使用数据框减去字典中列表的值 [英] subtracting values of a list in a dictionary using a dataframe
本文介绍了使用数据框减去字典中列表的值的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我在下面有一个数据框,其中包含用户购买的产品.
I have a dataframe below, with the products purchased by users.
数据集:
user age maritalstatus product
A Young married 111
B young married 222
C young Single 111
D old single 222
E old married 111
F teen married 222
G teen married 555
H adult single 444
I adult single 333
字典:
{A:[111,222], B:[111,222], C:[111], D:[222], G:[222,555], X:[222,444] }
预期输出:
{A:[222], B:[111], C:[], D:[], G:[222], X:[222,444] }
字典应查看数据框并删除用户已经购买的产品.
The dictionary should look into the dataframe and remove the products already purchased by the users.
推荐答案
您可以使用字典理解:
{k:[e for e in v if e not in df.loc[df.user.eq(k), 'product'].tolist()] for k,v in d.items()}
Out[292]: {'A': [222], 'B': [111], 'C': [], 'D': [], 'G': [222], 'X': [222, 444]}
稍微冗长的解决方案,以便于理解:
A slightly more verbose solution for easier understanding:
首先构建用户产品字典:
First to build a user-product dict:
user_prod = df.groupby('user')['product'].apply(list).to_dict()
{'A': [111],
'B': [222],
'C': [111],
'D': [222],
'E': [111],
'F': [222],
'G': [555],
'H': [444],
'I': [333]}
然后,使用dict理解删除user_prod dict中的元素.
Then, use a dict comprehension to remove elements which are in the user_prod dict.
{k:[e for e in v if e not in user_prod.get(k,[])] for k,v in d.items()}
Out[319]: {'A': [222], 'B': [111], 'C': [], 'D': [], 'G': [222], 'X': [222, 444]}
使用user_prod.get是必要的,因为用户可能不存在,并且.get通过返回一个空列表来避免出现异常.
The use of user_prod.get is necessary because the user may not exist and .get will avoid an exception by returning an empty list.
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