dataframe的合并（append, merge, concat）

Posted 2020-12-29

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1，pd.concat：拼接
1.1，axis
df1 = pd.DataFrame(np.ones((3,4))*0, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])
df2 = pd.DataFrame(np.ones((3,4))*1, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])
df3 = pd.DataFrame(np.ones((3,4))*2, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])

a b c d
0 0.0 0.0 0.0 0.0
1 0.0 0.0 0.0 0.0
2 0.0 0.0 0.0 0.0

a b c d
0 1.0 1.0 1.0 1.0
1 1.0 1.0 1.0 1.0
2 1.0 1.0 1.0 1.0

a b c d
0 2.0 2.0 2.0 2.0
1 2.0 2.0 2.0 2.0
2 2.0 2.0 2.0 2.0

result = pd.concat([df1, df2, df3], axis = 0) # 0表示竖向的合并

a b c d
0 0.0 0.0 0.0 0.0
1 0.0 0.0 0.0 0.0
2 0.0 0.0 0.0 0.0
0 1.0 1.0 1.0 1.0
1 1.0 1.0 1.0 1.0
2 1.0 1.0 1.0 1.0
0 2.0 2.0 2.0 2.0
1 2.0 2.0 2.0 2.0
2 2.0 2.0 2.0 2.0

result=pd.concat([df1,df2,df3], axis=1) # 1表示横向的合并
a b c d a b c d a b c d
0 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0
1 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0
2 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0

1.2，ignore_index
df1 = pd.DataFrame(np.ones((3,4))*0, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])
df2 = pd.DataFrame(np.ones((3,4))*1, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])
df3 = pd.DataFrame(np.ones((3,4))*2, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])
result = pd.concat([df1, df2, df3], axis = 0, ignore_index=True) # ignore_index=True表示忽略原来index(axis=0)/column(axis=1)
a b c d
0 0.0 0.0 0.0 0.0
1 0.0 0.0 0.0 0.0
2 0.0 0.0 0.0 0.0
3 1.0 1.0 1.0 1.0
4 1.0 1.0 1.0 1.0
5 1.0 1.0 1.0 1.0
6 2.0 2.0 2.0 2.0
7 2.0 2.0 2.0 2.0
8 2.0 2.0 2.0 2.0

result = pd.concat([df1, df2, df3], axis = 0, ignore_index=True) # ignore_index=True表示忽略原来index
0 1 2 3 4 5 6 7 8 9 10 11
0 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0
1 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0
2 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 2.0 2.0 2.0 2.0

1.3，join
df1 = pd.DataFrame(np.ones((3,4))*0, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘])
df2 = pd.DataFrame(np.ones((3,4))*1, columns = [‘b‘, ‘c‘, ‘d‘, ‘e‘])
result=pd.concat([df1,df2], axis=0, join=‘inner‘, ignore_index=True) # join=‘inner‘表示只保留column名相同的部分
b c d
0 0.0 0.0 0.0
1 0.0 0.0 0.0
2 0.0 0.0 0.0
3 1.0 1.0 1.0
4 1.0 1.0 1.0
5 1.0 1.0 1.0

1.4, join_axes
df1 = pd.DataFrame(np.ones((3,4))*0, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘], index=[1, 2, 3])
df2 = pd.DataFrame(np.ones((3,4))*1, columns = [‘b‘, ‘c‘, ‘d‘, ‘e‘], index=[2, 3, 4])

result=pd.concat([df1,df2], axis=1)
a b c d b c d e
1 0.0 0.0 0.0 0.0 NaN NaN NaN NaN
2 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0
3 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0
4 NaN NaN NaN NaN 1.0 1.0 1.0 1.0

result=pd.concat([df1,df2], axis=1, join_axes=[df1.index]) # join_axes=[df1.index]表示按照df1的index进行合并
a b c d b c d e
1 0.0 0.0 0.0 0.0 NaN NaN NaN NaN
2 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0
3 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0

2，pd.append()
2.1，append
pd.append只有竖向往下加数据，没有axis属性
df1 = pd.DataFrame(np.ones((3,4))*0, columns = [‘a‘, ‘b‘, ‘c‘, ‘d‘], index=[1, 2, 3])
df2 = pd.DataFrame(np.ones((3,4))*1, columns = [‘b‘, ‘c‘, ‘d‘, ‘e‘], index=[2, 3, 4])
df3 = pd.DataFrame(np.ones((3,4))*2, columns = [‘b‘, ‘c‘, ‘d‘, ‘e‘], index=[2, 3, 4])

df1 = df1.append(df2)
a b c d e
1 0.0 0.0 0.0 0.0 NaN
2 0.0 0.0 0.0 0.0 NaN
3 0.0 0.0 0.0 0.0 NaN
2 NaN 1.0 1.0 1.0 1.0
3 NaN 1.0 1.0 1.0 1.0
4 NaN 1.0 1.0 1.0 1.0

df1 = df1.append([df2, df3], ignore_index=True) # 可以一次加多个df，也可以忽略index
a b c d e
0 0.0 0.0 0.0 0.0 NaN
1 0.0 0.0 0.0 0.0 NaN
2 0.0 0.0 0.0 0.0 NaN
3 NaN 1.0 1.0 1.0 1.0
4 NaN 1.0 1.0 1.0 1.0
5 NaN 1.0 1.0 1.0 1.0
6 NaN 2.0 2.0 2.0 2.0
7 NaN 2.0 2.0 2.0 2.0
8 NaN 2.0 2.0 2.0 2.0

s1=pd.Series([1,2,3,4],index=[‘a‘,‘b‘,‘c‘,‘d‘])
df1 = df1.append(s1, ignore_index=True)   # 可以加series，ignore_index=True必须加上，s1个数必须和df1的列数相同
     a    b    c    d
0 0.0 0.0 0.0 0.0   # 注意ignore_index = True以后，df1的index变成从0开始了！！！
1 0.0 0.0 0.0 0.0
2 0.0 0.0 0.0 0.0
3 1.0 2.0 3.0 4.0

l1=[1,2,3,4]
df1.loc[4] = l1
a b c d
1 0.0 0.0 0.0 0.0 # df1的index还是按照原来从1开始
2 0.0 0.0 0.0 0.0
3 0.0 0.0 0.0 0.0
4 1.0 2.0 3.0 4.0

2.2，append举例
df1 = pd.DataFrame({‘B‘: [‘B0‘, ‘B1‘, ‘B2‘, ‘B3‘,‘B4‘],
                 ‘D‘: [‘D0‘, ‘D1‘, ‘D2‘, ‘D3‘,‘D4‘],
                 ‘A‘: [‘A0‘, ‘A1‘, ‘A2‘, ‘A3‘,‘A4‘],
                 ‘E‘: [‘E0‘, ‘E1‘, ‘E2‘, ‘E3‘,‘E4‘]},
                  index=[0, 1, 2, 3,4],columns=[‘B‘,‘D‘,‘A‘,‘E‘])

df2 = pd.DataFrame({‘F‘: [‘F4‘, ‘F5‘, ‘F6‘, ‘F7‘,‘F8‘],
                ‘A‘: [‘A4‘, ‘A5‘, ‘A6‘, ‘A7‘,‘A8‘],
                ‘B‘: [‘B4‘, ‘B5‘, ‘B6‘, ‘B7‘,‘B8‘],
                ‘C‘: [‘C4‘, ‘C5‘, ‘C6‘, ‘C7‘,‘C8‘]},
                index=[5, 9, 6, 7,10])

输入df3=df1.append(df2)后，df3变为：
        A    B     C     D     E     F
  0   A0 B0 NaN   D0   E0 NaN
  1   A1 B1 NaN   D1   E1 NaN
  2   A2 B2 NaN   D2   E2 NaN
  3   A3 B3 NaN   D3   E3 NaN
  4   A4 B4 NaN   D4   E4 NaN
  5   A4 B4   C4 NaN NaN   F4
  9   A5 B5   C5 NaN NaN   F5
  6   A6 B6   C6 NaN NaN   F6
  7   A7 B7   C7 NaN NaN   F7
  10 A8 B8   C8 NaN NaN   F8

列项重新排列了，如果想要列项按照df1的走，即df1的次序，并且如果df1没有的列，就直接扔掉：

df4=pd.concat([df1,df2],axis=0)

df4.reindex(columns=df1.columns)

3，pd.merge # 合并
3.1，1个key
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‘]})

res=pd.merge(left,right)
res=pd.merge(left,right,on=[‘key‘]) # 默认会自动选择共同的column进行合并，如果没有共同的column会报错
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

3.2，多个key，how可取[‘left‘, ‘right‘, ‘inner‘, ‘outer‘]
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‘]})
    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

res=pd.merge(left,right,on=[‘key1‘,‘key2‘])   # 需要key1，key2完全相同才合并，left有一行K1/K0，right有两行K1/K0，则打印2遍
res=pd.merge(left,right,on=[‘key1‘,‘key2‘], how = ‘inner‘) # 默认是how=‘inner‘，即key的value要相同
     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

res=pd.merge(left,right,on=[‘key1‘,‘key2‘], how = ‘outer‘) # 根据key1，key2取值全部合并，如果left或者right没有的取NaN
      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

res=pd.merge(left,right,on=[‘key1‘,‘key2‘], how = ‘left‘) # 根据left的key合并，重复出现了2行K1K0（因为right有2行）
    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

res=pd.merge(left,right,on=[‘key1‘,‘key2‘],how=‘right‘) # 根据right的key合并
     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

3.3, indicator
默认false，true表示显示合并方式，是left还是right，还是both

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