我对python和pandas相当陌生,因此对以后的误会深表歉意。
我有一个带有每小时值的pandas DataFrame,看起来像这样:
2014-04-01 09:00:00 52.9 41.1 36.3
2014-04-01 10:00:00 56.4 41.6 70.8
2014-04-01 11:00:00 53.3 41.2 49.6
2014-04-01 12:00:00 50.4 39.5 36.6
2014-04-01 13:00:00 51.1 39.2 33.3
2016-11-30 16:00:00 16.0 13.5 36.6
2016-11-30 17:00:00 19.6 17.4 44.3
现在我需要计算从2014-04-01 12:00到2014-04-02 11:00开始的每列24小时平均值 所以我想每天从中午到中午。
不幸的是,我不知道该怎么做。我已经阅读了一些有关使用groupby的建议,但是我真的不知道如何...
非常感谢您!任何帮助表示赞赏!
答案 0 :(得分:7)
base
参数。一天是24小时,因此以12为基数的分组将从中午开始-中午。重新采样为您提供了介于两天之间的所有时间,因此如果您不需要完整的基础,可以.dropna(how='all')
。 (我假设您有一个DatetimeIndex
,如果没有,则可以使用resample的on
参数指定日期时间列。)
df.resample('24H', base=12).mean()
#df.groupby(pd.Grouper(level=0, base=12, freq='24H')).mean() # Equivalent
1 2 3
0
2014-03-31 12:00:00 54.20 41.30 52.233333
2014-04-01 12:00:00 50.75 39.35 34.950000
2014-04-02 12:00:00 NaN NaN NaN
2014-04-03 12:00:00 NaN NaN NaN
2014-04-04 12:00:00 NaN NaN NaN
... ... ... ...
2016-11-26 12:00:00 NaN NaN NaN
2016-11-27 12:00:00 NaN NaN NaN
2016-11-28 12:00:00 NaN NaN NaN
2016-11-29 12:00:00 NaN NaN NaN
2016-11-30 12:00:00 17.80 15.45 40.450000
答案 1 :(得分:3)
您可以减去时间和分组依据:
df.groupby((df.index - pd.to_timedelta('12:00:00')).normalize()).mean()
答案 2 :(得分:0)
您可以将小时数更改12小时,然后按天重新采样。
from io import StringIO
import pandas as pd
data = """
2014-04-01 09:00:00,52.9,41.1,36.3
2014-04-01 10:00:00,56.4,41.6,70.8
2014-04-01 11:00:00,53.3,41.2,49.6
2014-04-01 12:00:00,50.4,39.5,36.6
2014-04-01 13:00:00,51.1,39.2,33.3
2016-11-30 16:00:00,16.0,13.5,36.6
2016-11-30 17:00:00,19.6,17.4,44.3
"""
df = pd.read_csv(StringIO(data), sep=',', header=None, index_col=0)
df.index = pd.to_datetime(df.index)
# shift by 12 hours
df.index = df.index - pd.Timedelta(hours=12)
# resample and drop na rows
df.resample('D').mean().dropna()