python - Pandas :合并两个索引列

Question

我有以下 Pandas 系列:

data = {(pd.Timestamp('2016-01-01 00:00:00'), datetime.time(0, 0)): 6.885,
        (pd.Timestamp('2016-01-01 00:00:00'), datetime.time(0, 5)): 6.363, 
        (pd.Timestamp('2016-01-01 00:00:00'), datetime.time(0, 10)): 6.093,
        (pd.Timestamp('2016-01-01 00:00:00'), datetime.time(0, 15)): 6.768, 
        (pd.Timestamp('2016-01-01 00:00:00'), datetime.time(0, 20)): 7.11}
s = pd.Series(data)

2016-01-01  00:00:00    6.885
            00:05:00    6.363
            00:10:00    6.093
            00:15:00    6.768
            00:20:00    7.110
dtype: float64

我怎样才能组合这两个索引列来创建一个 DatetimeIndex，如下所示:

2016-01-01 00:00:00    6.885
2016-01-01 00:05:00    6.363
2016-01-01 00:10:00    6.093
2016-01-01 00:15:00    6.768
2016-01-01 00:20:00    7.110
dtype: float64

Answer 1

直观的答案
使用 pd.Index.map 和 pd.Timedelta

s.index = s.index.map(lambda t: t[0] + pd.Timedelta(str(t[1])))
s

2016-01-01 00:00:00    6.885
2016-01-01 00:05:00    6.363
2016-01-01 00:10:00    6.093
2016-01-01 00:15:00    6.768
2016-01-01 00:20:00    7.110
dtype: float64

---

快速回答
如果速度是你的追求，试试这个

t = np.array(
    [t.hour * 60 + t.minute for t in s.index.get_level_values(1)],
    'timedelta64[m]'
)

s.index = s.index.get_level_values(0) + t

2016-01-01 00:00:00    6.885
2016-01-01 00:05:00    6.363
2016-01-01 00:10:00    6.093
2016-01-01 00:15:00    6.768
2016-01-01 00:20:00    7.110
dtype: float64

---

时间测试

请注意，这仅在您关心优化的情况下。否则，请使用您认为正确的选择。

jez = lambda s: s.index.get_level_values(0) + pd.to_timedelta(s.index.get_level_values(1).astype(str))
pir1 = lambda s: s.index.map(lambda t: t[0] + pd.Timedelta(str(t[1])))
pir2 = lambda s: s.index.get_level_values(0) + np.array([t.hour * 60 + t.minute for t in s.index.get_level_values(1)], 'timedelta64[m]')

res = pd.DataFrame(
    np.nan, [10, 30, 100, 300, 1000, 3000, 10000, 30000],
    'jez pir1 pir2'.split()
)

for i in res.index:
    s_ = pd.concat([s] * i)
    for j in res.columns:
        stmt = f'{j}(s_)'
        setp = f'from __main__ import {j}, s_'
        res.at[i, j] = timeit(stmt, setp, number=100)

res.plot(loglog=True)

res.div(res.min(1), 0)

             jez       pir1  pir2
10      2.400808   3.530032   1.0
30      4.045287   8.378484   1.0
100     6.337601  18.610263   1.0
300     8.664829  30.363422   1.0
1000   11.593935  44.210358   1.0
3000   11.899037  47.425953   1.0
10000  12.226166  49.546467   1.0
30000  12.543602  50.730653   1.0