Question

对于每一天，我希望得到的值的平均值在上午8点到下午5点之间。对于那些日平均值，我想为范围周期（例如一个月或一年或自定义选择范围）创建新的平均值。我怎么能在熊猫中做到这一点？

例如，2011年8月至2011年11月期间的平均值，每日上午8点至下午5点。

Time                   T_Sanyo_Gesloten

2010-08-31 12:30:00    33.910
2010-08-31 12:40:00    33.250
2010-08-31 12:50:00    30.500
2010-08-31 13:00:00    27.065
2010-08-31 13:10:00    25.610
...

2013-06-07 02:10:00    16.970
2013-06-07 02:20:00    16.955
2013-06-07 02:30:00    17.000
2013-06-07 02:40:00    17.015
2013-06-07 02:50:00    16.910

Answer 1

import datetime as DT
import numpy as np
import pandas as pd

np.random.seed(2013)
N = 10**4
df = pd.DataFrame(
    np.cumsum(np.random.random(N) - 0.5),
    index=pd.date_range('2010-8-31', freq='10T', periods=N))
#                             0
# 2010-08-31 00:00:00  0.175448
# 2010-08-31 00:10:00  0.631796
# 2010-08-31 00:20:00  0.399373
# 2010-08-31 00:30:00  0.499184
# 2010-08-31 00:40:00  0.631005
# ...
# 2010-11-08 09:50:00 -3.474801
# 2010-11-08 10:00:00 -3.172819
# 2010-11-08 10:10:00 -2.988451
# 2010-11-08 10:20:00 -3.101262
# 2010-11-08 10:30:00 -3.477685

eight_to_five = df.ix[df.index.indexer_between_time(DT.time(8), DT.time(17))]
#                             0
# 2010-08-31 08:00:00  1.440543
# 2010-08-31 08:10:00  1.450957
# 2010-08-31 08:20:00  1.746454
# 2010-08-31 08:30:00  1.443941
# 2010-08-31 08:40:00  1.845446
# ...
# 2010-11-08 09:50:00 -3.474801
# 2010-11-08 10:00:00 -3.172819
# 2010-11-08 10:10:00 -2.988451
# 2010-11-08 10:20:00 -3.101262
# 2010-11-08 10:30:00 -3.477685

# daily_mean = eight_to_five.groupby()
daily_mean = eight_to_five.resample('D', how='mean')
#                    0
# 2010-08-31  0.754004
# 2010-09-01  0.203610
# 2010-09-02  5.219528
# 2010-09-03  6.337688
# 2010-09-04  2.765504

monthly_mean = daily_mean.resample('M', how='mean')
#                    0
# 2010-08-31  0.754004
# 2010-09-30 -0.437582
# 2010-10-31  3.533525
# 2010-11-30  4.356728

yearly_mean = daily_mean.groupby(daily_mean.index.year).mean()
#              0
# 2010  1.885995

要获得自定义均值，您需要更改传递给groupby的参数。

Python - 熊猫：在更长的时间段内获得一组每日范围的平均值

1 个答案: