2 回答

• 0

  它可以在 iterrows() 函数的帮助下完成 .

  这是代码：

  first table creation ：

  value=[(0,11,10,20),(1,22,11,21),(2,33,12,22),(3,44,13,23),(4,55,14,24), 
  (5,66,15,25),(6,77,16,26),(7,88,17,27),(8,99,18,28)]
  header=["index","time","vel","yaw"]
  df1 = pd.DataFrame.from_records(value, columns=header)
  

  

  second table creation ：

  value=[(0,67,"nan","nan"),(1,75,"nan" ,"nan" ),(2,87,"nan" ,"nan" ) 
  (3,99,"nan" ,"nan" )]
  header=["index","time","vel","yaw"]
  df2 = pd.DataFrame.from_records(value, columns=header)
  

  

  storing the result 在df2本身：

  for index, row in df2.iterrows():
      min=10000000
      for indexer, rows in df1.iterrows():
          if abs(row['time']-rows['time'])<min:
              min = abs(row['time']-rows['time'])
              #storing the position 
              pos = indexer
      df2.loc[index,'vel'] = df1['vel'][pos]
      df2.loc[index,'yaw'] = df1['yaw'][pos]
  

  

  回复于 2024-04-19T10:22:28+08:00
• 0

  创建笛卡尔积，然后进行过滤 -

  df1 = pd.DataFrame({'time': [11, 22, 33,44, 55, 66,77,88,99], 'vel':[10, 11,12,13,14,15,16,17,18],
                     'yaw' : [20, 21, 22,23,24,25,26,27,28]})
  df2 = pd.DataFrame({'time' : [67, 75, 87, 99]})
  
  df1['key'] = 1
  df2['key'] = 1
  df1.rename(index=str, columns ={'time' : 'time_x'}, inplace=True)
  
  df = df2.merge(df1, on='key', how ='left').reset_index()
  df['diff'] = df.apply(lambda x: abs(x['time']  - x['time_x']), axis=1)
  df.sort_values(by=['time', 'diff'], inplace=True)
  
  df=df.groupby(['time']).first().reset_index()[['time', 'vel', 'yaw']]
  

  回复于 2024-04-19T10:22:28+08:00