Question

我有这个DataFrame：

    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

我需要的一列用于统计每个用户在购买产品之前看到了多少次。

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

有人可以帮我吗？

Answer 1

您可能需要为groupby创建一个附加密钥，方法是在cumsum之后使用shfit

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

Answer 2

一种方法是：

首先获得所有用户和产品

In [5]: df = pd.DataFrame(np.arange(6).reshape(3,2), columns=['a', 'b'])

In [6]: df
Out[6]: 
   a  b
0  0  1
1  2  3
2  4  5

In [7]: df.compound()
Out[7]: 
a    14
b    47
dtype: int32

为用户产品组合创建字典，以跟踪每个用户看到过哪些产品

users=list(df.users.unique())
products=list(df.products.unique())

初始化空列

see_dict={users[i]:{products[j]:0 for j in range(len(products))} for i in range(len(users))}

#{'A': {1: 0, 2: 0}, 'B': {1: 0, 2: 0}}

现在，对于每一行，如果是see动作，则更新字典（增量）并分配值。如果是购买操作，则仅分配值并重置计数器

df["see_before_buy"]=None

输出

for i in range(len(df)):
    user=df.loc[i,"users"]
    product=df.loc[i,"products"]
    if(df.loc[i,"action"]=="see"): #if the action is see
        see_dict[user][product]+=1 #increment the see dictionary
        df.loc[i,"see_before_buy"]=see_dict[user][product] #assign this value for this row
    else: #buy action
        df.loc[i,"see_before_buy"]=see_dict[user][product] #assign the current value
        see_dict[user][product]=0 #reset the counter

有条件分组的CumCount熊猫

2 个答案: