python数据分析——pandas的拼接操作
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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
1. 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
In?[1]:
import numpy as np
import pandas as pd
from pandas import Series,DataFrame
In?[2]:
df1 = DataFrame(data=np.random.randint(0,100,size=(3,3)),index=['a','b','c'],columns=['A','B','C'])
df2 = DataFrame(data=np.random.randint(0,100,size=(3,3)),index=['a','d','c'],columns=['A','d','C'])
In?[7]:
pd.concat((df1,df1),axis=0,join='inner')
Out[7]:
| A | B | C |
a |
59 |
40 |
89 |
b |
71 |
5 |
76 |
c |
29 |
34 |
87 |
a |
59 |
40 |
89 |
b |
71 |
5 |
76 |
c |
29 |
34 |
87 |
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:
In?[11]:
pd.concat((df1,df2),axis=1,join='outer')
Out[11]:
| A | B | C | A | d | C |
a |
59.0 |
40.0 |
89.0 |
50.0 |
26.0 |
45.0 |
b |
71.0 |
5.0 |
76.0 |
NaN |
NaN |
NaN |
c |
29.0 |
34.0 |
87.0 |
31.0 |
82.0 |
35.0 |
d |
NaN |
NaN |
NaN |
23.0 |
95.0 |
94.0 |
3) 使用df.append()函数添加
由于在后面级联的使用非常普遍,因此有一个函数append专门用于在后面添加
2. 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:
- 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
In?[12]:
df1 = DataFrame({'employee':['Bob','Jake','Lisa'],
'group':['Accounting','Engineering','Engineering'],
})
df1
Out[12]:
| employee | group |
0 |
Bob |
Accounting |
1 |
Jake |
Engineering |
2 |
Lisa |
Engineering |
In?[13]:
df2 = DataFrame({'employee':['Lisa','Bob','Jake'],
'hire_date':[2004,2008,2012],
})
df2
Out[13]:
| employee | hire_date |
0 |
Lisa |
2004 |
1 |
Bob |
2008 |
2 |
Jake |
2012 |
In?[14]:
pd.merge(df1,df2,how='outer')
Out[14]:
| employee | group | hire_date |
0 |
Bob |
Accounting |
2008 |
1 |
Jake |
Engineering |
2012 |
2 |
Lisa |
Engineering |
2004 |
2) 多对一合并
In?[15]:
df3 = DataFrame({
'employee':['Lisa','Jake'],
'group':['Accounting','Engineering'],
'hire_date':[2004,2016]})
df3
Out[15]:
| employee | group | hire_date |
0 |
Lisa |
Accounting |
2004 |
1 |
Jake |
Engineering |
2016 |
In?[16]:
df4 = DataFrame({'group':['Accounting','Engineering','Engineering'],
'supervisor':['Carly','Guido','Steve']
})
df4
Out[16]:
| group | supervisor |
0 |
Accounting |
Carly |
1 |
Engineering |
Guido |
2 |
Engineering |
Steve |
In?[17]:
pd.merge(df3,df4)
Out[17]:
| 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
In?[18]:
df1 = DataFrame({'employee':['Bob','Jake','Lisa'],
'group':['Accounting','Engineering','Engineering']})
df1
Out[18]:
| employee | group |
0 |
Bob |
Accounting |
1 |
Jake |
Engineering |
2 |
Lisa |
Engineering |
In?[19]:
df5 = DataFrame({'group':['Engineering','Engineering','HR'],
'supervisor':['Carly','Guido','Steve']
})
df5
Out[19]:
| group | supervisor |
0 |
Engineering |
Carly |
1 |
Engineering |
Guido |
2 |
HR |
Steve |
In?[21]:
pd.merge(df1,df5,how='outer')
Out[21]:
| 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 |
- 加载excl数据:pd.read_excel('excl_path',sheetname=1)
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
In?[10]:
df1 = DataFrame({'employee':['Jack',"Summer","Steve"],
'group':['Accounting','Finance','Marketing']})
In?[11]:
df2 = DataFrame({'employee':['Jack','Bob',"Jake"],
'hire_date':[2003,2009,2012],
'group':['Accounting','sell','ceo']})
In?[22]:
display(df1,df2)
?
| employee | group |
0 |
Bob |
Accounting |
1 |
Jake |
Engineering |
2 |
Lisa |
Engineering |
?
| employee | hire_date |
0 |
Lisa |
2004 |
1 |
Bob |
2008 |
2 |
Jake |
2012 |
- 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
In?[12]:
df1 = DataFrame({'employee':['Bobs','Linda','Bill'],
'group':['Accounting','Product','Marketing'],
'hire_date':[1998,2017,2018]})
In?[13]:
df5 = DataFrame({'name':['Lisa','Bobs','Bill'],
'hire_dates':[1998,2016,2007]})
In?[23]:
display(df1,df5)
?
| employee | group |
0 |
Bob |
Accounting |
1 |
Jake |
Engineering |
2 |
Lisa |
Engineering |
?
| group | supervisor |
0 |
Engineering |
Carly |
1 |
Engineering |
Guido |
2 |
HR |
Steve |
5) 内合并与外合并:out取并集 inner取交集
In?[25]:
df6 = DataFrame({'name':['Peter','Paul','Mary'],
'food':['fish','beans','bread']}
)
df7 = DataFrame({'name':['Mary','Joseph'],
'drink':['wine','beer']})
In?[26]:
display(df6,df7)
?
| name | food |
0 |
Peter |
fish |
1 |
Paul |
beans |
2 |
Mary |
bread |
?
| name | drink |
0 |
Mary |
wine |
1 |
Joseph |
beer |
In?[27]:
df6 = DataFrame({'name':['Peter','Paul','Mary'],
'food':['fish','beans','bread']}
)
df7 = DataFrame({'name':['Mary','Joseph'],
'drink':['wine','beer']})
display(df6,df7)
pd.merge()
?
| name | food |
0 |
Peter |
fish |
1 |
Paul |
beans |
2 |
Mary |
bread |
?
| name | drink |
0 |
Mary |
wine |
1 |
Joseph |
beer |
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