我有一本字典如下：

{'result': '{"maid":{"0":"0365206d-e97d-4ab0-aa63-64091e66a1a4","1":"0955bbcc-3a83-4c64-8170-f5deb799a5ba","2":"0570ba29-1ee8-4bc6-a12c-c2b706d805c8"},"category":{"0":"EventHall","1":"SuperMarket","2":"Bank"},"geo_behavior":{"0":null,"1":null,"2":null},"polygonid":{"0":2332,"1":2332,"2":2332},"places":{"0":"Shri Sai Dj","1":"D Mart","2":"Bank of Baroda"},"age":{"0":38.0,"1":18.0,"2":37.0},"gender":{"0":0.0,"1":0.0,"2":0.0},"mobile":{"0":null,"1":null,"2":null},"make":{"0":"oppo","1":"vivo","2":"oneplus"},"deviceprice":{"0":160.0,"1":190.0,"2":392.0},"weight":{"0":2.5684166749,"1":2.0,"2":2.0},"__index_level_0__":{"0":0,"1":1,"2":2}}'}

字符串
我试图将这个字典解析为一个数据框架。我尝试了以下方法：

l=k.get("result")
m=json.loads(l)
maid=m.get('maid')
cate=m.get('category')
geobh=m.get('geo_behavior')
pid=m.get('polygonid')
places=m.get('places')
age=m.get('age')
gender=m.get('gender')
mobile=m.get('mobile')
make=m.get('make')
deviceprice=m.get('deviceprice')
weight=m.get('weight')

型
这给了我单独的字典合并以形成数据框架。我也试过

pd.json_normalize
pd.DataFrame.from_records(

型
但帮不上忙我想知道是否有更简单的方法将result转换为字典。