这个用python怎么写呢

b、查找如下字典中所有的_id，输出结果为字典
输入：
{ "Result":{ "FailureCode":0,"StatusCode":0,"UserInfo":{ "EmployeeID":1214, "EnterpriseAccount":"158833" } }, "Value":{ "describe":{ "tenant_id":"158833", "api_name":"AccountObj", "_id":"5a320c9c2848bdbc06a58a88", "fields":{ "lock_rule":{ "label":"锁定规则", "type":"lock_rule", "is_required":false, "api_name":"lock_rule", "_id":"5f7301970ea8660001c8ebc3" } "UDMoney1__c":{ "type":"currency", "label":"成交金额", "api_name":"UDMoney1__c", "_id":"5d27f2a22848bdfa3adb4c32" } }, "isHaveDetailObj":true, "isInApprovalWhiteList":false, "supportTag":true } }}

输出：
{"Value._id":"5a320c9c2848bdbc06a58a88","Value.fields.lock_rule._id":"5f7301970ea8660001c8ebc3","Value.fields.lock_rule.UDMoney1__c":"5d27f2a22848bdfa3adb4c32"
c、对给定的两个字典dist1和字典dist2进行如下处理，输出字典dist1中存在，字典dist2中不存在的key，并输出字典dist2中存在，字典dist1中不存在的key。
dist1={"layoutVersion":"V3","objectDataId":"9b5302ffe1394cafb1f5b1c3b53e803c","objectDescribeApiName":"AccountObj","fromRecycleBin":false,"management":false,"serializeEmpty1":false}
dist2={"layoutVersion2":"V3","objectDataId":"9b5302ffe1394cafb1f5b1c3b53e803c","objectDescribeApiName":"AccountObj","fromRecycleBin":false,"management":false,"serializeEmpty":false}

你题目b应该用递归,代码如下：（如有帮助，望采纳！谢谢! 点击我这个回答右上方的【采纳】按钮）

import json
jsonstr='''{
    "Result":{
        "FailureCode":0,
        "StatusCode":0,
        "UserInfo":{
            "EmployeeID":1214,
            "EnterpriseAccount":"158833"
        }
    },
    "Value":{
        "describe":{
            "tenant_id":"158833",
            "api_name":"AccountObj",
            "_id":"5a320c9c2848bdbc06a58a88",
            "fields":{
                "lock_rule":{
                    "label":"锁定规则",
                    "type":"lock_rule",
                    "is_required":false,
                    "api_name":"lock_rule",
                    "_id":"5f7301970ea8660001c8ebc3"
                },
                "UDMoney1__c":{
                    "type":"currency",
                    "label":"成交金额",
                    "api_name":"UDMoney1__c",
                    "_id":"5d27f2a22848bdfa3adb4c32"
                }
            },
            "isHaveDetailObj":true,
            "isInApprovalWhiteList":false,
            "supportTag":true
        }
    }
}'''

dict1=json.loads(jsonstr)
dict2={}
def op(dic,path):
    for k,v in dic.items():
        if k=="_id":
            dict2[path+k] = v
        if type(v)==dict:
            op(v,path+k+".")
op(dict1,"")
print(dict2)

你题目C是用set集合，求差集：

dist1={"layoutVersion":"V3","objectDataId":"9b5302ffe1394cafb1f5b1c3b53e803c","objectDescribeApiName":"AccountObj","fromRecycleBin":False,"management":False,"serializeEmpty1":False}
dist2={"layoutVersion2":"V3","objectDataId":"9b5302ffe1394cafb1f5b1c3b53e803c","objectDescribeApiName":"AccountObj","fromRecycleBin":False,"management":False,"serializeEmpty":False}

s1=set(dist1)
s2=set(dist2)
print("字典dist1中存在，字典dist2中不存在的key:", s1-s2)
print("字典dist2中存在，字典dist1中不存在的key:", s2-s1)

参考如下文档。
https://www.cnblogs.com/paranoia/p/6178665.html