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Pandas的拼接操作

时间:2019-10-20 19:56:51      阅读:88      评论:0      收藏:0      [点我收藏+]

标签:rand   rom   连接   pytho   左右   wine   import   右连接   group   

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

Pandas的拼接操作

标签:rand   rom   连接   pytho   左右   wine   import   右连接   group   

原文地址:https://www.cnblogs.com/zyyhxbs/p/11708522.html

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