pandas的级联操作

级联操作

  • pd.concat, pd.append
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
  • 匹配级联
df1 = DataFrame({'employee':['Bobs','Linda','Bill'],
                'group':['Accounting','Product','Marketing'],
               'hire_date':[1998,2017,2018]})
df1
employee group hire_date
0 Bobs Accounting 1998
1 Linda Product 2017
2 Bill Marketing 2018
pd.concat((df1,df1),axis=0)
employee group hire_date
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
employee groupps hire_date
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
employee group groupps hire_date
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')
employee hire_date
0 Bobs 1998
1 Linda 2017
2 Bill 2018
0 Bobs 1998
1 Linda 2017
2 Bill 2018
  • append函数的使用: append只可以进行纵向的级联

    employee group groupps hire_date
    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

合并操作

  • merge与concat的区别在于,merge需要依据某一共同列来进行合并

  • 使用pd.merge()合并时,会自动根据两者相同column名称的那一列,作为key来进行合并。

  • 注意每一列元素的顺序不要求一致

一对一合并

df1 = DataFrame({'employee':['Bob','Jake','Lisa'],
                'group':['Accounting','Engineering','Engineering'],
                })
df1
employee group
0 Bob Accounting
1 Jake Engineering
2 Lisa Engineering
df2 = DataFrame({'employee':['Lisa','Bob','Jake'],
                'hire_date':[2004,2008,2012],
                })
df2
employee hire_date
0 Lisa 2004
1 Bob 2008
2 Jake 2012
pd.merge(df1,df2,on='employee')
employee group hire_date
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
employee group hire_date
0 Lisa Accounting 2004
1 Jake Engineering 2016
df4 = DataFrame({'group':['Accounting','Engineering','Engineering'],
                       'supervisor':['Carly','Guido','Steve']
                })
df4
group supervisor
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
employee group hire_date supervisor
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
employee group
0 Bob Accounting
1 Jake Engineering
2 Lisa Engineering
df5 = DataFrame({'group':['Engineering','Engineering','HR'],
                'supervisor':['Carly','Guido','Steve']
                })
df5
group supervisor
0 Engineering Carly
1 Engineering Guido
2 HR Steve
pd.merge(df1,df5,how='outer')
employee group supervisor
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
employee group
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
employee group hire_date
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
employee_x group employee_y hire_date
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
employee group hire_date
0 Bobs Accounting 1998
1 Linda Product 2017
2 Bill Marketing 2018
df5 = DataFrame({'name':['Lisa','Bobs','Bill'],
                'hire_dates':[1998,2016,2007]})
df5
hire_dates name
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,
employee group hire_date hire_dates name
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

原文地址:https://www.cnblogs.com/zhufanyu/p/12031641.html