如何解决根据条件分组+新列+值抢先行
我有这个设置
df = pd.DataFrame({'user':[1,1,2,3,4,4],'date':['1995-09-01','1995-09-02','1995-10-03','1995-10-04','1995-10-05','1995-11-07','1995-11-08','1995-11-09','1995-11-10','1995-11-15','1995-12-18','1995-12-19','1995-12-20','1995-12-23','1995-12-26','1995-12-27'],'dc':['1995-09-02','1995-10-02','1995-11-05','1995-12-10','1995-12-23'],'tp':['s','c','f','s','f'],'vt':['0','1','0','0'],'c1':['1','5','2','3','9','6','4','c2':['3','8','7','c3':['5','7']})
df
给出：
user date dc tp vt c1 c2 c3
1 1995-09-01 1995-09-02 s 0 1 3 5
1 1995-09-02 1995-09-02 c 1 5 4 5
1 1995-10-03 1995-10-02 f 0 0 0 2
2 1995-10-04 1995-10-05 s 0 2 2 5
2 1995-10-05 1995-10-05 c 1 3 5 6
2 1995-11-07 1995-11-05 c 0 9 3 4
2 1995-11-08 1995-11-05 f 0 3 8 2
3 1995-11-09 1995-11-10 s 0 2 4 4
3 1995-11-10 1995-11-10 c 1 0 0 4
3 1995-11-15 1995-11-10 s 0 5 6 6
3 1995-12-18 1995-12-10 f 0 5 2 3
4 1995-12-19 1995-12-23 s 0 6 7 4
4 1995-12-20 1995-12-23 s 0 4 0 3
4 1995-12-23 1995-12-23 c 1 0 0 8
4 1995-12-26 1995-12-23 s 0 6 8 2
4 1995-12-27 1995-12-23 f 0 0 0 7
我要创建新列，请创建新列df ['dc2']，其中groupby用户的列df ['dc2'] = df ['dc']。 但是，如果df ['dc']满足条件'tp'='c'＆'vt'= 1＆'c1'= 0＆'c2'= 0， 然后获取前一个条目的日期（用户原始）
#ie。对于用户3，在df ['dc']列上，如果我们查看条目'tp'=' c '＆'vtb'= 1 ，我们可以看到它具有'c1'= 0 和'c2'= 0 ， ＃因此df ['dc2']的值将（对于用户3）为' 1995-11-09 '而不是'1995-11-10'
#ie。对于用户4，在列df ['dc']上，如果我们查看条目'tp'=' c '＆'vtb'= 1 ，我们可以看到它具有'c1'= 0 和'c2'= 0 ， 在这种情况下，df ['dc2']应该是（对于用户4）“ 1995-12-20 ”而不是“ 1995-12-23”
这是预期的结果：
user date dc dc2 tp vt c1 c2 c3
1 1995-09-01 1995-09-02 1995-09-02 s 0 1 3 5
1 1995-09-02 1995-09-02 1995-09-02 c 1 5 4 5
1 1995-10-03 1995-10-02 1995-10-02 f 0 0 0 2
2 1995-10-04 1995-10-05 1995-10-05 s 0 2 2 5
2 1995-10-05 1995-10-05 1995-10-05 c 1 3 5 6
2 1995-11-07 1995-11-05 1995-11-05 c 0 9 3 4
2 1995-11-08 1995-11-05 1995-11-05 f 0 3 8 2
3 1995-11-09 1995-11-10 1995-11-09 s 0 2 4 4
3 1995-11-10 1995-11-10 1995-11-09 c 1 0 0 4
3 1995-11-15 1995-11-10 1995-11-09 s 0 5 6 6
3 1995-12-18 1995-12-10 1995-12-09 f 0 5 2 3
4 1995-12-19 1995-12-23 1995-12-20 s 0 6 7 4
4 1995-12-20 1995-12-23 1995-12-20 s 0 4 0 3
4 1995-12-23 1995-12-23 1995-12-20 c 1 0 0 8
4 1995-12-26 1995-12-23 1995-12-20 s 0 6 8 2
4 1995-12-27 1995-12-23 1995-12-20 f 0 0 0 7
解决方法
让我们创建一个表示条件tp=c和vt=1和c1=0和c2=0的布尔掩码，然后在列user上进行分组，并应用自定义转换函数f根据以下条件选择前一行的值：
m = df['tp'].eq('c') & df['vt'].eq('1')\
& df['c1'].eq('0') & df['c2'].eq('0')
f = lambda s: s.mask(~m.shift(-1,fill_value=False)).ffill().bfill()
df['dc2'] = df.groupby('user')['date'].apply(f).fillna(df['dc'])
user date dc tp vt c1 c2 c3 dc2
0 1 1995-09-01 1995-09-02 s 0 1 3 5 1995-09-02
1 1 1995-09-02 1995-09-02 c 1 5 4 5 1995-09-02
2 1 1995-10-03 1995-10-02 f 0 0 0 2 1995-10-02
3 2 1995-10-04 1995-10-05 s 0 2 2 5 1995-10-05
4 2 1995-10-05 1995-10-05 c 1 3 5 6 1995-10-05
5 2 1995-11-07 1995-11-05 c 0 9 3 4 1995-11-05
6 2 1995-11-08 1995-11-05 f 0 3 8 2 1995-11-05
7 3 1995-11-09 1995-11-10 s 0 2 4 4 1995-11-09
8 3 1995-11-10 1995-11-10 c 1 0 0 4 1995-11-09
9 3 1995-11-15 1995-11-10 s 0 5 6 6 1995-11-09
10 3 1995-12-18 1995-12-10 f 0 5 2 3 1995-11-09
11 4 1995-12-19 1995-12-23 s 0 6 7 4 1995-12-20
12 4 1995-12-20 1995-12-23 s 0 4 0 3 1995-12-20
13 4 1995-12-23 1995-12-23 c 1 0 0 8 1995-12-20
14 4 1995-12-26 1995-12-23 s 0 6 8 2 1995-12-20
15 4 1995-12-27 1995-12-23 f 0 0 0 7 1995-12-20
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