在Python中使用Pandas查找每日最大小时数 [英] Finding hour of daily max using Pandas in Python
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问题描述
我正在尝试在需求时间序列中找到每天的最大需求小时.
I am trying to find the hour of max demand every day in my demand time series.
我创建了一个看起来像..p的数据框.
I have created a dataframe that looks like..
power
2011-01-01 00:00:00 1015.70
2011-01-01 01:00:00 1015.70
2011-01-01 02:00:00 1010.30
2011-01-01 03:00:00 1010.90
2011-01-01 04:00:00 1021.10
2011-01-01 05:00:00 1046.00
2011-01-01 06:00:00 1054.60
...
和一组分组的序列,以使用.max()找出每天的最大值
and a grouped series to find the max value from each day using .max()
grouped = df.groupby(pd.TimeGrouper('D'))
grouped['power'].max()
输出
2011-01-01 1367.30
2011-01-02 1381.90
2011-01-03 1289.00
2011-01-04 1323.50
2011-01-05 1372.70
2011-01-06 1314.40
2011-01-07 1310.60
...
但是我也需要小时的最大值.像这样:
However I need the hour of the max value also. So something like:
2011-01-01 18 1367.30
2011-01-02 5 1381.90
2011-01-03 22 1289.00
2011-01-04 10 1323.50
...
我尝试使用idxmax(),但我不断收到ValueError
I have tried using idxmax() but I keep getting a ValueError
推荐答案
自2018-09-19起更新:
FutureWarning:不建议使用pd.TimeGrouper并将其删除; 请使用pd.Grouper(freq = ...)
FutureWarning: pd.TimeGrouper is deprecated and will be removed; Please use pd.Grouper(freq=...)
解决方案:
In [295]: df.loc[df.groupby(pd.Grouper(freq='D')).idxmax().iloc[:, 0]]
Out[295]:
power
2011-01-01 06:00:00 1054.6
2011-01-02 06:00:00 2054.6
旧答案:
尝试一下:
In [376]: df.loc[df.groupby(pd.TimeGrouper('D')).idxmax().iloc[:, 0]]
Out[376]:
power
2011-01-01 06:00:00 1054.6
2011-01-02 06:00:00 2054.6
数据:
In [377]: df
Out[377]:
power
2011-01-01 00:00:00 1015.7
2011-01-01 01:00:00 1015.7
2011-01-01 02:00:00 1010.3
2011-01-01 03:00:00 1010.9
2011-01-01 04:00:00 1021.1
2011-01-01 05:00:00 1046.0
2011-01-01 06:00:00 1054.6
2011-01-02 00:00:00 2015.7
2011-01-02 01:00:00 2015.7
2011-01-02 02:00:00 2010.3
2011-01-02 03:00:00 2010.9
2011-01-02 04:00:00 2021.1
2011-01-02 05:00:00 2046.0
2011-01-02 06:00:00 2054.6
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