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python - 有条件的 groupby CumCount pandas

转载 作者:太空宇宙 更新时间:2023-11-04 08:31:55 25 4
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我有这个数据框:

    dic = {'users' : ['A','A','B','A','A','B','A','A','A','A','A','B','A'],
'product' : [1,1,2,2,1,2,1,2,1,1,2,1,1],
'action' : ['see', 'see', 'see', 'see', 'buy', 'buy', 'see', 'see', 'see', 'see', 'buy', 'buy', 'buy']
}

df = pd.DataFrame(dic, columns=dic.keys())

df


users product action
0 A 1 see
1 A 1 see
2 B 2 see
3 A 2 see
4 A 1 buy
5 B 2 buy
6 A 1 see
7 A 2 see
8 A 1 see
9 A 1 see
10 A 2 buy
11 B 1 buy
12 A 1 buy

我需要的是一个列,用于计算每个用户在购买产品之前看到产品的次数。

结果应该是这样的:

dic = {'users' : ['A','A','B','A','A','B','A','A','A','A','A','B','A'],
'product' : [1,1,2,2,1,2,1,2,1,1,2,1,1],
'action' : ['see', 'see', 'see', 'see', 'buy', 'buy', 'see', 'see', 'see', 'see', 'buy', 'buy', 'buy'],
'see_before_buy' : [1,2,1,1,2,1,1,2,2,3,2,0,3]
}

users product action see_before_buy
0 A 1 see 1
1 A 1 see 2
2 B 2 see 1
3 A 2 see 1
4 A 1 buy 2
5 B 2 buy 1
6 A 1 see 1
7 A 2 see 2
8 A 1 see 2
9 A 1 see 3
10 A 2 buy 2
11 B 1 buy 0
12 A 1 buy 3

有人能帮帮我吗?

最佳答案

您可能需要在 shift

之后使用 cumsumgroupby 创建一个附加键

addkey=df.groupby(['user','#product']).action.apply(lambda x : x.eq('buy').shift().fillna(0).cumsum())
df['seebefore']=df.action.eq('see').groupby([df.user,df['#product'],addkey]).cumsum()
df
Out[131]:
index user #product action seebefore
0 0 A 1 see 1.0
1 1 A 1 see 2.0
2 2 B 2 see 1.0
3 3 A 2 see 1.0
4 4 A 1 buy 2.0
5 5 B 2 buy 1.0
6 6 A 1 see 1.0
7 7 A 2 see 2.0
8 8 A 1 see 2.0
9 9 A 1 see 3.0
10 10 A 2 buy 2.0
11 11 B 1 buy 0.0
12 12 A 1 buy 3.0

关于python - 有条件的 groupby CumCount pandas,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52250345/

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