📅  最后修改于: 2020-11-06 05:47:46             🧑  作者: Mango
Pandas具有与SQL等关系数据库非常相似的功能齐全的高性能内存中连接操作。
Pandas提供单个函数merge作为DataFrame对象之间所有标准数据库联接操作的入口点-
pd.merge(left, right, how='inner', on=None, left_on=None, right_on=None,
left_index=False, right_index=False, sort=True)
在这里,我们使用了以下参数-
左-一个DataFrame对象。
右-另一个DataFrame对象。
在-列(名)加入上。必须在左右DataFrame对象中都找到。
left_on-左侧DataFrame中的列用作键。可以是列名,也可以是长度等于DataFrame长度的数组。
right_on-右侧DataFrame中的列用作键。可以是列名,也可以是长度等于DataFrame长度的数组。
left_index-如果为True,则使用左侧DataFrame的索引(行标签)作为其连接键。如果DataFrame具有MultiIndex(分层),则级别数必须与右侧DataFrame中的连接键数匹配。
right_index -相同的使用作为left_index为正确的数据帧。
怎么样-“左”,“右”,“外”,“内”之一。默认为内部。每种方法已在下面描述。
排序-按该结果数据框中加入字典顺序按键。默认情况下为True,在许多情况下,设置为False会大大提高性能。
现在让我们创建两个不同的DataFrame并对其执行合并操作。
# import the pandas library
import pandas as pd
left = pd.DataFrame({
'id':[1,2,3,4,5],
'Name': ['Alex', 'Amy', 'Allen', 'Alice', 'Ayoung'],
'subject_id':['sub1','sub2','sub4','sub6','sub5']})
right = pd.DataFrame(
{'id':[1,2,3,4,5],
'Name': ['Billy', 'Brian', 'Bran', 'Bryce', 'Betty'],
'subject_id':['sub2','sub4','sub3','sub6','sub5']})
print left
print right
其输出如下-
Name id subject_id
0 Alex 1 sub1
1 Amy 2 sub2
2 Allen 3 sub4
3 Alice 4 sub6
4 Ayoung 5 sub5
Name id subject_id
0 Billy 1 sub2
1 Brian 2 sub4
2 Bran 3 sub3
3 Bryce 4 sub6
4 Betty 5 sub5
import pandas as pd
left = pd.DataFrame({
'id':[1,2,3,4,5],
'Name': ['Alex', 'Amy', 'Allen', 'Alice', 'Ayoung'],
'subject_id':['sub1','sub2','sub4','sub6','sub5']})
right = pd.DataFrame({
'id':[1,2,3,4,5],
'Name': ['Billy', 'Brian', 'Bran', 'Bryce', 'Betty'],
'subject_id':['sub2','sub4','sub3','sub6','sub5']})
print pd.merge(left,right,on='id')
其输出如下-
Name_x id subject_id_x Name_y subject_id_y
0 Alex 1 sub1 Billy sub2
1 Amy 2 sub2 Brian sub4
2 Allen 3 sub4 Bran sub3
3 Alice 4 sub6 Bryce sub6
4 Ayoung 5 sub5 Betty sub5
import pandas as pd
left = pd.DataFrame({
'id':[1,2,3,4,5],
'Name': ['Alex', 'Amy', 'Allen', 'Alice', 'Ayoung'],
'subject_id':['sub1','sub2','sub4','sub6','sub5']})
right = pd.DataFrame({
'id':[1,2,3,4,5],
'Name': ['Billy', 'Brian', 'Bran', 'Bryce', 'Betty'],
'subject_id':['sub2','sub4','sub3','sub6','sub5']})
print pd.merge(left,right,on=['id','subject_id'])
其输出如下-
Name_x id subject_id Name_y
0 Alice 4 sub6 Bryce
1 Ayoung 5 sub5 Betty
合并的how参数指定如何确定要在结果表中包括哪些键。如果左侧或右侧表中均未出现组合键,则联接表中的值为NA。
这里的如何选择和他们的SQL等价的名字摘要-
Merge Method | SQL Equivalent | Description |
---|---|---|
left | LEFT OUTER JOIN | Use keys from left object |
right | RIGHT OUTER JOIN | Use keys from right object |
outer | FULL OUTER JOIN | Use union of keys |
inner | INNER JOIN | Use intersection of keys |
import pandas as pd
left = pd.DataFrame({
'id':[1,2,3,4,5],
'Name': ['Alex', 'Amy', 'Allen', 'Alice', 'Ayoung'],
'subject_id':['sub1','sub2','sub4','sub6','sub5']})
right = pd.DataFrame({
'id':[1,2,3,4,5],
'Name': ['Billy', 'Brian', 'Bran', 'Bryce', 'Betty'],
'subject_id':['sub2','sub4','sub3','sub6','sub5']})
print pd.merge(left, right, on='subject_id', how='left')
其输出如下-
Name_x id_x subject_id Name_y id_y
0 Alex 1 sub1 NaN NaN
1 Amy 2 sub2 Billy 1.0
2 Allen 3 sub4 Brian 2.0
3 Alice 4 sub6 Bryce 4.0
4 Ayoung 5 sub5 Betty 5.0
import pandas as pd
left = pd.DataFrame({
'id':[1,2,3,4,5],
'Name': ['Alex', 'Amy', 'Allen', 'Alice', 'Ayoung'],
'subject_id':['sub1','sub2','sub4','sub6','sub5']})
right = pd.DataFrame({
'id':[1,2,3,4,5],
'Name': ['Billy', 'Brian', 'Bran', 'Bryce', 'Betty'],
'subject_id':['sub2','sub4','sub3','sub6','sub5']})
print pd.merge(left, right, on='subject_id', how='right')
其输出如下-
Name_x id_x subject_id Name_y id_y
0 Amy 2.0 sub2 Billy 1
1 Allen 3.0 sub4 Brian 2
2 Alice 4.0 sub6 Bryce 4
3 Ayoung 5.0 sub5 Betty 5
4 NaN NaN sub3 Bran 3
import pandas as pd
left = pd.DataFrame({
'id':[1,2,3,4,5],
'Name': ['Alex', 'Amy', 'Allen', 'Alice', 'Ayoung'],
'subject_id':['sub1','sub2','sub4','sub6','sub5']})
right = pd.DataFrame({
'id':[1,2,3,4,5],
'Name': ['Billy', 'Brian', 'Bran', 'Bryce', 'Betty'],
'subject_id':['sub2','sub4','sub3','sub6','sub5']})
print pd.merge(left, right, how='outer', on='subject_id')
其输出如下-
Name_x id_x subject_id Name_y id_y
0 Alex 1.0 sub1 NaN NaN
1 Amy 2.0 sub2 Billy 1.0
2 Allen 3.0 sub4 Brian 2.0
3 Alice 4.0 sub6 Bryce 4.0
4 Ayoung 5.0 sub5 Betty 5.0
5 NaN NaN sub3 Bran 3.0
连接将在索引上执行。联接操作接受调用它的对象。因此, a.join(b)不等于b.join(a) 。
import pandas as pd
left = pd.DataFrame({
'id':[1,2,3,4,5],
'Name': ['Alex', 'Amy', 'Allen', 'Alice', 'Ayoung'],
'subject_id':['sub1','sub2','sub4','sub6','sub5']})
right = pd.DataFrame({
'id':[1,2,3,4,5],
'Name': ['Billy', 'Brian', 'Bran', 'Bryce', 'Betty'],
'subject_id':['sub2','sub4','sub3','sub6','sub5']})
print pd.merge(left, right, on='subject_id', how='inner')
其输出如下-
Name_x id_x subject_id Name_y id_y
0 Amy 2 sub2 Billy 1
1 Allen 3 sub4 Brian 2
2 Alice 4 sub6 Bryce 4
3 Ayoung 5 sub5 Betty 5