Python:转换深度嵌套字典或数组中的类型

from datetime import datetime from bson import ObjectId def clean_dict_helper(d): if not d: return None if isinstance(d, list): # For those db functions which return list if len(d) == 0: return [] return [clean_dict_helper(x) for x in d] for k, v in d.items(): if isinstance(v, dict): v = clean_dict_helper(v) d.update({k: v}) else: if isinstance(v, ObjectId): v = str(v) elif isinstance(v, datetime): v = str(v) d.update({k: v}) return d test_case = {'id': bson.ObjectId("5e126ddf276ab18820e00ddf"), 'arr': [{'value': 4, 'id': bson.ObjectId("5e126ddf276ab18820e00ddf")}]} result = clean_dict_helper(test_case) print(result) >>> {'arr': [{'id': ObjectId('5e126ddf276ab18820e00ddf'), 'value': 4}], 'id': '5e126ddf276ab18820e00ddf'}

我认为如果将转换放在开头,允许递归调用传递 ObjectId 并将转换后的字符串直接与函数一起返回,而不是尝试在函数的循环内进行转换。然后您可以将该值传递到递归中,并将字符串取回。

类似于:

from datetime import datetime

def clean_dict_helper(d):
    if isinstance(d, ObjectId) or isinstance(d, datetime):
        return str(d)

    if isinstance(d, list):  # For those db functions which return list
        return [clean_dict_helper(x) for x in d]

    if isinstance(d, dict):
        for k, v in d.items():
            d.update({k: clean_dict_helper(v)})

    # return anything else, like a string or number
    return d

打电话:

clean_dict_helper({
    1: {
        1: ObjectId('5e0e83a6d0fbe7238c960ea0'),
        2: [ObjectId('5e0e83a6d0fbe7238c960ea0'), ObjectId('5e0e83a6d0fbe7238c960ea0')],
        3: datetime(2020, 1, 1),
        4: [{1: ObjectId('5e0e83a6d0fbe7238c960ea0')}],
        5: 'test'
    },
    2: [ObjectId('5e0e83a6d0fbe7238c960ea0'), ObjectId('5e0e83a6d0fbe7238c960ea0')]
})

然后将返回:

{1: {1: '5e0e83a6d0fbe7238c960ea0',
  2: ['5e0e83a6d0fbe7238c960ea0', '5e0e83a6d0fbe7238c960ea0'],
  3: '2020-01-01 00:00:00',
  4: [{1: '5e0e83a6d0fbe7238c960ea0'}],
  5: 'test'},
 2: ['5e0e83a6d0fbe7238c960ea0', '5e0e83a6d0fbe7238c960ea0']}

另外,由于您正在创建一个具有理解力的新列表,因此返回一个新的dict可能是有意义的,而不是用如下内容对传入值进行变异:

    if isinstance(d, dict):
        return {k:clean_dict_helper(v) for k, v in d.items() }