1

我想以按列分组的频率data使用前向填充对列进行重新采样：ffill1mindfid

df：

          id   timestamp                data  

      1    1   2017-01-02 13:14:53.040  10.0
      2    1   2017-01-02 16:04:43.240  11.0  
                           ...
      4    2   2017-01-02 15:22:06.540   1.0  
      5    2   2017-01-03 13:55:34.240   2.0  
                           ...

预期输出：

           id   timestamp                data  

      1    1   2017-01-02 13:14:53.040  10.0
               2017-01-02 13:14:54.040  10.0
               2017-01-02 13:14:55.040  10.0
               2017-01-02 13:14:56.040  10.0
                           ...

      2    1   2017-01-02 16:04:43.240  11.0  
               2017-01-02 16:04:44.240  11.0
               2017-01-02 16:04:45.240  11.0
               2017-01-02 16:04:46.240  11.0
                           ...

      4    2   2017-01-02 15:22:06.540   1.0  
               2017-01-02 15:22:07.540   1.0
               2017-01-02 15:22:08.540   1.0
               2017-01-02 15:22:09.540   1.0
                           ...

      5    2   2017-01-03 13:55:34.240   2.0
               2017-01-03 13:55:35.240   2.0
               2017-01-03 13:55:36.240   2.0
               2017-01-03 13:55:37.240   2.0
                           ...

类似这篇文章的东西，但我试过了：

df.set_index('timestamp').groupby('id').resample('1min').asfreq().drop(['id'], 1).reset_index()

并且data列仅返回NaN值：


    id  timestamp               data
0   1   2017-01-02 13:14:53.040 NaN
1   1   2017-01-02 13:14:54.040 NaN
2   1   2017-01-02 13:14:55.040 NaN
3   1   2017-01-02 13:14:56.040 NaN
4   1   2017-01-02 13:14:57.040 NaN
... ... ... ...

编辑：

  1. 第二行df timestamp从2017-01-02 12:04:43.240变为2017-01-02 16:04:43.240，即属于同一行的行id应该被排序。
  2. 我误认为预期输出中的第二个是最小的，但@jezrael 的答案是正确的。
4

1 回答 1

1

Timedelta使用自定义函数通过withdate_range和定义需要多少新行DataFrame.reindex：

def f(x):
    new = x.index[0] + pd.Timedelta(5, unit='Min')
    r = pd.date_range(x.index[0], new, freq='Min')
    
    return x.reindex(r, method='ffill')


df = (df.reset_index()
        .set_index('timestamp')
        .groupby(['index','id'], sort=False)['data']
        .apply(f)
        .reset_index(level=0, drop=True)
        .rename_axis(['id','timestamp'])
        .reset_index()
        )

print (df)
    id               timestamp  data
0    1 2017-01-02 13:14:53.040  10.0
1    1 2017-01-02 13:15:53.040  10.0
2    1 2017-01-02 13:16:53.040  10.0
3    1 2017-01-02 13:17:53.040  10.0
4    1 2017-01-02 13:18:53.040  10.0
5    1 2017-01-02 13:19:53.040  10.0
6    1 2017-01-02 12:04:43.240  11.0
7    1 2017-01-02 12:05:43.240  11.0
8    1 2017-01-02 12:06:43.240  11.0
9    1 2017-01-02 12:07:43.240  11.0
10   1 2017-01-02 12:08:43.240  11.0
11   1 2017-01-02 12:09:43.240  11.0
12   2 2017-01-02 15:22:06.540   1.0
13   2 2017-01-02 15:23:06.540   1.0
14   2 2017-01-02 15:24:06.540   1.0
15   2 2017-01-02 15:25:06.540   1.0
16   2 2017-01-02 15:26:06.540   1.0
17   2 2017-01-02 15:27:06.540   1.0
18   2 2017-01-03 13:55:34.240   2.0
19   2 2017-01-03 13:56:34.240   2.0
20   2 2017-01-03 13:57:34.240   2.0
21   2 2017-01-03 13:58:34.240   2.0
22   2 2017-01-03 13:59:34.240   2.0
23   2 2017-01-03 14:00:34.240   2.0

因为如果使用ffill输出不同：

df = df.set_index('timestamp').groupby('id', sort=False)['data'].resample('1min').ffill()
print (df)
id  timestamp          
1   2017-01-02 12:04:00     NaN
    2017-01-02 12:05:00    11.0
    2017-01-02 12:06:00    11.0
    2017-01-02 12:07:00    11.0
    2017-01-02 12:08:00    11.0

2   2017-01-03 13:51:00     1.0
    2017-01-03 13:52:00     1.0
    2017-01-03 13:53:00     1.0
    2017-01-03 13:54:00     1.0
    2017-01-03 13:55:00     1.0
Name: data, Length: 1425, dtype: float64
于 2021-03-12T05:43:42.277 回答