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pandas _合并 merge

pandas _合并 merge

作者: Ledestin | 来源:发表于2017-05-21 13:55 被阅读128次

    pandas中的merge和concat类似,但主要是用于两组有key column的数据,统一索引的数据. 通常也被用在Database的处理当中.
    1.依据一组key合并
    2.依据两组key合并
    3.indicator
    4.依据index合并
    5.解决overlapping的问题


    Demo.py

    #依据一组key合并
    import pandas as pd
    #定义资料集并打印出
    left = pd.DataFrame({'key': ['K0', 'K1', 'K2', 'K3'],
                                 'A': ['A0', 'A1', 'A2', 'A3'],
                                 'B': ['B0', 'B1', 'B2', 'B3']})
    right = pd.DataFrame({'key': ['K0', 'K1', 'K2', 'K3'],
                                  'C': ['C0', 'C1', 'C2', 'C3'],
                                  'D': ['D0', 'D1', 'D2', 'D3']})
    print(left)
    print(right)
    #依据key column合并,并打印出
    res = pd.merge(left, right, on='key')
    print(res)
    
    #依据两组key合并
    #合并时有4种方法how = ['left', 'right', 'outer', 'inner'],预设值how='inner'
    import pandas as pd
    #定义资料集并打印出
    left = pd.DataFrame({'key1': ['K0', 'K0', 'K1', 'K2'],
                          'key2': ['K0', 'K1', 'K0', 'K1'],
                          'A': ['A0', 'A1', 'A2', 'A3'],
                          'B': ['B0', 'B1', 'B2', 'B3']})
    right = pd.DataFrame({'key1': ['K0', 'K1', 'K1', 'K2'],
                           'key2': ['K0', 'K0', 'K0', 'K0'],
                           'C': ['C0', 'C1', 'C2', 'C3'],
                           'D': ['D0', 'D1', 'D2', 'D3']})
    
    print(left)
    print(right)
    #依据key1与key2 columns进行合并,并打印出四种结果['left', 'right', 'outer', 'inner']
    res = pd.merge(left, right, on=['key1', 'key2'], how='inner')
    print(res)
    res = pd.merge(left, right, on=['key1', 'key2'], how='outer')
    print(res)
    res = pd.merge(left, right, on=['key1', 'key2'], how='left')
    print(res)
    res = pd.merge(left, right, on=['key1', 'key2'], how='right')
    print(res)
    
    #Indicator
    #indicator=True会将合并的记录放在新的一列。
    import pandas as pd
    #定义资料集并打印出
    df1 = pd.DataFrame({'col1':[0,1], 'col_left':['a','b']})
    df2 = pd.DataFrame({'col1':[1,2,2],'col_right':[2,2,2]})
    print(df1)
    print(df2)
    # 依据col1进行合并,并启用indicator=True,最后打印出
    res = pd.merge(df1, df2, on='col1', how='outer', indicator=True)
    print(res)
    # 自定indicator column的名称,并打印出
    res = pd.merge(df1, df2, on='col1', how='outer', indicator='indicator_column')
    print(res)
    
    #依据index合并
    import pandas as pd
    #定义资料集并打印出
    left = pd.DataFrame({'A': ['A0', 'A1', 'A2'],
                         'B': ['B0', 'B1', 'B2']},
                         index=['K0', 'K1', 'K2'])
    right = pd.DataFrame({'C': ['C0', 'C2', 'C3'],
                          'D': ['D0', 'D2', 'D3']},
                         index=['K0', 'K2', 'K3'])
    print(left)
    print(right)
    #依据左右资料集的index进行合并,how='outer',并打印出
    res = pd.merge(left, right, left_index=True, right_index=True, how='outer')
    print(res)
    #依据左右资料集的index进行合并,how='inner',并打印出
    res = pd.merge(left, right, left_index=True, right_index=True, how='inner')
    print(res)
    
    #解决overlapping的问题
    import pandas as pd
    #定义资料集
    boys = pd.DataFrame({'k': ['K0', 'K1', 'K2'], 'age': [1, 2, 3]})
    girls = pd.DataFrame({'k': ['K0', 'K0', 'K3'], 'age': [4, 5, 6]})
    #使用suffixes解决overlapping的问题
    res = pd.merge(boys, girls, on='k', suffixes=['_boy', '_girl'], how='inner')
    print(res)
    

    结果:

        A   B key
    0  A0  B0  K0
    1  A1  B1  K1
    2  A2  B2  K2
    3  A3  B3  K3
        C   D key
    0  C0  D0  K0
    1  C1  D1  K1
    2  C2  D2  K2
    3  C3  D3  K3
        A   B key   C   D
    0  A0  B0  K0  C0  D0
    1  A1  B1  K1  C1  D1
    2  A2  B2  K2  C2  D2
    3  A3  B3  K3  C3  D3
        A   B key1 key2
    0  A0  B0   K0   K0
    1  A1  B1   K0   K1
    2  A2  B2   K1   K0
    3  A3  B3   K2   K1
        C   D key1 key2
    0  C0  D0   K0   K0
    1  C1  D1   K1   K0
    2  C2  D2   K1   K0
    3  C3  D3   K2   K0
        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
         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
        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
         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
    3  NaN  NaN   K2   K0  C3  D3
       col1 col_left
    0     0        a
    1     1        b
       col1  col_right
    0     1          2
    1     2          2
    2     2          2
       col1 col_left  col_right      _merge
    0     0        a        NaN   left_only
    1     1        b        2.0        both
    2     2      NaN        2.0  right_only
    3     2      NaN        2.0  right_only
       col1 col_left  col_right indicator_column
    0     0        a        NaN        left_only
    1     1        b        2.0             both
    2     2      NaN        2.0       right_only
    3     2      NaN        2.0       right_only
         A   B
    K0  A0  B0
    K1  A1  B1
    K2  A2  B2
         C   D
    K0  C0  D0
    K2  C2  D2
    K3  C3  D3
          A    B    C    D
    K0   A0   B0   C0   D0
    K1   A1   B1  NaN  NaN
    K2   A2   B2   C2   D2
    K3  NaN  NaN   C3   D3
         A   B   C   D
    K0  A0  B0  C0  D0
    K2  A2  B2  C2  D2
       age_boy   k  age_girl
    0        1  K0         4
    1        1  K0         5
    

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