我需要合并多个字典,这是我有例如:

dict1 = {1:{"a":{A}}, 2:{"b":{B}}}

dict2 = {2:{"c":{C}}, 3:{"d":{D}}}

A、B、C和D是树的叶子,比如{"info1":"value", "info2":"value2"}

字典的级别(深度)未知,可能是{2:{"c":{"z":{"y":{c}}}}}

在我的例子中,它表示一个目录/文件结构,节点是文档,叶子是文件。

我想将它们合并得到:

 dict3 = {1:{"a":{A}}, 2:{"b":{B},"c":{C}}, 3:{"d":{D}}}

我不确定如何用Python轻松做到这一点。


当前回答

下面的函数将b合并为a。

def mergedicts(a, b):
    for key in b:
        if isinstance(a.get(key), dict) or isinstance(b.get(key), dict):
            mergedicts(a[key], b[key])
        else:
            a[key] = b[key]
    return a

其他回答

这个简单的递归过程将一个字典合并到另一个字典,同时覆盖冲突的键:

#!/usr/bin/env python2.7

def merge_dicts(dict1, dict2):
    """ Recursively merges dict2 into dict1 """
    if not isinstance(dict1, dict) or not isinstance(dict2, dict):
        return dict2
    for k in dict2:
        if k in dict1:
            dict1[k] = merge_dicts(dict1[k], dict2[k])
        else:
            dict1[k] = dict2[k]
    return dict1

print (merge_dicts({1:{"a":"A"}, 2:{"b":"B"}}, {2:{"c":"C"}, 3:{"d":"D"}}))
print (merge_dicts({1:{"a":"A"}, 2:{"b":"B"}}, {1:{"a":"A"}, 2:{"b":"C"}}))

输出:

{1: {'a': 'A'}, 2: {'c': 'C', 'b': 'B'}, 3: {'d': 'D'}}
{1: {'a': 'A'}, 2: {'b': 'C'}}
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}}

基于@andrew cooke。这个版本处理字典的嵌套列表,还允许选项更新值

def merge(a, b, path=None, update=True):
    "http://stackoverflow.com/questions/7204805/python-dictionaries-of-dictionaries-merge"
    "merges b into a"
    if path is None: path = []
    for key in b:
        if key in a:
            if isinstance(a[key], dict) and isinstance(b[key], dict):
                merge(a[key], b[key], path + [str(key)])
            elif a[key] == b[key]:
                pass # same leaf value
            elif isinstance(a[key], list) and isinstance(b[key], list):
                for idx, val in enumerate(b[key]):
                    a[key][idx] = merge(a[key][idx], b[key][idx], path + [str(key), str(idx)], update=update)
            elif update:
                a[key] = b[key]
            else:
                raise Exception('Conflict at %s' % '.'.join(path + [str(key)]))
        else:
            a[key] = b[key]
    return a

我还没有对此进行广泛的测试,因此,鼓励您的反馈。

from collections import defaultdict

dict1 = defaultdict(list)

dict2= defaultdict(list)

dict3= defaultdict(list)


dict1= dict(zip(Keys[ ],values[ ]))

dict2 = dict(zip(Keys[ ],values[ ]))


def mergeDict(dict1, dict2):

    dict3 = {**dict1, **dict2}

    for key, value in dict3.items():

        if key in dict1 and key in dict2:

           dict3[key] = [value , dict1[key]]

    return dict3

dict3 = mergeDict(dict1, dict2)

#sort keys alphabetically.

dict3.keys()

合并两个字典并添加公共键的值

这应该有助于将所有项从dict2合并到dict1:

for item in dict2:
    if item in dict1:
        for leaf in dict2[item]:
            dict1[item][leaf] = dict2[item][leaf]
    else:
        dict1[item] = dict2[item]

请测试一下,告诉我们这是否是你想要的。

编辑:

上述解决方案只合并了一个级别,但正确地解决了op给出的例子。如果合并多个级别,应该使用递归。