在Python中,我的JSON格式如下:
lists = [
{
"id": 1,
"data": {
"id": "1",
"title": "Title1",
"list_category": "green",
"list_group": ["foo", "bar"],
"tag_Number": ["One","Three","Four","Seven"],
"tag_ABCD": ["B", "C", "D", "E"]
}
},
{
"id": 2,
"data": {
"id": "32",
"title": "Title32",
"list_category": "blue",
"list_group": ["foo2", "bar2"],
"tag_Number": ["One","Three","Four","Seven"],
"tag_ABCD": ["B", "C", "D", "E"]
},
...
我想编写不同的功能以从这些列表中提取不同种类的数据。例如,要提取标题和ID,我编写了以下函数:
def getTitle():
index_list = []
index_dict = dict()
for x in checklists:
val = ast.literal_eval(x["data"])
Index = dict(id= x['id'], title = val["title"])
index_dict[x['id']]=Index
return index_dict
我的问题是,当我尝试提取其他值(例如tag_ABCD
)时,我尝试通过以下方式更改函数:
def getTag():
index_list = []
index_dict = dict()
for x in checklists:
val = ast.literal_eval(x["data"])
_val = ast.literal_eval(x["val"])
Index = dict(id= x['id'], tag = _val["tag_ABCD"])
index_dict[x['id']]=Index
return index_dict
但是它不会返回正确的元素。
另一个问题(附带问题..): 当我想向字典添加第三个元素时,为什么不能返回我的数据?
def getTitle():
index_list = []
index_dict = dict()
for x in checklists:
val = ast.literal_eval(x["data"])
Index = dict(id= x['id'], title = val["title"], list_category = val["list_category"])
index_dict[x['id']]=Index
return index_dict
对于我的每一个variatoins,我都会遇到相同的错误:
KeyError:'list_category'
答案 0 :(得分:1)
编写列表所需的功能的一种巧妙方法是使用列表理解和dict理解。有关这些工作原理的信息,请参见此处:https://www.smallsurething.com/list-dict-and-set-comprehensions-by-example/
代码看起来像这样:
def title_id(the_list):
return {id: title for (id, title) in [(i["id"], i["title"]) for i in the_list]}
至于您在list_category周围的错误,我敢打赌,您列表中的字典之一没有该键。要查看发生了什么,请在出现关键错误的情况下使用try-except和print val。