python - 按两列分组， Pandas python 中第三列的最大值

Question

我有一个包含 PERIOD_START_TIME、ID、更多列和 VALUE 列的数据框。我需要的是按 PERIOD_START_TIME 和 ID 分组(因为按时间和 ID 有重复的行)并取 VALUE 列的最大值。df:

PERIOD_START_TIME     ID       VALUE
06.01.2017 02:00:00   55  ...   35
06.01.2017 02:00:00   55  ...   22
06.01.2017 03:00:00   55  ...   63
06.01.2017 03:00:00   55  ...   33
06.01.2017 04:00:00   55  ...   63
06.01.2017 04:00:00   55  ...   45
06.01.2017 02:00:00   65  ...   10
06.01.2017 02:00:00   65  ...   5
06.01.2017 03:00:00   65  ...   22
06.01.2017 03:00:00   65  ...   5
06.01.2017 04:00:00   65  ...   12
06.01.2017 04:00:00   65  ...   15

期望的输出:

PERIOD_START_TIME     ID  ...  VALUE
06.01.2017 02:00:00   55  ...   35
06.01.2017 03:00:00   55  ...   63
06.01.2017 04:00:00   55  ...   63
06.01.2017 02:00:00   65  ...   10
06.01.2017 03:00:00   65  ...   22
06.01.2017 04:00:00   65  ...   15

Answer 1

使用groupby和聚合max :

print (df)
      PERIOD_START_TIME  ID  A  VALUE
0   06.01.2017 02:00:00  55  8     35
1   06.01.2017 02:00:00  55  8     22
2   06.01.2017 03:00:00  55  8     63
3   06.01.2017 03:00:00  55  8     33
4   06.01.2017 04:00:00  55  8     63
5   06.01.2017 04:00:00  55  8     45
6   06.01.2017 02:00:00  65  8     10
7   06.01.2017 02:00:00  65  8      5
8   06.01.2017 03:00:00  65  8     22
9   06.01.2017 03:00:00  65  8      5
10  06.01.2017 04:00:00  65  8     12
11  06.01.2017 04:00:00  65  8     15

df = df.groupby(['PERIOD_START_TIME','ID'], as_index=False)['VALUE'].max()            

或者:

df = df.groupby(['PERIOD_START_TIME','ID'])['VALUE'].max().reset_index()

---

print (df)
     PERIOD_START_TIME  ID  VALUE
0  06.01.2017 02:00:00  55     35
1  06.01.2017 02:00:00  65     10
2  06.01.2017 03:00:00  55     63
3  06.01.2017 03:00:00  65     22
4  06.01.2017 04:00:00  55     63
5  06.01.2017 04:00:00  65     15

更多栏目需要idxmax并选择 loc :

df = df.loc[df.groupby(['PERIOD_START_TIME','ID'])['VALUE'].idxmax()]  
print (df)
      PERIOD_START_TIME  ID  A  VALUE
0   06.01.2017 02:00:00  55  8     35
6   06.01.2017 02:00:00  65  8     10
2   06.01.2017 03:00:00  55  8     63
8   06.01.2017 03:00:00  65  8     22
4   06.01.2017 04:00:00  55  8     63
11  06.01.2017 04:00:00  65  8     15 

备选方案:

cols = ['PERIOD_START_TIME','ID']
df = df.sort_values(cols).groupby(cols, as_index=False).first()
print (df)
     PERIOD_START_TIME  ID  A  VALUE
0  06.01.2017 02:00:00  55  8     35
1  06.01.2017 02:00:00  65  8     10
2  06.01.2017 03:00:00  55  8     63
3  06.01.2017 03:00:00  65  8     22
4  06.01.2017 04:00:00  55  8     63
5  06.01.2017 04:00:00  65  8     12