Question

通过创建似乎与this question重复的内容道歉。我的数据框形状或多或少与下面的形状相同：

df_lenght = 240
df = pd.DataFrame(np.random.randn(df_lenght,2), columns=['a','b'] )
df['datetime'] = pd.date_range('23/06/2017', periods=df_lenght, freq='H')

unique_jobs = ['job1','job2','job3',]
job_id = [unique_jobs for i in range (1, int((df_lenght/len(unique_jobs))+1) ,1) ]
df['job_id'] = sorted( [val for sublist in job_id for val in sublist] )

df.set_index(['job_id','datetime'], append=True, inplace=True)

print(df[:5])返回：

                                     a         b
  job_id datetime                               
0 job1   2017-06-23 00:00:00 -0.067011 -0.516382
1 job1   2017-06-23 01:00:00 -0.174199  0.068693
2 job1   2017-06-23 02:00:00 -1.227568 -0.103878
3 job1   2017-06-23 03:00:00 -0.847565 -0.345161
4 job1   2017-06-23 04:00:00  0.028852  3.111738

我需要重新取样df['a']以获得每日滚动均值，即应用.resample('D').mean().rolling(window=2).mean()。

我尝试了两种方法：

1 - 按照建议here

进行拆散和堆叠

df.unstack('job_id','datetime').resample('D').mean().rolling(window=2).mean().stack('job_id', 'datetime')

这会返回错误

2 - 按建议使用pd.Grouper here

level_values = df.index.get_level_values
result = df.groupby( [ level_values(i) for i in [0,1] ] + [ pd.Grouper(freq='D', level=2) ] ).mean().rolling(window=2).mean()

这不会返回错误但似乎没有适当地重新取样/分组df。结果似乎包含每小时数据点，而不是每天：

print(result[:5])
                            a         b
  job_id datetime                      
0 job1   2017-06-23       NaN       NaN
1 job1   2017-06-23  0.831609  1.348970
2 job1   2017-06-23 -0.560047  1.063316
3 job1   2017-06-23 -0.641936 -0.199189
4 job1   2017-06-23  0.254402 -0.328190

Answer 1

首先让我们定义一个重采样器功能：

def resampler(x):    
    return x.set_index('datetime').resample('D').mean().rolling(window=2).mean()

然后，我们通过job_id分组并应用重新采样器功能：

 df.reset_index(level=2).groupby(level=1).apply(resampler)

Out[657]: 
                          a         b
job_id datetime                      
job1   2017-06-23       NaN       NaN
       2017-06-24  0.053378  0.004727
       2017-06-25  0.265074  0.234081
       2017-06-26  0.192286  0.138148
job2   2017-06-26       NaN       NaN
       2017-06-27 -0.016629 -0.041284
       2017-06-28 -0.028662  0.055399
       2017-06-29  0.113299 -0.204670
job3   2017-06-29       NaN       NaN
       2017-06-30  0.233524 -0.194982
       2017-07-01  0.068839 -0.237573
       2017-07-02 -0.051211 -0.069917

如果这就是您的目的，请告诉我。

Answer 2

IIUC，您希望按job_id和（每日）datetime进行分组，并希望忽略DataFrame索引的第一级。因此，而不是按

分组

( [ level_values(i) for i in [0,1] ] + [ pd.Grouper(freq='D', level=2) ] )

你想要分组

[df.index.get_level_values(1), pd.Grouper(freq='D', level=2)]

import numpy as np
import pandas as pd
np.random.seed(2017)

df_length = 240
df = pd.DataFrame(np.random.randn(df_length,2), columns=['a','b'] )
df['datetime'] = pd.date_range('23/06/2017', periods=df_length, freq='H')

unique_jobs = ['job1','job2','job3',]
job_id = [unique_jobs for i in range (1, int((df_length/len(unique_jobs))+1) ,1) ]
df['job_id'] = sorted( [val for sublist in job_id for val in sublist] )

df.set_index(['job_id','datetime'], append=True, inplace=True)

grouped = df.groupby([df.index.get_level_values(1), pd.Grouper(freq='D', level=2)])
result = grouped.mean().rolling(window=2).mean()

print(result)

产量

                          a         b
job_id datetime                      
job1   2017-06-23       NaN       NaN
       2017-06-24 -0.203083  0.176141
       2017-06-25 -0.077083  0.072510
       2017-06-26 -0.237611 -0.493329
job2   2017-06-26 -0.297775 -0.370543
       2017-06-27  0.005124  0.052603
       2017-06-28  0.226142 -0.015584
       2017-06-29 -0.065595  0.210628
job3   2017-06-29 -0.186865  0.347683
       2017-06-30  0.051508  0.029909
       2017-07-01  0.005341  0.075378
       2017-07-02 -0.027131  0.132192

使用包含多索引的时间序列重新采样pandas数据帧

2 个答案: