Python Pandas - 找出两个数据帧之间的差异

3 回答

• 0

  使用 drop_duplicates

  pd.concat([df1,df2]).drop_duplicates(keep=False)
  

  ---

  Update :

  上述方法仅适用于那些数据帧，它们本身没有重复，例如

  df1=pd.DataFrame({'A':[1,2,3,3],'B':[2,3,4,4]})
  df2=pd.DataFrame({'A':[1],'B':[2]})
  

  它输出如下，这是错误的

  输出错误：

  pd.concat([df1, df2]).drop_duplicates(keep=False)
  Out[655]: 
     A  B
  1  2  3
  

  正确的输出

  Out[656]: 
     A  B
  1  2  3
  2  3  4
  3  3  4
  

  ---

  如何实现这一目标？

  使用 isin 与 tuple

  df1[~df1.apply(tuple,1).isin(df2.apply(tuple,1))]
  Out[657]: 
     A  B
  1  2  3
  2  3  4
  3  3  4
  

  回复于 2024-04-29T10:40:52+08:00
• 14

  对于行，请尝试此操作，并将 cols 设置为您要比较的列列表：

  m = df1.merge(df2, on=cols, how='outer', suffixes=['', '_'], indicator=True)
  

  对于列，请尝试以下方法：

  set(df1.columns).symmetric_difference(df2.columns)
  

  回复于 2024-04-29T10:40:52+08:00
• 3

  import pandas as pd
  # given
  df1 = pd.DataFrame({'Name':['John','Mike','Smith','Wale','Marry','Tom','Menda','Bolt','Yuswa',],
      'Age':[23,45,12,34,27,44,28,39,40]})
  df2 = pd.DataFrame({'Name':['John','Smith','Wale','Tom','Menda','Yuswa',],
      'Age':[23,12,34,44,28,40]})
  
  # find elements in df1 that are not in df2
  df_1notin2 = df1[~(df1['Name'].isin(df2['Name']) & df1['Age'].isin(df2['Age']))].reset_index(drop=True)
  
  # output:
  print('df1\n', df1)
  print('df2\n', df2)
  print('df_1notin2\n', df_1notin2)
  
  # df1
  #     Age   Name
  # 0   23   John
  # 1   45   Mike
  # 2   12  Smith
  # 3   34   Wale
  # 4   27  Marry
  # 5   44    Tom
  # 6   28  Menda
  # 7   39   Bolt
  # 8   40  Yuswa
  # df2
  #     Age   Name
  # 0   23   John
  # 1   12  Smith
  # 2   34   Wale
  # 3   44    Tom
  # 4   28  Menda
  # 5   40  Yuswa
  # df_1notin2
  #     Age   Name
  # 0   45   Mike
  # 1   27  Marry
  # 2   39   Bolt
  

  回复于 2024-04-29T10:40:52+08:00