级联操作
import pandas as pd
from pandas import DataFrame
import numpy as np
pandas使用pd.concat函数,与np.concatenate函数类似,只是多了一些参数:
objs
axis=0
keys
join='outer' / 'inner':表示的是级联的方式,outer会将所有的项进行级联(忽略匹配和不匹配),而inner只会将匹配的项级联到一起,不匹配的不级联
ignore_index=False
df1 = DataFrame({'employee':['Bobs','Linda','Bill'],
'group':['Accounting','Product','Marketing'],
'hire_date':[1998,2017,2018]})
df1
0 | Bobs | Accounting | 1998 |
1 | Linda | Product | 2017 |
2 | Bill | Marketing | 2018 |
pd.concat((df1,df1),axis=0)
0 | Bobs | Accounting | 1998 |
1 | Linda | Product | 2017 |
2 | Bill | Marketing | 2018 |
0 | Bobs | Accounting | 1998 |
1 | Linda | Product | 2017 |
2 | Bill | Marketing | 2018 |
- 不匹配级联
- 不匹配指的是级联的维度的索引不一致。例如纵向级联时列索引不一致,横向级联时行索引不一致
- 有2种连接方式:
- 外连接:补NaN(默认模式)
- 内连接:只连接匹配的项
df2 = df1.copy()
df2.columns = ['employee','groupps','hire_date']
df2
0 | Bobs | Accounting | 1998 |
1 | Linda | Product | 2017 |
2 | Bill | Marketing | 2018 |
pd.concat((df1,df2),axis=0)
# 按列级联,发现不同列就加在表中,用NaN补全空的字段
0 | Bobs | Accounting | NaN | 1998 |
1 | Linda | Product | NaN | 2017 |
2 | Bill | Marketing | NaN | 2018 |
0 | Bobs | NaN | Accounting | 1998 |
1 | Linda | NaN | Product | 2017 |
2 | Bill | NaN | Marketing | 2018 |
- join:
- inner:只对可以匹配的项进行级联
- outer:可以级联所有的项
pd.concat((df1,df2),axis=0,join='inner')
0 | Bobs | 1998 |
1 | Linda | 2017 |
2 | Bill | 2018 |
0 | Bobs | 1998 |
1 | Linda | 2017 |
2 | Bill | 2018 |
合并操作
一对一合并
df1 = DataFrame({'employee':['Bob','Jake','Lisa'],
'group':['Accounting','Engineering','Engineering'],
})
df1
0 | Bob | Accounting |
1 | Jake | Engineering |
2 | Lisa | Engineering |
df2 = DataFrame({'employee':['Lisa','Bob','Jake'],
'hire_date':[2004,2008,2012],
})
df2
0 | Lisa | 2004 |
1 | Bob | 2008 |
2 | Jake | 2012 |
pd.merge(df1,df2,on='employee')
0 | Bob | Accounting | 2008 |
1 | Jake | Engineering | 2012 |
2 | Lisa | Engineering | 2004 |
一对多合并
df3 = DataFrame({
'employee':['Lisa','Jake'],
'group':['Accounting','Engineering'],
'hire_date':[2004,2016]})
df
0 | Lisa | Accounting | 2004 |
1 | Jake | Engineering | 2016 |
df4 = DataFrame({'group':['Accounting','Engineering','Engineering'],
'supervisor':['Carly','Guido','Steve']
})
df4
0 | Accounting | Carly |
1 | Engineering | Guido |
2 | Engineering | Steve |
pd.merge(df3,df4)
# 会默认以两个表的共同字段 group 进行关联,将两张表中的全部数据进行合并
0 | Lisa | Accounting | 2004 | Carly |
1 | Jake | Engineering | 2016 | Guido |
2 | Jake | Engineering | 2016 | Steve |
多对多合并
df1 = DataFrame({'employee':['Bob','Jake','Lisa'],
'group':['Accounting','Engineering','Engineering']})
df1
0 | Bob | Accounting |
1 | Jake | Engineering |
2 | Lisa | Engineering |
df5 = DataFrame({'group':['Engineering','Engineering','HR'],
'supervisor':['Carly','Guido','Steve']
})
df5
0 | Engineering | Carly |
1 | Engineering | Guido |
2 | HR | Steve |
pd.merge(df1,df5,how='outer')
0 | Bob | Accounting | NaN |
1 | Jake | Engineering | Carly |
2 | Jake | Engineering | Guido |
3 | Lisa | Engineering | Carly |
4 | Lisa | Engineering | Guido |
5 | NaN | HR | Steve |
key的规范化
- 当列冲突时,即有多个列名称相同时,需要使用on=来指定哪一个列作为key,配合suffixes指定冲突列名
df1 = DataFrame({'employee':['Jack',"Summer","Steve"],
'group':['Accounting','Finance','Marketing']})
df1
0 | Jack | Accounting |
1 | Summer | Finance |
2 | Steve | Marketing |
df2 = DataFrame({'employee':['Jack','Bob',"Jake"],
'hire_date':[2003,2009,2012],
'group':['Accounting','sell','ceo']})
df2
0 | Jack | Accounting | 2003 |
1 | Bob | sell | 2009 |
2 | Jake | ceo | 2012 |
pd.merge(df1,df2,on='group')
# 指定按照 group 这一列来进行分组,会将两个表中相同的字段名都加在合并后的表中,两个字段会改变
# 也可以通过这个 suffixes=('_x', '_y') 参数进行一个修改
0 | Jack | Accounting | Jack | 2003 |
- 当两张表没有可进行连接的列时,可使用left_on和right_on手动指定merge中左右两边的哪一列作为连接的列
df1 = DataFrame({'employee':['Bobs','Linda','Bill'],
'group':['Accounting','Product','Marketing'],
'hire_date':[1998,2017,2018]})
df1
0 | Bobs | Accounting | 1998 |
1 | Linda | Product | 2017 |
2 | Bill | Marketing | 2018 |
df5 = DataFrame({'name':['Lisa','Bobs','Bill'],
'hire_dates':[1998,2016,2007]})
df5
0 | 1998 | Lisa |
1 | 2016 | Bobs |
2 | 2007 | Bill |
pd.merge(df1,df5,left_on='employee',right_on='name',how='outer')
# 分别指定df1表中的employee字段和df5表中的name字段进行一个比对连接,
0 | Bobs | Accounting | 1998.0 | 2016.0 | Bobs |
1 | Linda | Product | 2017.0 | NaN | NaN |
2 | Bill | Marketing | 2018.0 | 2007.0 | Bill |
3 | NaN | NaN | NaN | 1998.0 | Lisa |
内合并与外合并:out取并集 inner取交集