Python - Pandas：在較長範圍內得到一組日均值的平均值

Question

對於每一天，我想要得到的值的平均值在上午8點到下午5點之間。對於那些每日平均值，我想爲例如一個月或一年的範圍期間或自定義的選定範圍做出新的平均值。我如何在熊貓中做到這一點？

例如，2011年8月到5月的每日範圍的2011年8月到2011年11月的平均值。

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 

爲了得到一個定製意味着你會改變傳遞給groupby的說法。