Question

我正在寻找一种将多个字典相互合并的方法，这些字典也包含嵌套的字典。嵌套字典的数量不是静态的，而是动态的。

最后的dict应该包含所有dict及其值的总和：

COUNTRY1 = {'a': {'X': 10, 'Y': 18, 'Z': 17}, 'b': {'AA':{'AAx':45,'AAy':22},'BB':{'BBx':45,'BBy':22}}, 'c': 100}
COUNTRY2 = {'a': {'U': 12, 'V': 34, 'W': 23}, 'b': {'AA':{'AAz':23,'AAa':26},'BB':{'BBz':11,'BBa':15}}, 'c': 115}
COUNTRY3 = {'a': {'Y': 15, 'Z': 14, 'X': 12}, 'b': {'AA':{'AAx':45,'AAz':22},'BB':{'BBy':45,'BBz':22}}, 'c': 232}

# After merging the dictionaries the result should look like:
ALL
>>> {'a': {'X': 22, 'Y': 33, 'Z': 31, 'U': 12, 'V': 34, 'W': 23}, 'b': {'AA':{'AAx':90,'AAy':22,'AAz':45,'AAa':26},'BB':{'BBx':45,'BBy':67, 'BBz':33,'BBa':15}}, 'c': 447}

我尝试了以下代码，该代码允许嵌套字典最多为3个嵌套字典。不幸的是，代码没有达到我的期望。因此，它看起来不是很干净，我觉得可以通过递归函数来完成，但是我找不到解决方法。

COUNTRIES = ['COUNTRY1','COUNTRY2', 'COUNTRY3']
ALL = {}
for COUNTRY_CODE in COUNTRIES:

    COUNTRY = pickle.load(open(COUNTRY_CODE+".p", "rb"))
    keys = COUNTRY.keys()
    for key in keys:
        try:
            keys2 = COUNTRY[key].keys()
            print(key, keys2)

            for key2 in keys2:
                try:
                    keys3 = COUNTRY[key][key2].keys()
                    print(key2, keys3)

                    for key3 in keys3:
                        try:
                            keys4 = COUNTRY[key][key2][key3].keys()
                            print(key3, keys4)
                        except:
                            print(key3, "NO KEY3")
                            if not key3 in ALL[key][key2]:
                                ALL[key][key2][key3] = COUNTRY[key][key2][key3]
                            else:
                                ALL[key][key2][key3] =+ COUNTRY[key][key2][key3]

                except:
                    print(key2, "NO KEY2")
                    if not key2 in ALL[key]:
                        ALL[key][key2] = COUNTRY[key][key2]
                    else:
                        ALL[key][key2] =+ COUNTRY[key][key2]

        except:
            print(key, "NO KEY")
            if not key in ALL:
                ALL[key] = COUNTRY[key]
            else:
                ALL[key] =+ COUNTRY[key]

print(ALL)

Answer 1

问题是您需要根据值的类型确定如何处理字典键。基本思想是：

输入是一对字典，输出是和字典
遍历两个输入词典
如果值是字典，则递归
如果值是数字，请将其添加到其他数字

这很容易理解：

def add_dicts(d1, d2):
    def sum(v1, v2):
        if v2 is None:
            return v1
        try:
            return v1 + v2
        except TypeError:
            return add_dicts(v1, v2)
    result = d2.copy()
    result.update({k: sum(v, d2.get(k)) for k, v in d1.items()})
    return result

该副本可确保将d2中所有不在d1中的键简单地复制过来。

您现在可以总结如下：

ALL = add_dicts(add_dicts(COUNTRY1, COUNTRY2), COUNTRY3)

通常，您可以使用functools.reduce来执行不确定数量的字典：

dicts = [COUNTRY1, COUNTRY2, COUNTRY3]
ALL = reduce(add_dicts, dicts)

Answer 2

执行以下两个功能：

def cal_sum(lst):
    final_dict = dict()
    for l in lst:
        sum(final_dict,l)
    return final_dict

def sum(final_dict,iter_dict):
    for k, v in iter_dict.items():
        if isinstance(v, dict):
            sum(final_dict.setdefault(k, dict()), v)
        elif isinstance(v, int):
            final_dict[k] = final_dict.get(k, 0) + v

按如下所示调用上面的代码会产生所需的输出：

>>> print(cal_sum([COUNTRY1, COUNTRY2, COUNTRY3]))
{'a': {'U': 12, 'W': 23, 'V': 34, 'Y': 33, 'X': 22, 'Z': 31}, 'c': 447, 'b': {'AA': {'AAa': 26, 'AAy': 22, 'AAx': 90, 'AAz': 45}, 'BB': {'BBa': 15, 'BBz': 33, 'BBy': 67, 'BBx': 45}}}

合并命令和和值的命令

2 个答案: