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]

这里有一个使用生成器的简单方法:

def mergedicts(dict1, dict2):
    for k in set(dict1.keys()).union(dict2.keys()):
        if k in dict1 and k in dict2:
            if isinstance(dict1[k], dict) and isinstance(dict2[k], dict):
                yield (k, dict(mergedicts(dict1[k], dict2[k])))
            else:
                # If one of the values is not a dict, you can't continue merging it.
                # Value from second dict overrides one in first and we move on.
                yield (k, dict2[k])
                # Alternatively, replace this with exception raiser to alert you of value conflicts
        elif k in dict1:
            yield (k, dict1[k])
        else:
            yield (k, dict2[k])

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

print dict(mergedicts(dict1,dict2))

这个打印:

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

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

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

如果有人想要另一种方法来解决这个问题，这是我的解决方案。

优点:简洁、声明性和函数式风格(递归，没有突变)。

潜在缺点:这可能不是你想要的合并。查阅文档字符串以了解语义。

def deep_merge(a, b):
    """
    Merge two values, with `b` taking precedence over `a`.

    Semantics:
    - If either `a` or `b` is not a dictionary, `a` will be returned only if
      `b` is `None`. Otherwise `b` will be returned.
    - If both values are dictionaries, they are merged as follows:
        * Each key that is found only in `a` or only in `b` will be included in
          the output collection with its value intact.
        * For any key in common between `a` and `b`, the corresponding values
          will be merged with the same semantics.
    """
    if not isinstance(a, dict) or not isinstance(b, dict):
        return a if b is None else b
    else:
        # If we're here, both a and b must be dictionaries or subtypes thereof.

        # Compute set of all keys in both dictionaries.
        keys = set(a.keys()) | set(b.keys())

        # Build output dictionary, merging recursively values with common keys,
        # where `None` is used to mean the absence of a value.
        return {
            key: deep_merge(a.get(key), b.get(key))
            for key in keys
        }

当然，代码将取决于您解决合并冲突的规则。这里有一个版本，它可以接受任意数量的参数，并递归地将它们合并到任意深度，而不使用任何对象突变。它使用以下规则来解决合并冲突:

字典优先于非字典值({"foo":{…}}优先于{"foo": "bar"}) 后面的参数优先于前面的参数(如果按顺序合并{"a": 1}， {"a"， 2}和{"a": 3}，结果将是{"a": 3})

try:
    from collections import Mapping
except ImportError:
    Mapping = dict

def merge_dicts(*dicts):                                                            
    """                                                                             
    Return a new dictionary that is the result of merging the arguments together.   
    In case of conflicts, later arguments take precedence over earlier arguments.   
    """                                                                             
    updated = {}                                                                    
    # grab all keys                                                                 
    keys = set()                                                                    
    for d in dicts:                                                                 
        keys = keys.union(set(d))                                                   

    for key in keys:                                                                
        values = [d[key] for d in dicts if key in d]                                
        # which ones are mapping types? (aka dict)                                  
        maps = [value for value in values if isinstance(value, Mapping)]            
        if maps:                                                                    
            # if we have any mapping types, call recursively to merge them          
            updated[key] = merge_dicts(*maps)                                       
        else:                                                                       
            # otherwise, just grab the last value we have, since later arguments    
            # take precedence over earlier arguments                                
            updated[key] = values[-1]                                               
    return updated