python – 如何找到两个数据帧的补充

给定两个大型数据帧,是否有任何简洁有效的代码(避免直接使用任何for循环),这使我能够获得这两个数据帧的补充？

对我来说最直接的方法是计算union-intersection,如下面的天真示例所示,但我不知道如何用优雅的pandas或np语言实现它

df1= pd.DataFrame({'key1': ['K0', 'K0', 'K1', 'K2'],
                     'key2': ['K0', 'K1', 'K0', 'K1'],
                   'A': ['A0', 'A1', 'A2', 'A3'],
                    'B': ['B0', 'B1', 'B2', 'B3']})     
df2= pd.DataFrame({'key1': ['K0', 'K1', 'K1', 'K2'],
                      'key2': ['K0', 'K0', 'K0', 'K0'],
                      'C': ['C0', 'C1', 'C2', 'C3'],
                      'D': ['D0', 'D1', 'D2', 'D3']})        
intersection= pd.merge(df1, df2, how='inner',on=['key1', 'key2'])
union=pd.merge(df1, df2, how='outer',on=['key1', 'key2'])       


complement=union-intersection

感谢您的任何意见和解答

解决方法:

从这开始：

df1= pd.DataFrame({'key1': ['K0', 'K0', 'K1', 'K2'],
                     'key2': ['K0', 'K1', 'K0', 'K1'],
                   'A': ['A0', 'A1', 'A2', 'A3'],
                    'B': ['B0', 'B1', 'B2', 'B3']})     
df2= pd.DataFrame({'key1': ['K0', 'K1', 'K1', 'K2'],
                      'key2': ['K0', 'K0', 'K0', 'K0'],
                      'C': ['C0', 'C1', 'C2', 'C3'],
                      'D': ['D0', 'D1', 'D2', 'D3']})        
intersection  = pd.merge(df1, df2, how='inner',on=['key1', 'key2'])
union         = pd.merge(df1, df2, how='outer',on=['key1', 'key2'])       

印刷联盟

     A    B key1 key2    C    D
0   A0   B0   K0   K0   C0   D0
1   A1   B1   K0   K1  NaN  NaN
2   A2   B2   K1   K0   C1   D1
3   A2   B2   K1   K0   C2   D2
4   A3   B3   K2   K1  NaN  NaN
5  NaN  NaN   K2   K0   C3   D3

印刷交叉口

    A   B key1 key2   C   D
0  A0  B0   K0   K0  C0  D0
1  A2  B2   K1   K0  C1  D1
2  A2  B2   K1   K0  C2  D2

union-intersection尝试这个：

union[union.isnull().any(axis=1)]

     A    B key1 key2    C    D
1   A1   B1   K0   K1  NaN  NaN
4   A3   B3   K2   K1  NaN  NaN
5  NaN  NaN   K2   K0   C3   D3