Pandas在系列/数据框架上有resample方法，但似乎无法自行重新取样DatetimeIndex？
具体来说，我每天Datetimeindex可能缺少日期，我想以小时频率重新取样，但只包括原始每日索引中的天数。
有没有比我下面的尝试更好的方法？
In [56]: daily_index = pd.period_range('01-Jan-2017', '31-Jan-2017', freq='B').asfreq('D')
In [57]: daily_index
Out[57]:
PeriodIndex(['2017-01-02', '2017-01-03', '2017-01-04', '2017-01-05',
'2017-01-06', '2017-01-09', '2017-01-10', '2017-01-11',
'2017-01-12', '2017-01-13', '2017-01-16', '2017-01-17',
'2017-01-18', '2017-01-19', '2017-01-20', '2017-01-23',
'2017-01-24', '2017-01-25', '2017-01-26', '2017-01-27',
'2017-01-30', '2017-01-31'],
dtype='int64', freq='D')
In [58]: daily_index.shape
Out[58]: (22,)
In [59]: hourly_index = pd.DatetimeIndex([]).union_many(
...: pd.date_range(day.to_timestamp('H','S'), day.to_timestamp('H','E'), freq='H')
...: for day in daily_index
...: )
In [60]: hourly_index
Out[60]:
DatetimeIndex(['2017-01-02 00:00:00', '2017-01-02 01:00:00',
'2017-01-02 02:00:00', '2017-01-02 03:00:00',
'2017-01-02 04:00:00', '2017-01-02 05:00:00',
'2017-01-02 06:00:00', '2017-01-02 07:00:00',
'2017-01-02 08:00:00', '2017-01-02 09:00:00',
...
'2017-01-31 14:00:00', '2017-01-31 15:00:00',
'2017-01-31 16:00:00', '2017-01-31 17:00:00',
'2017-01-31 18:00:00', '2017-01-31 19:00:00',
'2017-01-31 20:00:00', '2017-01-31 21:00:00',
'2017-01-31 22:00:00', '2017-01-31 23:00:00'],
dtype='datetime64[ns]', length=528, freq=None)
In [61]: 22*24
Out[61]: 528
In [62]: %%timeit
...: hourly_index = pd.DatetimeIndex([]).union_many(
...: pd.date_range(day.to_timestamp('H','S'), day.to_timestamp('H','E'), freq='H')
...: for day in daily_index
...: )
100 loops, best of 3: 13.7 ms per loop
更新
我的@NTAWolf答案略有不同，其答案相似，但如果输入日期没有排序则不会重新排序
def resample_index(index, freq):
"""Resamples each day in the daily `index` to the specified `freq`.
Parameters
----------
index : pd.DatetimeIndex
The daily-frequency index to resample
freq : str
A pandas frequency string which should be higher than daily
Returns
-------
pd.DatetimeIndex
The resampled index
"""
assert isinstance(index, pd.DatetimeIndex)
start_date = index.min()
end_date = index.max() + pd.DateOffset(days=1)
resampled_index = pd.date_range(start_date, end_date, freq=freq)[:-1]
series = pd.Series(resampled_index, resampled_index.floor('D'))
return pd.DatetimeIndex(series.loc[index].values)
In [184]: %%timeit
...: hourly_index3 = pd.date_range(daily_index.start_time.min(),
...: daily_index.end_time.max() + 1,
...: normalize=True, freq='H')
...: hourly_index3 = hourly_index3[hourly_index3.floor('D').isin(daily_index.start_time)]
100 loops, best of 3: 2.97 ms per loop
In [185]: %timeit resample_index(daily_index.to_timestamp('D','S'), freq='H')
100 loops, best of 3: 2.93 ms per loop
答案 0 :(得分：4)
| Method | Time | Relative | |---------------------------------|---------|----------| | OP's updated approach | 1.31 ms | 17.6 % | | Generate daterange, np.in1d | 1.75 ms | 23.5 % | | Generate daterange, Series.isin | 1.90 ms | 25.5 % | | Resample with dummy Series | 4.37 ms | 58.7 % | | OP's initial approach | 7.45 ms | 100.0 % |
np.in1d 再次，@ IanS激发了更多优化！这可读性稍差，但速度要快一些：
%%timeit -r 10
hourly_index4 = pd.date_range(daily_index.start_time.min(),
daily_index.end_time.max() + pd.DateOffset(days=1),
normalize=True, freq='H')
overlap = np.in1d(np.array(hourly_index4.values, dtype='datetime64[D]'),
np.array(daily_index.start_time.values, dtype='datetime64[D]'))
hourly_index4 = hourly_index4[overlap]
1000 loops, best of 10: 1.75 ms per loop
在这里，通过将两个系列的值转换为相同的numpy日期时间类型（过程中的地板hourly_index）来获得加速。将.values传递给numpy加速了一点点。
Series.isin 比初始出价更快的方法，受@ IanS方法的启发：为您的数据中的每小时生成完整日期范围的日期范围，并仅选择与数据中现有日期匹配的条目：
%%timeit
hourly_index3 = pd.date_range(daily_index.start_time.min(),
# The following line should use
# +pd.DateOffset(days=1) in place of +1
# but is left as is to show the option.
daily_index.end_time.max() + 1,
normalize=True, freq='H')
hourly_index3 = hourly_index3[hourly_index3.floor('D').isin(daily_index.start_time)]
100 loops, best of 3: 1.9 ms per loop
缩短了约75％的处理时间。
使用虚拟系列，可以避免循环。在我的电脑上，它减少了大约40％的运行时间。
我的方法得到以下时间：
In [14]: %%timeit -o -r 10
....: hourly_index = pd.DatetimeIndex([]).union_many(
....: pd.date_range(day.to_timestamp('H','S'), day.to_timestamp('H','E'), freq='H')
....: for day in daily_index
....: )
....:
100 loops, best of 10: 7.45 ms per loop
更快的方法：
In [13]: %%timeit -o -r 10
s = pd.Series(0, index=daily_index)
s = s.resample('H').last()
s = s[s.index.start_time.floor('D').isin(daily_index.start_time)]
hourly_index2 = s.index.start_time
....:
100 loops, best of 10: 4.37 ms per loop
请注意，我们并不关心系列中的价值;在这里，我只是默认为int。
表达式s.index.start_time.floor('D').isin(daily_index.start_time)为我们提供了一个布尔向量，s.index中的值与daily_index中的天数匹配。
答案 1 :(得分：3)
另一种选择是直接生成小时索引，然后删除非工作日：
hourly_index = pd.date_range('01-Jan-2017', '31-Jan-2017', freq='H')
hourly_index = hourly_index[hourly_index.dayofweek < 5]
效果比较：
10 loops, best of 3: 44.2 ms per loop 1000 loops, best of 3: 1.46 ms per loop 1000 loops, best of 3: 598 µs per loop