关于python实现json/字典数据中所有key路径拼接组合问题

import jmespath
 
data = {
    "id": 1,
    "name": '张三',
    "age": 25,
    "grade": {
        "Chinese": 99,
        "Math": 100
    },
    "records": [
        {
            "Chinese": 98,
            "Math": 99
        },
        {
            "Chinese": 97,
            "Math": 97
        }
    ]
}
 
# 方法一：
# 下面写法是最笨的方法，当迭代次数非常多时，要写好多循环，代码量也增加了
my_list = []
for key, value in data.items():
    my_list.append(key)
    if type(value) is dict:
        for key2, value2 in data[key].items():
            my_list.append(key + '.' + key2)
print(my_list)
 
 
# 方法二（递归实现）：
def get_keys_path(dict_data, enable_index=None):
    def dfs(dict_data, parent=''):
        if not isinstance(dict_data, dict):
            return
        for key, value in dict_data.items():
            if f"{key}.{value}" not in tmp_list:
                tmp_list.append(f"{parent}.{key}")
            if isinstance(value, dict):
                dfs(value, parent=f"{parent}.{key}")
            elif isinstance(value, list):
                num = 0
                for i in value:
                    if enable_index:
                        dfs(i, parent=f"{parent}.{key}[{num}]")
                        # dfs(i, parent=f"{parent}.{key}.{num}")
                    else:
                        dfs(i, parent=f"{parent}.{key}")
                    num += 1
            else:
                pass
 
    tmp_list = []
    dfs(dict_data, parent='')
    if enable_index:
        return [i[1:] for i in tmp_list]
    else:
        return [i[1:] for i in list(set(tmp_list))]
 
 
# 获取key路径组合
print(get_keys_path(data, 1))
 
# 获取key路径值
print(jmespath.search('records[1].Chinese', data))