GROUP BY+新列+基于条件的前一行抓取值 [英] Group by + New Column + Grab value former row based on conditionals

问题描述

我有这套

df = pd.DataFrame({'user':[1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,4],
                  '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','1995-09-02','1995-10-02','1995-10-05','1995-10-05','1995-11-05','1995-11-05','1995-11-10','1995-11-10','1995-11-10','1995-12-10','1995-12-23','1995-12-23','1995-12-23','1995-12-23','1995-12-23'],
                  'tp':['s','c','f','s','c','c','f','s','c','s','f','s','s','c','s','f'],
                  'vt':['0','1','0','0','1','0','0','0','1','0','0','0','0','1','0','0'],
                  'c1':['1','5','0','2','3','9','3','2','0','5','5','6','4','0','6','0'],
                  'c2':['3','4','0','2','5','3','8','4','0','6','2','7','0','0','8','0'],
                  'c3':['5','5','2','5','6','4','2','4','4','6','3','4','3','8','2','7']})
df

这提供了：

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

我想创建新列CREATE新列df[‘dc2’]，其中groupby user，列df[‘dc2’]=df[‘dc’]。 然而，如果df[‘dc’]满足条件‘tp’=‘c’&；‘Vt’=1&；‘c1’=0&；‘c2’=0， 然后抓取前一个条目的日期(用户的原始数据)

#ie.对于用户3，在df[‘dc’]列上，如果我们查看条目‘tp’=‘c’&；‘vt’=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df[‘dc’]列中，如果我们查看条目‘tp’=‘c’&；‘vt’=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(对于用户4)是‘1995-12-20’，而不是‘1995-12-23’

以下是所需结果：

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

推荐答案

我们创建一个表示条件tp=c&；vt=1&；c1=0&；c2=0的布尔掩码，然后对列user应用GROUPBY，并应用自定义转换函数f，该函数根据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

这篇关于GROUP BY+新列+基于条件的前一行抓取值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！