我有一个迭代的解决方案-工作得更好的大字典&很多(例如jsons等):

import collections


def merge_dict_with_subdicts(dict1: dict, dict2: dict) -> dict:
    """
    similar behaviour to builtin dict.update - but knows how to handle nested dicts
    """
    q = collections.deque([(dict1, dict2)])
    while len(q) > 0:
        d1, d2 = q.pop()
        for k, v in d2.items():
            if k in d1 and isinstance(d1[k], dict) and isinstance(v, dict):
                q.append((d1[k], v))
            else:
                d1[k] = v

    return dict1

注意，这将使用d2中的值来覆盖d1，以防它们都不是字典。(与python的dict.update()相同)

一些测试:

def test_deep_update():
    d = dict()
    merge_dict_with_subdicts(d, {"a": 4})
    assert d == {"a": 4}

    new_dict = {
        "b": {
            "c": {
                "d": 6
            }
        }
    }
    merge_dict_with_subdicts(d, new_dict)
    assert d == {
        "a": 4,
        "b": {
            "c": {
                "d": 6
            }
        }
    }

    new_dict = {
        "a": 3,
        "b": {
            "f": 7
        }
    }
    merge_dict_with_subdicts(d, new_dict)
    assert d == {
        "a": 3,
        "b": {
            "c": {
                "d": 6
            },
            "f": 7
        }
    }

    # test a case where one of the dicts has dict as value and the other has something else
    new_dict = {
        'a': {
            'b': 4
        }
    }
    merge_dict_with_subdicts(d, new_dict)
    assert d['a']['b'] == 4

我已经测试了大约1200个字典——这种方法花了0.4秒，而递归的解决方案花了2.5秒。

我一直在测试你的解决方案，并决定在我的项目中使用这个:

def mergedicts(dict1, dict2, conflict, no_conflict):
    for k in set(dict1.keys()).union(dict2.keys()):
        if k in dict1 and k in dict2:
            yield (k, conflict(dict1[k], dict2[k]))
        elif k in dict1:
            yield (k, no_conflict(dict1[k]))
        else:
            yield (k, no_conflict(dict2[k]))

dict1 = {1:{"a":"A"}, 2:{"b":"B"}}
dict2 = {2:{"c":"C"}, 3:{"d":"D"}}

#this helper function allows for recursion and the use of reduce
def f2(x, y):
    return dict(mergedicts(x, y, f2, lambda x: x))

print dict(mergedicts(dict1, dict2, f2, lambda x: x))
print dict(reduce(f2, [dict1, dict2]))

将函数作为参数传递是将jterrace解决方案扩展为所有其他递归解决方案的关键。

andrew cookes的回答有一个小问题:在某些情况下，当你修改返回的dict时，它会修改第二个参数b。具体来说是因为这句话:

if key in a:
    ...
else:
    a[key] = b[key]

如果b[key]是一个字典，它将被简单地赋给a，这意味着对该字典的任何后续修改将同时影响a和b。

a={}
b={'1':{'2':'b'}}
c={'1':{'3':'c'}}
merge(merge(a,b), c) # {'1': {'3': 'c', '2': 'b'}}
a # {'1': {'3': 'c', '2': 'b'}} (as expected)
b # {'1': {'3': 'c', '2': 'b'}} <----
c # {'1': {'3': 'c'}} (unmodified)

为了解决这个问题，这一行必须用这个替换:

if isinstance(b[key], dict):
    a[key] = clone_dict(b[key])
else:
    a[key] = b[key]

其中clone_dict为:

def clone_dict(obj):
    clone = {}
    for key, value in obj.iteritems():
        if isinstance(value, dict):
            clone[key] = clone_dict(value)
        else:
            clone[key] = value
    return

不动。这显然没有考虑到list, set和其他东西，但我希望它说明了合并字典时的陷阱。

为了完整起见，这里是我的版本，在那里你可以传递它多个字典:

def merge_dicts(*args):
    def clone_dict(obj):
        clone = {}
        for key, value in obj.iteritems():
            if isinstance(value, dict):
                clone[key] = clone_dict(value)
            else:
                clone[key] = value
        return

    def merge(a, b, 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
                else:
                    raise Exception('Conflict at `{path}\''.format(path='.'.join(path + [str(key)])))
            else:
                if isinstance(b[key], dict):
                    a[key] = clone_dict(b[key])
                else:
                    a[key] = b[key]
        return a
    return reduce(merge, args, {})

这个问题的一个问题是字典的值可以是任意复杂的数据块。基于这些和其他答案，我得出了以下代码:

class YamlReaderError(Exception):
    pass

def data_merge(a, b):
    """merges b into a and return merged result

    NOTE: tuples and arbitrary objects are not handled as it is totally ambiguous what should happen"""
    key = None
    # ## debug output
    # sys.stderr.write("DEBUG: %s to %s\n" %(b,a))
    try:
        if a is None or isinstance(a, str) or isinstance(a, unicode) or isinstance(a, int) or isinstance(a, long) or isinstance(a, float):
            # border case for first run or if a is a primitive
            a = b
        elif isinstance(a, list):
            # lists can be only appended
            if isinstance(b, list):
                # merge lists
                a.extend(b)
            else:
                # append to list
                a.append(b)
        elif isinstance(a, dict):
            # dicts must be merged
            if isinstance(b, dict):
                for key in b:
                    if key in a:
                        a[key] = data_merge(a[key], b[key])
                    else:
                        a[key] = b[key]
            else:
                raise YamlReaderError('Cannot merge non-dict "%s" into dict "%s"' % (b, a))
        else:
            raise YamlReaderError('NOT IMPLEMENTED "%s" into "%s"' % (b, a))
    except TypeError, e:
        raise YamlReaderError('TypeError "%s" in key "%s" when merging "%s" into "%s"' % (e, key, b, a))
    return a

我的用例是合并YAML文件，其中我只需要处理可能的数据类型的子集。因此我可以忽略元组和其他对象。对我来说，合理的合并逻辑意味着

取代标量 添加列表 通过添加缺失键和更新现有键来合并字典

其他任何事情和不可预见的事情都会导致错误。

由于dictviews支持集合操作，我能够极大地简化jterrace的答案。

def merge(dict1, dict2):
    for k in dict1.keys() - dict2.keys():
        yield (k, dict1[k])

    for k in dict2.keys() - dict1.keys():
        yield (k, dict2[k])

    for k in dict1.keys() & dict2.keys():
        yield (k, dict(merge(dict1[k], dict2[k])))

任何将dict与非dict(技术上讲，一个带有'keys'方法的对象和一个没有'keys'方法的对象)组合的尝试都会引发AttributeError。这包括对函数的初始调用和递归调用。这正是我想要的，所以我离开了。您可以很容易地捕获递归调用抛出的AttributeErrors，然后生成您想要的任何值。

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'}}