用Pandas标记每组的第n行

Question

我有一个数据框与客户信息与他们的购买细节。我试图添加一个新的列，表明每第三次购买由同一个客户完成。你知道吗

下面给出的是数据帧

customer_name,bill_no,date
Mark,101,2018-10-01
Scott,102,2018-10-01
Pete,103,2018-10-02
Mark,104,2018-10-02
Mark,105,2018-10-04
Scott,106,2018-10-21
Julie,107,2018-10-03
Kevin,108,2018-10-07
Steve,109,2018-10-02
Mark,110,2018-10-06
Mark,111,2018-10-02
Mark,112,2018-10-05
Mark,113,2018-10-05

我写这封信是为了过滤同一位顾客每三次购买的商品。所以在这种情况下，我想为下面的bill\u no添加一个标志

Mark,105,2018-10-04
Mark,112,2018-10-05

基本上是为同一个客户生成的3个账单的每一个倍数。你知道吗

Answer 1

使用^{}：

n = 3
df['flag'] = df.groupby('customer_name').cumcount() + 1
df['flag'] = ((df['flag'] % n) == 0).astype(int)

print(df)
   customer_name  bill_no        date  flag
0           Mark      101  2018-10-01     0
1          Scott      102  2018-10-01     0
2           Pete      103  2018-10-02     0
3           Mark      104  2018-10-02     0
4           Mark      105  2018-10-04     1
5          Scott      106  2018-10-21     0
6          Julie      107  2018-10-03     0
7          Kevin      108  2018-10-07     0
8          Steve      109  2018-10-02     0
9           Mark      110  2018-10-06     0
10          Mark      111  2018-10-02     0
11          Mark      112  2018-10-05     1
12          Mark      113  2018-10-05     0

Answer 2

{{}如果实际使用的是cd2上的重要索引^}

n = 3
idx = df.groupby('customer_name', group_keys=False).apply(
    lambda x: x.index[n-1::n].to_series())
# So you can query these rows easily.
df.loc[idx]

   customer_name  bill_no        date
4           Mark      105  2018-10-04
11          Mark      112  2018-10-05

现在，使用索引标记它们：

df['flag'] = 0
df.loc[idx, 'flag'] = 1

df
   customer_name  bill_no        date  flag
0           Mark      101  2018-10-01     0
1          Scott      102  2018-10-01     0
2           Pete      103  2018-10-02     0
3           Mark      104  2018-10-02     0
4           Mark      105  2018-10-04     1
5          Scott      106  2018-10-21     0
6          Julie      107  2018-10-03     0
7          Kevin      108  2018-10-07     0
8          Steve      109  2018-10-02     0
9           Mark      110  2018-10-06     0
10          Mark      111  2018-10-02     0
11          Mark      112  2018-10-05     1
12          Mark      113  2018-10-05     0

如果性能很重要，请使用Sandeep的解决方案。你知道吗