时间戳

首页 > 解决方案 > Group by + New Column + 根据条件获取前一行的值

python - Group by + New Column + 根据条件获取前一行的值

问题描述

我有这一套

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

这使：

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

我想创建新列创建新列 df['dc2']，其中 groupby 用户，列 df['dc2']= df['dc']。但是如果 df['dc'] 满足条件 'tp'='c' & 'vt'=1 & 'c1'=0 & 'c2'=0，则抓取前一个条目的日期（用户的原始)

＃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'] 的值将是（对于用户 3）“ 1995-11-09 ”而不是“1995-11-10”

＃IE。对于用户 4，在 df['dc'] 列上，如果我们查看条目 'tp'= ' c ' & 'vt'= 1，我们可以看到它有 'c1'= 0和 'c2'= 0，在这个case df['dc2'] 应该是（对于用户 4）' 1995-12-20 '而不是'1995-12-23'

这是期望的结果：

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

标签： pythonpandasif-statementconditional-statementspandas-groupby

解决方案

让我们创建一个表示条件tp=c& vt=1& c1=0&的布尔掩码c2=0，然后在列上分组user并应用自定义转换函数，该函数f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

推荐阅读