按数据帧分组的值只有不到一秒的时间 - 熊猫

2 回答

• 1

  （假设你的意思是前两个具有相同的分钟值 . ）

  我不知道如何使用groupby，但这里有相同的结果：

  df=pd.DataFrame({'dt':pd.to_datetime(['2018-12-10 16:35:34.246',
                                        '2018-12-10 16:35:34.243',
                                        '2018-12-10 16:38:34.216',
                                        '2018-12-10 16:42:34.123']),
                                        'value':[1,2,3,4]})
  
              # Select the rows that are greater than a second less
              # And less than a second more
              # Get their value columns and sum them
  df['val'] = [df[(df.dt>t-pd.Timedelta(seconds=1))&
                  (df.dt<t+pd.Timedelta(seconds=1))]['value'].sum()
               for t in df.dt]
  
                         dt  value  val
  0 2018-12-10 16:35:34.246      1    3
  1 2018-12-10 16:35:34.243      2    3
  2 2018-12-10 16:38:34.216      3    3
  3 2018-12-10 16:42:34.123      4    4
  

  ---

  作为旁注，我研究了 groupby 做同样的事情，但我无法弄清楚如何让它发挥作用 . 您可以将函数传递给 groupby 方法 . 如果您选择走这条路线，请注意该功能是接收Dataframe的索引 . 让's me think it would be hard to use the groupby since I don' t知道一行可以属于多个组......

  回复于 2024-04-23T15:03:07+08:00
• 1

  一个强力解决方案是获取 datetime 系列和每个 datetime 值之间的绝对差值，然后与阈值进行比较：

  # data from @StephenCowley
  
  threshold = pd.Timedelta(seconds=1)
  
  df['val'] = [df.loc[(df['dt'] - t).abs() < threshold, 'value'].sum()
               for t in df['dt']]
  
  print(df)
  
                         dt  value  val
  0 2018-12-10 16:35:34.246      1    3
  1 2018-12-10 16:35:34.243      2    3
  2 2018-12-10 16:38:34.216      3    3
  3 2018-12-10 16:42:34.123      4    4
  

  回复于 2024-04-23T15:03:07+08:00