python - 找到每个 pandas 组最接近的时间值

Question

import pandas as pd
df = pd.DataFrame({'date': ['2014-06-22 17:46:00', '2014-06-24 16:52:00', '2014-06-25 20:02:00', '2014-06-25 17:55:00', '2014-07-02 11:36:00', '2014-07-06 12:40:00', '2014-07-05 12:46:00', '2014-07-27 15:12:00'],
    'type': ['A', 'A', 'A', 'B', 'B', 'C', 'C', 'C']})

>>> df
                  date type
0  2014-06-22 17:46:00    A
1  2014-06-24 16:52:00    A
2  2014-06-25 20:02:00    A
3  2014-06-25 17:55:00    B
4  2014-07-02 11:36:00    B
5  2014-07-06 12:40:00    C
6  2014-07-05 12:46:00    C
7  2014-07-27 15:12:00    C

如何获取最接近时间(例如 17:00)(不考虑日期)的每个组元素的索引？期望的结果是:

>>> df.groupby('type').date. ???
type
A    1
B    3
C    7
Name: date, dtype: int64

此外，如果我想找到最接近但早于给定时间的时间怎么办？再次到 17:00 时，需要返回:

>>> df.groupby('type').date. ???
type
A    1
B    4
C    7
Name: date, dtype: int64

Answer 1

获取默认日期，添加时间s并获取与时间t的差值:

首先通过 DataFrameGroupBy.idxmin 获取每组绝对值的最小索引，对于第二个解决方案，通过将正值替换为 DataFrameGroupBy.idxmax 的 NaN 来获取每组的最大负值和 mask :

df = pd.DataFrame({'date': ['2014-06-22 17:46:00', '2014-06-22 16:52:00', 
                            '2014-06-25 20:02:00', '2014-06-25 17:55:00', 
                            '2014-07-02 11:36:00', '2014-07-06 12:40:00', 
                            '2014-07-05 12:46:00', '2014-07-27 15:12:00'],
    'type': ['A', 'A', 'A', 'B', 'B', 'C', 'C', 'C']})

---

#convert column to datetimes
df['date'] = pd.to_datetime(df.date)

t = '17:00:00'
a = pd.to_datetime(df['date'].dt.strftime('%H:%M:%S')) - pd.to_datetime(t)
print (a)
0            00:46:00
1   -1 days +23:52:00
2            03:02:00
3            00:55:00
4   -1 days +18:36:00
5   -1 days +19:40:00
6   -1 days +19:46:00
7   -1 days +22:12:00
Name: date, dtype: timedelta64[ns]


b = a.abs().groupby(df['type']).idxmin()
print (b)
type
A    1
B    3
C    7
Name: date, dtype: int64

c = a.mask(a > pd.Timedelta(0)).groupby(df['type']).idxmax()
print (c)
type
A    1
B    4
C    7
Name: date, dtype: int64

详细信息:

df1 = pd.concat([df, a, a.abs(), a.mask(a >  pd.Timedelta(0))], axis=1)
df1.columns = ['date','type','diff','absolute diff','max negative']
print (df1)
                 date type              diff absolute diff      max negative
0 2014-06-22 17:46:00    A          00:46:00      00:46:00               NaT
1 2014-06-22 16:52:00    A -1 days +23:52:00      00:08:00 -1 days +23:52:00
2 2014-06-25 20:02:00    A          03:02:00      03:02:00               NaT
3 2014-06-25 17:55:00    B          00:55:00      00:55:00               NaT
4 2014-07-02 11:36:00    B -1 days +18:36:00      05:24:00 -1 days +18:36:00
5 2014-07-06 12:40:00    C -1 days +19:40:00      04:20:00 -1 days +19:40:00
6 2014-07-05 12:46:00    C -1 days +19:46:00      04:14:00 -1 days +19:46:00
7 2014-07-27 15:12:00    C -1 days +22:12:00      01:48:00 -1 days +22:12:00