我有一个DataFrame，其数据类似于以下
import pandas as pd; import numpy as np; import datetime; from datetime import timedelta;
df = pd.DataFrame(index=pd.date_range(start='20160102', end='20170301', freq='5min'))
df['value'] = np.random.randn(df.index.size)
df.index += pd.Series([timedelta(seconds=np.random.randint(-60, 60))
for _ in range(df.index.size)])
看起来像这样
In[37]: df
Out[37]:
value
2016-01-02 00:00:33 0.546675
2016-01-02 00:04:52 1.080558
2016-01-02 00:10:46 -1.551206
2016-01-02 00:15:52 -1.278845
2016-01-02 00:19:04 -1.672387
2016-01-02 00:25:36 -0.786985
2016-01-02 00:29:35 1.067132
2016-01-02 00:34:36 -0.575365
2016-01-02 00:39:33 0.570341
2016-01-02 00:44:56 -0.636312
...
2017-02-28 23:14:57 -0.027981
2017-02-28 23:19:51 0.883150
2017-02-28 23:24:15 -0.706997
2017-02-28 23:30:09 -0.954630
2017-02-28 23:35:08 -1.184881
2017-02-28 23:40:20 0.104017
2017-02-28 23:44:10 -0.678742
2017-02-28 23:49:15 -0.959857
2017-02-28 23:54:36 -1.157165
2017-02-28 23:59:10 0.527642
现在，我的目标是在24小时内每5分钟获得一次平均值 - 而不考虑这些值实际来自哪一天。
如何有效地执行此操作？我想我可以以某种方式从索引中删除实际日期，然后使用类似pd.TimeGrouper的内容，但我还没有想出如何做到这一点。
我的不太好的解决方案
到目前为止，我的解决方案是在这样的循环中使用between_time，只使用任意一天。
aggregates = []
start_time = datetime.datetime(1990, 1, 1, 0, 0, 0)
while start_time < datetime.datetime(1990, 1, 1, 23, 59, 0):
aggregates.append(
(
start_time,
df.between_time(start_time.time(),
(start_time + timedelta(minutes=5)).time(),
include_end=False).value.mean()
)
)
start_time += timedelta(minutes=5)
result = pd.DataFrame(aggregates, columns=['time', 'value'])
按预期工作
In[68]: result
Out[68]:
time value
0 1990-01-01 00:00:00 0.032667
1 1990-01-01 00:05:00 0.117288
2 1990-01-01 00:10:00 -0.052447
3 1990-01-01 00:15:00 -0.070428
4 1990-01-01 00:20:00 0.034584
5 1990-01-01 00:25:00 0.042414
6 1990-01-01 00:30:00 0.043388
7 1990-01-01 00:35:00 0.050371
8 1990-01-01 00:40:00 0.022209
9 1990-01-01 00:45:00 -0.035161
.. ... ...
278 1990-01-01 23:10:00 0.073753
279 1990-01-01 23:15:00 -0.005661
280 1990-01-01 23:20:00 -0.074529
281 1990-01-01 23:25:00 -0.083190
282 1990-01-01 23:30:00 -0.036636
283 1990-01-01 23:35:00 0.006767
284 1990-01-01 23:40:00 0.043436
285 1990-01-01 23:45:00 0.011117
286 1990-01-01 23:50:00 0.020737
287 1990-01-01 23:55:00 0.021030
[288 rows x 2 columns]
但这并不像是一个非常适合熊猫的解决方案。
答案 0 :(得分：3)
IIUC以下应该有效：
In [62]:
df.groupby(df.index.floor('5min').time).mean()
Out[62]:
value
00:00:00 -0.038002
00:05:00 -0.011646
00:10:00 0.010701
00:15:00 0.034699
00:20:00 0.041164
00:25:00 0.151187
00:30:00 -0.006149
00:35:00 -0.008256
00:40:00 0.021389
00:45:00 0.016851
00:50:00 -0.074825
00:55:00 0.012861
01:00:00 0.054048
01:05:00 0.041907
01:10:00 -0.004457
01:15:00 0.052428
01:20:00 -0.021518
01:25:00 -0.019010
01:30:00 0.030887
01:35:00 -0.085415
01:40:00 0.002386
01:45:00 -0.002189
01:50:00 0.049720
01:55:00 0.032292
02:00:00 -0.043642
02:05:00 0.067132
02:10:00 -0.029628
02:15:00 0.064098
02:20:00 0.042731
02:25:00 -0.031113
... ...
21:30:00 -0.018391
21:35:00 0.032155
21:40:00 0.035014
21:45:00 -0.016979
21:50:00 -0.025248
21:55:00 0.027896
22:00:00 -0.117036
22:05:00 -0.017970
22:10:00 -0.008494
22:15:00 -0.065303
22:20:00 -0.014623
22:25:00 0.076994
22:30:00 -0.030935
22:35:00 0.030308
22:40:00 -0.124668
22:45:00 0.064853
22:50:00 0.057913
22:55:00 0.002309
23:00:00 0.083586
23:05:00 -0.031043
23:10:00 -0.049510
23:15:00 0.003520
23:20:00 0.037135
23:25:00 -0.002231
23:30:00 -0.029592
23:35:00 0.040335
23:40:00 -0.021513
23:45:00 0.104421
23:50:00 -0.022280
23:55:00 -0.021283
[288 rows x 1 columns]
这里我floor索引到'5分钟'的时间间隔，然后对时间属性进行分组并汇总mean