:
            # Extract all dictionaries by key and enter in recursion
            dicts_to_merge = list(filter(lambda obj: obj != None, [dictionary.get(key, None) for dictionary in dictionaries]))
            merged_dict[key] = DictHelper.merge_dictionaries(*dicts_to_merge)

        else:
            # if override => get value from last dictionary else make a list of all values
            values = list(filter(lambda obj: obj != None, [dictionary.get(key, None) for dictionary in dictionaries]))
            merged_dict[key] = values[-1] if override else values

    return dict(merged_dict)



if __name__ == '__main__':
  d1 = {'aaaaaaaaa': ['to short', 'to long'], 'bbbbb': ['to short', 'to long'], "cccccc": ["the is a test"]}
  d2 = {'aaaaaaaaa': ['field is not a bool'], 'bbbbb': ['field is not a bool']}
  d3 = {'aaaaaaaaa': ['filed is not a string', "to short"], 'bbbbb': ['field is not an integer']}
  print(DictHelper.merge_dictionaries(d1, d2, d3))

  d4 = {"a": {"x": 1, "y": 2, "z": 3, "d": {"x1": 10}}}
  d5 = {"a": {"x": 10, "y": 20, "d": {"x2": 20}}}
  print(DictHelper.merge_dictionaries(d4, d5))

输出:

{'bbbbb': {'to long', 'field is not an integer', 'to short', 'field is not a bool'}, 
'aaaaaaaaa': {'to long', 'to short', 'filed is not a string', 'field is not a bool'}, 
'cccccc': {'the is a test'}}

{'a': {'y': 20, 'd': {'x1': 10, 'x2': 20}, 'z': 3, 'x': 10}}

Short-n-sweet:

from collections.abc import MutableMapping as Map

def nested_update(d, v):
"""
Nested update of dict-like 'd' with dict-like 'v'.
"""

for key in v:
    if key in d and isinstance(d[key], Map) and isinstance(v[key], Map):
        nested_update(d[key], v[key])
    else:
        d[key] = v[key]

这类似于(并且构建在)Python的字典上。更新方法。它返回None(如果你喜欢，你总是可以添加返回d)，因为它在原地更新dict d。v中的键将覆盖d中任何现有的键(它不会尝试解释字典的内容)。

它也适用于其他(“类字典”)映射。

以下是来自@andrew cooke的回答。 它以更好的方式处理嵌套列表。

def deep_merge_lists(original, incoming):
    """
    Deep merge two lists. Modifies original.
    Recursively call deep merge on each correlated element of list. 
    If item type in both elements are
     a. dict: Call deep_merge_dicts on both values.
     b. list: Recursively call deep_merge_lists on both values.
     c. any other type: Value is overridden.
     d. conflicting types: Value is overridden.

    If length of incoming list is more that of original then extra values are appended.
    """
    common_length = min(len(original), len(incoming))
    for idx in range(common_length):
        if isinstance(original[idx], dict) and isinstance(incoming[idx], dict):
            deep_merge_dicts(original[idx], incoming[idx])

        elif isinstance(original[idx], list) and isinstance(incoming[idx], list):
            deep_merge_lists(original[idx], incoming[idx])

        else:
            original[idx] = incoming[idx]

    for idx in range(common_length, len(incoming)):
        original.append(incoming[idx])


def deep_merge_dicts(original, incoming):
    """
    Deep merge two dictionaries. Modifies original.
    For key conflicts if both values are:
     a. dict: Recursively call deep_merge_dicts on both values.
     b. list: Call deep_merge_lists on both values.
     c. any other type: Value is overridden.
     d. conflicting types: Value is overridden.

    """
    for key in incoming:
        if key in original:
            if isinstance(original[key], dict) and isinstance(incoming[key], dict):
                deep_merge_dicts(original[key], incoming[key])

            elif isinstance(original[key], list) and isinstance(incoming[key], list):
                deep_merge_lists(original[key], incoming[key])

            else:
                original[key] = incoming[key]
        else:
            original[key] = incoming[key]

在不影响输入字典的情况下返回一个合并。

def _merge_dicts(dictA: Dict = {}, dictB: Dict = {}) -> Dict:
    # it suffices to pass as an argument a clone of `dictA`
    return _merge_dicts_aux(dictA, dictB, copy(dictA))


def _merge_dicts_aux(dictA: Dict = {}, dictB: Dict = {}, result: Dict = {}, path: List[str] = None) -> Dict:

    # conflict path, None if none
    if path is None:
        path = []

    for key in dictB:

        # if the key doesn't exist in A, add the B element to A
        if key not in dictA:
            result[key] = dictB[key]

        else:
            # if the key value is a dict, both in A and in B, merge the dicts
            if isinstance(dictA[key], dict) and isinstance(dictB[key], dict):
                _merge_dicts_aux(dictA[key], dictB[key], result[key], path + [str(key)])

            # if the key value is the same in A and in B, ignore
            elif dictA[key] == dictB[key]:
                pass

            # if the key value differs in A and in B, raise error
            else:
                err: str = f"Conflict at {'.'.join(path + [str(key)])}"
                raise Exception(err)

    return result

灵感来自@andrew cooke的解决方案

嘿，我也有同样的问题，但我想出了一个解决方案，我会把它贴在这里，以防它对其他人也有用，基本上合并嵌套字典和添加值，对我来说，我需要计算一些概率，所以这一个工作得很好:

#used to copy a nested dict to a nested dict
def deepupdate(target, src):
    for k, v in src.items():
        if k in target:
            for k2, v2 in src[k].items():
                if k2 in target[k]:
                    target[k][k2]+=v2
                else:
                    target[k][k2] = v2
        else:
            target[k] = copy.deepcopy(v)

通过使用上述方法，我们可以合并:

目标={6 6:{“63”:1},“63,4:{4 4:1},4,4:{“4 3”:1},“63”:{63,4:1}}

src ={5 4:{4 4: 1}, 5、5:{“5、4”:1},4,4:{“4 3”:1}}

这将变成: {', 5 ':{“5、4”:1},“5、4”:{4 4:1},“6 6”:{“63”:1},“63,4:{4 4:1},4,4:{“4 3”:2},“63”:{63,4:1}}

还要注意这里的变化:

目标={6 6:{“63”:1},“63”:{63,4:1},4,4:{“4 3”:1},“63,4:{4 4:1}}

src ={5 4:{4 4: 1},“4 3”:{“3、4”:1},4,4:{“4、9”:1},3、4:{4 4:1},5、5:{“5、4”:1}}

merge =可不,‘五,四’:可不,‘4、4’:一个出于美观,‘4、三’:可不,‘3、4”:一个有关联,“6、63”:可不,‘63倍或四’:一个出于美观,‘5、5:可不,' 5、4”:一个有关联,“6、6”:可不,‘6、63’:一个出于美观,‘3,4‘:可不,‘四,四’:一个出于美观,‘63倍或四’一‘::可不,‘四,四出于美观,‘4,4:可不,’‘四,三’:一,‘4 9,‘:一个出于美观出于美观。

别忘了还添加导入copy:

import copy