5 dataframe级联与合并操作

import pandas as pd
import numpy as np

级联操作

pd.concat, pd.append

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 = pd.DataFrame(data=np.random.randint(0,100,size=(5,3)),columns=['A','B','C'])
df2 = pd.DataFrame(data=np.random.randint(0,100,size=(5,3)),columns=['A','D','C'])

pd.concat((df1,df1),axis=1) #行列索引都一致的级联叫做匹配级联

	A	B	C	A	B	C
0	26	63	95	26	63	95
1	66	86	35	66	86	35
2	74	3	4	74	3	4
3	85	0	67	85	0	67
4	59	28	65	59	28	65

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

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

/Users/bobo/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:1: FutureWarning: Sorting because non-concatenation axis is not aligned. A future version
of pandas will change to not sort by default.

To accept the future behavior, pass 'sort=False'.

To retain the current behavior and silence the warning, pass 'sort=True'.

  """Entry point for launching an IPython kernel.

	A	B	C	D
0	26	63.0	95	NaN
1	66	86.0	35	NaN
2	74	3.0	4	NaN
3	85	0.0	67	NaN
4	59	28.0	65	NaN
0	98	NaN	87	2.0
1	24	NaN	3	52.0
2	68	NaN	61	65.0
3	57	NaN	29	29.0
4	99	NaN	41	51.0

pd.concat((df1,df2),axis=0,join='inner') #inner直把可以级联的级联不能级联不处理

	A	C
0	26	95
1	66	35
2	74	4
3	85	67
4	59	65
0	98	87
1	24	3
2	68	61
3	57	29
4	99	41

如果想要保留数据的完整性必须使用outer（外连接）
append函数的使用

df1.append(df1)

	A	B	C
0	26	63	95
1	66	86	35
2	74	3	4
3	85	0	67
4	59	28	65
0	26	63	95
1	66	86	35
2	74	3	4
3	85	0	67
4	59	28	65

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

from pandas import DataFrame

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]})
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)#on如果不写，默认情况下使用两表中公有的列作为合并条件

	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='right')

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

pd.merge(df1,df5,how='right')

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

key的规范化

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

df5 = DataFrame({'name':['Lisa','Bobs','Bill'],
                'hire_dates':[1998,2016,2007]})
df5

	name	hire_dates
0	Lisa	1998
1	Bobs	2016
2	Bill	2007

pd.merge(df1,df5,left_on='employee',right_on='name')

	employee	group	hire_date	name	hire_dates
0	Bobs	Accounting	1998	Bobs	2016
1	Bill	Marketing	2018	Bill	2007

内合并与外合并:out取并集 inner取交集

df6 = DataFrame({'name':['Peter','Paul','Mary'],
               'food':['fish','beans','bread']}
               )
df7 = DataFrame({'name':['Mary','Joseph'],
                'drink':['wine','beer']})

df6 = DataFrame({'name':['Peter','Paul','Mary'],
               'food':['fish','beans','bread']}
               )
df7 = DataFrame({'name':['Mary','Joseph'],
                'drink':['wine','beer']})

作者：华王博客：https://www.cnblogs.com/huahuawang/

	A	B	C
0	26	63	95
1	66	86	35
2	74	3	4
3	85	0	67
4	59	28	65
0	26	63	95
1	66	86	35
2	74	3	4
3	85	0	67
4	59	28	65

	A	B	C
0	26	63	95
1	66	86	35
2	74	3	4
3	85	0	67
4	59	28	65
0	26	63	95
1	66	86	35
2	74	3	4
3	85	0	67
4	59	28	65

	A	B	C
0	26	63	95
1	66	86	35
2	74	3	4
3	85	0	67
4	59	28	65
0	26	63	95
1	66	86	35
2	74	3	4
3	85	0	67
4	59	28	65