2017-01-02 60 views
4

我一直在网上搜索很长时间,但无济于事。请帮助或试图给一些想法如何实现这一如何在Python3中将数据帧转换为字典

我用熊猫来读取MovieLens csv文件

ratings = pd.read_table('ml-latest-small/ratings.csv') 

然后我得到一个表是这样的:

userId movieId rating timestamp 
1  31  2.5  1260759144 
1  1029 3.0  1260759179 
1  1061 3.0  1260759182 
1  1129 2.0  1260759185 
1  1172 4.0  1260759205 
2  31  3.0  1260759134 
2  1111 4.5  1260759256 

我想改造它与dict像

{userId:{movieId:rating}} 

{ 
1:{31:2.5,1029:3.0,1061,3.0,1129:2.0,1172:4.0}, 
2:{31:3.0,1111:4.5} 
} 

我试过这个代码,但没有成功

for user in ratings['userId']: 
for movieid in ratings['movieId']: 
    di_rating.setdefault(user,{}) 
    di_rating[user][movieid]=ratings['rating'][ratings['userId'] == user][ratings['movieId'] == movieid] 

有人可以帮我吗?

回答

4

您可以使用groupbyiterrows

d = df.groupby('userId').apply(lambda y: {int(x.movieId): x.rating for i, x in y.iterrows()}) 
     .to_dict() 
print (d) 
{ 
1: {1129: 2.0, 1061: 3.0, 1172: 4.0, 1029: 3.0, 31: 2.5}, 
2: {1111: 4.5, 31: 3.0} 
} 

另一种解决方案缺失了答案:

d1 = df.groupby('userId').apply(lambda x: dict(zip(x['movieId'], x['rating']))).to_dict() 
print (d1) 
{ 
1: {1129: 2.0, 1061: 3.0, 1172: 4.0, 1029: 3.0, 31: 2.5}, 
2: {1111: 4.5, 31: 3.0} 
} 
+0

非常感谢!但似乎'movieId'转换为float类型 – Alfred

+0

您可以投射到'int' - 'd = df.groupby('userId')。apply(lambda:{int(x.movi​​eId):x.rating for i ,x in y.iterrows()})。to_dict()' – jezrael