3-6 merge操做

 

In [1]:

import pandas as pd

In [6]:

left =pd.DataFrame({ 'A':['A0','A1','A2','A3'],
                   'B':['B0','B1','B2','B3'],
                   'key':['K0','K1','K2','K3'],})
right =pd.DataFrame({ 'C':['C0','C1','C2','C3'],
                   'D':['D0','D1','D2','D3'],
                    'key':['K0','K1','K2','K3'],})

In [7]:

left

Out[7]:

 

	A	B	key
0	A0	B0	K0
1	A1	B1	K1
2	A2	B2	K2
3	A3	B3	K3

In [8]:

right

Out[8]:

 

	C	D	key
0	C0	D0	K0
1	C1	D1	K1
2	C2	D2	K2
3	C3	D3	K3

 

merge:合并

In [10]:

pd.merge(left,right)#直接合并，重复的就再也不显示

Out[10]:

 

	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

In [12]:

pd.merge(left,right,on='key')#以key为界进行合并

Out[12]:

 

	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

In [13]:

left =pd.DataFrame({ 'A':['A0','A1','A2','A3'],
                   'B':['B0','B1','B2','B3'],
                   'key1':['K0','K1','K2','K3'],
                    'key2':['K0','K1','K2','K3']})
right =pd.DataFrame({ 'C':['C0','C1','C2','C3'],
                   'D':['D0','D1','D2','D3'],
                    'key1':['K0','K1','K2','K3'],
                    'key2':['K0','K1','K2','K3']})

In [14]:

left

Out[14]:

 

	A	B	key1	key2
0	A0	B0	K0	K0
1	A1	B1	K1	K1
2	A2	B2	K2	K2
3	A3	B3	K3	K3

In [16]:

right

Out[16]:

 

	C	D	key1	key2
0	C0	D0	K0	K0
1	C1	D1	K1	K1
2	C2	D2	K2	K2
3	C3	D3	K3	K3

In [17]:

pd.merge(left,right)#直接合并，重复的就再也不显示

Out[17]:

 

	A	B	key1	key2	C	D
0	A0	B0	K0	K0	C0	D0
1	A1	B1	K1	K1	C1	D1
2	A2	B2	K2	K2	C2	D2
3	A3	B3	K3	K3	C3	D3

In [18]:

pd.merge(left,right,on='key1')#以key1为界进行合并,key2自动分组

Out[18]:

 

	A	B	key1	key2_x	C	D	key2_y
0	A0	B0	K0	K0	C0	D0	K0
1	A1	B1	K1	K1	C1	D1	K1
2	A2	B2	K2	K2	C2	D2	K2
3	A3	B3	K3	K3	C3	D3	K3

In [19]:

pd.merge(left,right,on=['key1','key2'])#以key1,key2为界进行合并

Out[19]:

 

	A	B	key1	key2	C	D
0	A0	B0	K0	K0	C0	D0
1	A1	B1	K1	K1	C1	D1
2	A2	B2	K2	K2	C2	D2
3	A3	B3	K3	K3	C3	D3

 

使key2的值不彻底同样，right改为K4

In [20]:

right =pd.DataFrame({ 'C':['C0','C1','C2','C3'],
                   'D':['D0','D1','D2','D3'],
                    'key1':['K0','K1','K2','K3'],
                    'key2':['K0','K1','K2','K4']})

In [21]:

pd.merge(left,right,on=['key1','key2'])#以key1,key2为界进行合并,可是key2不一样的那一行就被删除

Out[21]:

 

	A	B	key1	key2	C	D
0	A0	B0	K0	K0	C0	D0
1	A1	B1	K1	K1	C1	D1
2	A2	B2	K2	K2	C2	D2

In [22]:

pd.merge(left,right,on=['key1','key2'],how='outer')#how='outer'为并集，可是默认是交集

Out[22]:

 

	A	B	key1	key2	C	D
0	A0	B0	K0	K0	C0	D0
1	A1	B1	K1	K1	C1	D1
2	A2	B2	K2	K2	C2	D2
3	A3	B3	K3	K3	NaN	NaN
4	NaN	NaN	K3	K4	C3	D3

In [23]:

pd.merge(left,right,on=['key1','key2'],how='outer',indicator=True)#指定当前的merge是交集仍是并集

Out[23]:

 

	A	B	key1	key2	C	D	_merge
0	A0	B0	K0	K0	C0	D0	both
1	A1	B1	K1	K1	C1	D1	both
2	A2	B2	K2	K2	C2	D2	both
3	A3	B3	K3	K3	NaN	NaN	left_only
4	NaN	NaN	K3	K4	C3	D3	right_only

In [24]:

pd.merge(left,right,on=['key1','key2'],how='left')#how='left'指定以left为基准，也能够指定其余为基准

Out[24]:

 

	A	B	key1	key2	C	D
0	A0	B0	K0	K0	C0	D0
1	A1	B1	K1	K1	C1	D1
2	A2	B2	K2	K2	C2	D2
3	A3	B3	K3	K3	NaN	NaN