Pandas的拼接操作

Posted 2021-03-25 zyyhxbs

pandas的拼接操作

pandas的拼接分为两种：

• 级联：pd.concat, pd.append
• 合并：pd.merge, pd.join

import pandas as pd
import numpy as np
from pandas import DataFrame,Series

一. 使用pd.concat()级联

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

1)匹配级联

行列索引均一致

df1 = DataFrame(data=np.random.randint(0,100,size=(3,4)))
df1

	0	1	2	3
0	61	89	68	51
1	46	79	1	55
2	52	4	72	18

df2 = DataFrame(data=np.random.randint(0,100,size=(3,4)))
df2

	0	1	2	3
0	15	62	20	78
1	60	79	70	58
2	71	87	20	95

pd.concat((df1,df2),axis=0)  # axis=0表示Y轴级联

	0	1	2	3
0	61	89	68	51
1	46	79	1	55
2	52	4	72	18
0	15	62	20	78
1	60	79	70	58
2	71	87	20	95

2) 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

df1 = DataFrame(data=np.random.randint(0,100,size=(3,4)))
df2 = DataFrame(data=np.random.randint(0,100,size=(3,3)))

pd.concat((df1,df2),axis=0)

	0	1	2	3
0	55	61	54	56.0
1	10	14	6	62.0
2	39	27	99	81.0
0	31	49	80	NaN
1	73	42	44	NaN
2	67	68	97	NaN

pd.concat((df1,df2),axis=0,join='inner')  # inner内连接,只级联匹配的项

	0	1	2
0	55	61	54
1	10	14	6
2	39	27	99
0	31	49	80
1	73	42	44
2	67	68	97

二. 使用pd.merge()合并

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

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

参数：

• how：outer取并集(外连接) inner取交集(内连接)

• on：当有多列相同的时候，可以使用on来指定使用那一列进行合并，on的值为一个列表

1) 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)  # 按照employee进行了合并

	employee	group	hire_date
0	Bob	Accounting	2008
1	Jake	Engineering	2012
2	Lisa	Engineering	2004

2) 多对一合并

df3 = DataFrame({
    'employee':['Lisa','Jake'],
    'group':['Accounting','Engineering'],
    'hire_date':[2004,2016]})
df3

	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)

	employee	group	hire_date	supervisor
0	Lisa	Accounting	2004	Carly
1	Jake	Engineering	2016	Guido
2	Jake	Engineering	2016	Steve

3) 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

df2 = DataFrame({'group':['Engineering','Engineering','HR'],
                'supervisor':['Carly','Guido','Steve']
                })
df2

	group	supervisor
0	Engineering	Carly
1	Engineering	Guido
2	HR	Steve

pd.merge(df1,df2,how='right')  # right表示右连接

	employee	group	supervisor
0	Jake	Engineering	Carly
1	Lisa	Engineering	Carly
2	Jake	Engineering	Guido
3	Lisa	Engineering	Guido
4	NaN	HR	Steve

4) 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='employee')  # 默认按照employee和group进行合并,可以指定列名

	employee	group_x	group_y	hire_date
0	Jack	Accounting	Accounting	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

df2 = DataFrame({'name':['Lisa','Bobs','Bill'],
                'hire_dates':[1998,2016,2007]})
df2

	hire_dates	name
0	1998	Lisa
1	2016	Bobs
2	2007	Bill

pd.merge(df1,df2,left_on='employee',right_on='name',how='outer')

	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

5) 内合并与外合并:out取并集 inner取交集

• 内合并：只保留两者都有的key（默认模式）

df6 = DataFrame({'name':['Peter','Paul','Mary'],
               'food':['fish','beans','bread']}
               )
df6

	food	name
0	fish	Peter
1	beans	Paul
2	bread	Mary

df7 = DataFrame({'name':['Mary','Joseph'],
                'drink':['wine','beer']})
df7

	drink	name
0	wine	Mary
1	beer	Joseph

pd.merge(df6, df7)

	food	name	drink
0	bread	Mary	wine

• 外合并 how=‘outer‘：补NaN

pd.merge(df6, df7, how='outer')

	food	name	drink
0	fish	Peter	NaN
1	beans	Paul	NaN
2	bread	Mary	wine
3	NaN	Joseph	beer