python，列表嵌套字典，获取字典(整个)数据

  • Z时代
  • 2024-02-26
  • 分类：IT

python，列表嵌套字典，获取字典(整个)数据

1、我有套列表嵌套字典的数据，我想把列表中的内容取出来，并且携带','。
2、数据模板如下：

list1 = [
  {
    "": "0",
    "id": "4d834ba0525f451d88767696774d3b4a",
    "name": "\u5347\u7ea7\u9879\u76ee005-OPT\u95ee\u9898\u4ee3\u7801",
    "project_start_date": "",
    "project_end_date": "2022-01-13",
    "data_start_date": "2021-11-02",
    "data_end_date": "2021-11-02",
    "status": "3",
    "create_user_id": "-200",
    "create_date": "2022-01-13 15:52:19",
    "update_date": "2022-01-13 16:08:53",
    "type": "1",
    "take_user_id": "-200",
    "take_date": "2022-01-13 16:07:01",
    "project_name_version": "0"
  },
  {
    "": "1",
    "id": "4d834ba0525f451d88767696774d3b4a",
    "name": "\u5347\u7ea7\u9879\u76ee005-OPT\u95ee\u9898\u4ee3\u7801",
    "project_start_date": "",
    "project_end_date": "2022-01-13",
    "data_start_date": "2021-11-02",
    "data_end_date": "2021-11-02",
    "status": "1",
    "create_user_id": "-200",
    "create_date": "2022-01-13 15:52:19",
    "update_date": "2022-01-13 16:08:53",
    "type": "2",
    "take_user_id": "-200",
    "take_date": "2022-01-13 16:07:01",
    "project_name_version": "0"
  },
  {
    "": "1",
    "id": "4d834ba0525f451d88767696774d3b4a",
    "name": "\u5347\u7ea7\u9879\u76ee005-OPT\u95ee\u9898\u4ee3\u7801",
    "project_start_date": "",
    "project_end_date": "2022-01-13",
    "data_start_date": "2021-11-02",
    "data_end_date": "2021-11-02",
    "status": "1",
    "create_user_id": "-200",
    "create_date": "2022-01-13 15:52:19",
    "update_date": "2022-01-13 16:08:53",
    "type": "2",
    "take_user_id": "-200",
    "take_date": "2022-01-13 16:07:01",
    "project_name_version": "0"
  }
]

我想获得的效果是取出里边的字典，因为后续需要字典值作为入参，引入后续流程。

考虑过用列表遍历，但是发现遍历后。虽然能获得数据，但是就变成了str，并且内容被拆分了

回答：

我没有看明白你需要的是什么数据
遍历列表得到的是dict对象,而不应该是str

python">list1 = [{'id':0,'name':'jack'},{'id':1,'name':'ma'}]
for i in list1:
    print(i)

输出为:

{'id': 0, 'name': 'jack'}
{'id': 1, 'name': 'ma'}

你要确定你的数据是list嵌套dict.而不是str类型的(json).你type(list1),type(list1[0])确定数据类型.
传递字典

def func(kw):
    for key,value in kw.items():
        print('key:%s---value:%s' %(key,value))
for i in list1:
     func(i)
---------
输出:
key:id---value:0
key:name---value:jack
key:id---value:1
key:name---value:ma

解包:

def unpack_func(id,name):
    print('id:%s---name:%s' %(id,name))
for i in list1:
    unpack_func(**i)
----------
输出:
id:0---name:jack
id:1---name:ma

回答：

for d in list1:
    print(type(d)) # 应为dict（前前后后加上的一大堆我忘了）
    print(d)

搞不懂就调试

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