Question

作为here我需要计算colums持续时间和km的平均值值== 1且值= 0的行。这次我希望聚合期是灵活的。

df
Out[20]: 
                          Date duration km   value
0   2015-03-28 09:07:00.800001    0      0    0
1   2015-03-28 09:36:01.819998    1      2    1
2   2015-03-30 09:36:06.839997    1      3    1 
3   2015-03-30 09:37:27.659997    nan    5    0 
4   2015-04-22 09:51:40.440003    3      7    0
5   2015-04-23 10:15:25.080002    0      nan  1

对于1天的聚合期，我可以使用之前建议的解决方案：

df.pivot_table(values=['duration','km'],columns=['value'],index=df['Date'].dt.date,aggfunc='mean'

ndf.columns = [i[0]+str(i[1]) for i in ndf.columns]

            duration0  duration1  km0  km1
Date                                      
2015-03-28        0.0        1.0  0.0  2.0
2015-03-30        NaN        1.0  5.0  3.0
2015-04-22        3.0        NaN  7.0  NaN
2015-04-23        NaN        0.0  NaN  NaN

但是，我不知道如何更改聚合期间，例如，我想将其作为函数的参数传递... 出于这个原因，pd.Grouper(freq=freq_aggregation)，freq_aggregation = 'd'或'60s'的方法将首选......

Answer 1

让我们使用pd.Grouper，unstack和列映射：

freq_str = '60s'
df_out = df.groupby([pd.Grouper(freq=freq_str, key='Date'),'value'])['duration','km'].agg('mean').unstack()

df_out.columns = df_out.columns.map('{0[0]}{0[1]}'.format)

df_out

输出：

                     duration0  duration1  km0  km1
Date                                               
2015-03-28 09:07:00        0.0        NaN  0.0  NaN
2015-03-28 09:36:00        NaN        1.0  NaN  2.0
2015-03-30 09:36:00        NaN        1.0  NaN  3.0
2015-03-30 09:37:00        NaN        NaN  5.0  NaN
2015-04-22 09:51:00        3.0        NaN  7.0  NaN
2015-04-23 10:15:00        NaN        0.0  NaN  NaN

现在，让我们将freq_str改为＆＃39; D＆＃39;：

freq_str = 'D'
df_out = df.groupby([pd.Grouper(freq=freq_str, key='Date'),'value'])['duration','km'].agg('mean').unstack()

df_out.columns = df_out.columns.map('{0[0]}{0[1]}'.format)

print(df_out)

输出：

            duration0  duration1  km0  km1
Date                                      
2015-03-28        0.0        1.0  0.0  2.0
2015-03-30        NaN        1.0  5.0  3.0
2015-04-22        3.0        NaN  7.0  NaN
2015-04-23        NaN        0.0  NaN  NaN

Answer 2

您可以将石斑鱼传递到数据透视表的索引。希望这就是你要找的东西，即

ndf = df.pivot_table(values=['duration','km'],columns=['value'],index=pd.Grouper(key='Date', freq='60s'),aggfunc='mean')
ndf.columns = [i[0]+str(i[1]) for i in ndf.columns]

输出：

                     duration0  duration1  km0  km1
Date                                               
2015-03-28 09:07:00        0.0        NaN  0.0  NaN
2015-03-28 09:36:00        NaN        1.0  NaN  2.0
2015-03-30 09:36:00        NaN        1.0  NaN  3.0
2015-03-30 09:37:00        NaN        NaN  5.0  NaN
2015-04-22 09:51:00        3.0        NaN  7.0  NaN
2015-04-23 10:15:00        NaN        0.0  NaN  NaN

如果频率为D则

         duration0  duration1  km0  km1
Date                                      
2015-03-28        0.0        1.0  0.0  2.0
2015-03-30        NaN        1.0  5.0  3.0
2015-04-22        3.0        NaN  7.0  NaN
2015-04-23        NaN        0.0  NaN  NaN

Answer 3

使用groupby

df = df.set_index('Date')    
df.groupby([pd.TimeGrouper('D'), 'value']).mean()

                 duration   km
Date       value               
2017-10-11 0      1.500000  4.0
           1      0.666667  2.5


df.groupby([pd.TimeGrouper('60s'), 'value']).mean()

                           duration   km
Date                value               
2017-10-11 09:07:00 0      0.0       0.0
2017-10-11 09:36:00 1      1.0       2.5
2017-10-11 09:37:00 0     NaN        5.0
2017-10-11 09:51:00 0      3.0       7.0
2017-10-11 10:15:00 1      0.0      NaN

如果你想要它取消堆叠，那么将它拆开。

df.groupby([pd.TimeGrouper('D'), 'value']).mean().unstack()

           duration        km     
value             0    1    0    1
Date                              
2017-10-11 1.50     0.67 4.00 2.50

具有灵活聚合周期的分组熊猫数据帧的平均值

3 个答案: