python - 查找每日最大值及其时间戳 (yyyy :mm:dd hh:mm:ss) in Python Pandas

Question

实际上，我有一个 150 MB 的每日分钟测量数据，持续了两年。我在这里给出了示例数据。我想创建一个新的数据框，其中包含每天的最大值及其时间戳。我的示例数据是:

    DateTime            Power
01-Aug-16 10:43:00.000  229.9607961
01-Aug-16 10:43:23.000  230.9030781
01-Aug-16 10:44:00.000  231.716212
01-Aug-16 10:45:00.000  232.4485882
01-Aug-16 10:46:00.000  233.2739154
02-Aug-16 09:42:00.000  229.6851724
02-Aug-16 09:43:00.000  230.9163998
02-Aug-16 09:43:06.000  230.9883337
02-Aug-16 09:44:00.000  231.2569098
02-Aug-16 09:49:00.000  229.5774805
02-Aug-16 09:50:00.000  229.8758693
02-Aug-16 09:51:00.000  229.9825204
03-Aug-16 10:09:00.000  231.3605982
03-Aug-16 10:10:00.000  231.6827163
03-Aug-16 10:11:00.000  231.1580262
03-Aug-16 10:12:00.000  230.4054286
03-Aug-16 10:13:00.000  229.6507959
03-Aug-16 10:13:02.000  229.6268353
03-Aug-16 10:14:00.000  230.4584964
03-Aug-16 10:15:00.000  230.9004206
03-Aug-16 10:16:00.000  231.189036

我现在的代码是:

max_per_day = df.groupby(pd.Grouper(key='time',freq='D')).max()
print(max_per_day)

我现在的输出是:

    time                  
2016-08-01  237.243835
2016-08-02  239.658539
2016-08-03  237.424683
2016-08-04  236.790695
2016-08-05  240.163910

目前它输出 yyyy:mm:dd 和值。但我什至想要 hh:mm(或 hh:mm:ss)对每个最大值。我尝试了以下代码:

max_pmpp_day = df.loc[df.groupby(pd.Grouper(freq='D')).idxmax().iloc[:,0]]

输出是:

 TypeError: Only valid with DatetimeIndex, TimedeltaIndex or PeriodIndex, but got an instance of 'Int64Index'

我试过@jezrael 的回答

df['DateTime'] = pd.to_datetime(df['time'])
s = df.groupby(pd.Grouper(key='DateTime', freq='D'))['Pmpp'].transform('max')
df = df[df['Pmpp'] == s]    
print(df)

输出为

                     time        Pmpp            DateTime
34    2016-08-01 11:11:00  237.243835 2016-08-01 11:11:00
434   2016-08-02 13:30:02  239.658539 2016-08-02 13:30:02
648   2016-08-03 12:39:00  237.424683 2016-08-03 12:39:00

Answer 1

您可以使用 GroupBy.transform或 Resampler.transform用于在新的 Series 中返回 max 值并与原始列进行比较:

df['DateTime'] = pd.to_datetime(df['DateTime'])
s = df.groupby(pd.Grouper(key='DateTime', freq='D'))['Power'].transform('max')
#alternative
#s = df.resample('D', on='DateTime')['Power'].transform('max')
df = df[df['Power'] == s]
print (df)
              DateTime       Power
4  2016-08-01 10:46:00  233.273915
8  2016-08-02 09:44:00  231.256910
13 2016-08-03 10:10:00  231.682716

或者创建DatetimeIndex 并在groupby 之后添加列以检查idxmax:

df['DateTime'] = pd.to_datetime(df['DateTime'])
df = df.set_index('DateTime')
df = df.loc[df.groupby(pd.Grouper(freq='D'))['Power'].idxmax()]
print (df)
                          Power
DateTime                       
2016-08-01 10:46:00  233.273915
2016-08-02 09:44:00  231.256910
2016-08-03 10:10:00  231.682716

@Jon Clements 的解决方案，谢谢:

df = (df.sort_values('Power')
        .groupby(df.DateTime.dt.to_period('D'))
        .last()
        .reset_index(drop=True))