Python 玩转数据 - Pandas 数据处理 拼接 pd.concat() axis join ignore

Python 玩转数据 - Pandas 数据处理 拼接 pd.concat() axis join ignore,第1张

Python 玩转数据 - Pandas 数据处理 拼接 pd.concat() axis join ignore 引言

本文主要介绍 Pandas 数据拼接 pd.concat()

内容提要:
数据拼接 pd.concat()举例 axis = 0 axis = 1
处理重复索引 ignore_index, verify_integrity
拼接 Dataframes 不同的列 join

数据拼接 pd.concat()

pd.concat( objs, axis=0, join=‘outer’, join_axes=None, ignore_index=False, keys=None, levels=None, names=None, verify_integrity=False, sort=None, copy=True)

两个或两个以上的 Pandas 对象 dataframe 或者 series 按某一轴(默认:axis = 0) 拼接在一起, 按行拼接 (axis=0), 按列拼接 (axis=1)

keys 和 names

当表格拼接完成后,就不能判断到底哪些数据是来自于哪一个表了,如果需要保留来源信息,就可以通过keys参数进行设置,而names参数可以给拼接后形成的数据结构添加名字。

串联两个Pandas Series


行拼接:

列拼接:

加上 Keys 和 Names 参数显示出数据来源:
行拼接

列拼接

代码:

import pandas as pd

s1 = pd.Series(['Ann', 'Fred'], index = [1, 2])
s2 = pd.Series (['John', 'Kate'], index = [3, 4])

print("s1:n{}".format(s1))
print("s2:n{}".format(s2))

s_concat_along_axis_0 = pd.concat ([s1, s2])
print("s_concat_along_axis_0:n{}".format(s_concat_along_axis_0))

s_concat_along_axis_1 = pd.concat ([s1, s2], axis=1)
print("s_concat_along_axis_1:n{}".format(s_concat_along_axis_1))

输出:

s1:
1     Ann
2    Fred
dtype: object
s2:
3    John
4    Kate
dtype: object
s_concat_along_axis_0:
1     Ann
2    Fred
3    John
4    Kate
dtype: object
s_concat_along_axis_1:
      0     1
1   Ann   NaN
2  Fred   NaN
3   NaN  John
4   NaN  Kate
串联两个 Pandas Dataframes

两个 Dataframe 都是默认的 index, 按行串联时,会有重复的 index


行拼接:

列拼接:

加上 Key 和 names显示数据来源:

代码:

import pandas as pd

idnumber = [1,2,5]
fname = ['Kate','John','Eli']
age = [10,20,50]
grade = ['A','B','C']

df1 = pd.Dataframe({'id':idnumber,'fname':fname})
df2 = pd.Dataframe({'age':age,'grade':grade})
df3=pd.Dataframe({'id':[3,4],'age':[30,40]})
print("df1:n{}".format(df1))
print("df2:n{}".format(df2))
print("df3:n{}".format(df3))

df_concat_along_axis_0 = pd.concat([df1, df2, df3], axis = 0,keys=["df1", "df2", "df3"], names=["From", "index"])
df_concat_along_axis_1 = pd.concat([df1, df2, df3], axis = 1,keys=["df1", "df2", "df3"], names=["From", "index"])

print("df_concat_along_axis_0:n{}".format(df_concat_along_axis_0))
print("df_concat_along_axis_1:n{}".format(df_concat_along_axis_1))

输出:

df1:       
   id fname
0   1  Kate
1   2  John
2   5   Eli
df2:
   age grade
0   10     A
1   20     B
2   50     C
df3:
   id  age
0   3   30
1   4   40
df_concat_along_axis_0:
             id fname   age grade
From index
df1  0      1.0  Kate   NaN   NaN
     1      2.0  John   NaN   NaN
     2      5.0   Eli   NaN   NaN
df2  0      NaN   NaN  10.0     A
     1      NaN   NaN  20.0     B
     2      NaN   NaN  50.0     C
df3  0      3.0   NaN  30.0   NaN
     1      4.0   NaN  40.0   NaN
df_concat_along_axis_1:
From  df1       df2        df3
index  id fname age grade   id   age
0       1  Kate  10     A  3.0  30.0
1       2  John  20     B  4.0  40.0
2       5   Eli  50     C  NaN   NaN

两个 Dataframe 自定义 Index,按行串联时,会有重复的 Index



代码:

import pandas as pd

idnumber = [1,2,5]
fname = ['Kate','John','Eli']
age = [10,20,50]
grade = ['A','B','C']

df1=pd.Dataframe({'id':idnumber,'fname':fname}, index = [3, 5, 7])
df2= pd.Dataframe({'age':age,'grade':grade}, index = [3, 6, 9])
print("df1:n{}".format(df1))
print("df2:n{}".format(df2))

df_concat_along_axis_0 = pd.concat([df1, df2], axis = 0)
df_concat_along_axis_1 = pd.concat([df1, df2], axis = 1)

print("df_concat_along_axis_0:n{}".format(df_concat_along_axis_0))
print("df_concat_along_axis_1:n{}".format(df_concat_along_axis_1))

输出:

df1:       
   id fname
3   1  Kate
5   2  John
7   5   Eli
df2:
   age grade
3   10     A
6   20     B
9   50     C
df_concat_along_axis_0:
    id fname   age grade
3  1.0  Kate   NaN   NaN
5  2.0  John   NaN   NaN
7  5.0   Eli   NaN   NaN
3  NaN   NaN  10.0     A
6  NaN   NaN  20.0     B
9  NaN   NaN  50.0     C
df_concat_along_axis_1:
    id fname   age grade
3  1.0  Kate  10.0     A
5  2.0  John   NaN   NaN
6  NaN   NaN  20.0     B
7  5.0   Eli   NaN   NaN
9  NaN   NaN  50.0     C
处理重复索引 Duplicate Indices

由上面的例子可以看出pd.concat 会保留索引 indices, 这样会导致重复的索引 indices

pd.concat() 提供几个选项来处理重复的索引:

  1. 当捕捉到重复的索引时抛出异常
    ● 验证是否存在重复的索引用 verify_integrity = True 选项
    ● 需要 catch 到重复索引的异常
  2. 忽略索引
    ● 如果 index 不重要,我们可以忽略掉,用 ignore_index = True 选项

下面两 Dataframe 为例

捕获异常:

忽略索引:

代码:

import pandas as pd

idnumber = [1,2,5]
fname = ['Kate','John','Eli']
age = [10,20,50]
grade = ['A','B','C']

df1=pd.Dataframe({'id':idnumber,'fname':fname}, index = [3, 5, 7])
df2= pd.Dataframe({'age':age,'grade':grade}, index = [3, 6, 9])
print("df1:n{}".format(df1))
print("df2:n{}".format(df2))

# catch repeat indices
try:
    df_concat_along_axis_0 = pd.concat([df1, df2], axis = 0, verify_integrity=True)
except ValueError as e:
    print("Value Error: ", e)    

# ignore index
df_concat_along_axis_0 = pd.concat([df1, df2], axis = 0, ignore_index=True)
df_concat_along_axis_1 = pd.concat([df1, df2], axis = 1, ignore_index=True)

print("df_concat_along_axis_0:n{}".format(df_concat_along_axis_0))
print("df_concat_along_axis_1:n{}".format(df_concat_along_axis_1))

输出:

df1:       
   id fname
3   1  Kate
5   2  John
7   5   Eli
df2:
   age grade
3   10     A
6   20     B
9   50     C
Value Error:  Indexes have overlapping values: Int64Index([3], dtype='int64')
df_concat_along_axis_0:
    id fname   age grade
0  1.0  Kate   NaN   NaN
1  2.0  John   NaN   NaN
2  5.0   Eli   NaN   NaN
3  NaN   NaN  10.0     A
4  NaN   NaN  20.0     B
5  NaN   NaN  50.0     C
df_concat_along_axis_1:
     0     1     2    3
3  1.0  Kate  10.0    A
5  2.0  John   NaN  NaN
6  NaN   NaN  20.0    B
7  5.0   Eli   NaN  NaN
9  NaN   NaN  50.0    C
串联 Dataframes 不同的列

对于简单的串联,两个 Pandas 对象正好共享相同的列。然而,现时情况,数据来源不同,可能存在不同的列名。为了处理不同列名情况,pd.concat() 提供了 join = 和 join_axis = 选项。其中 join 参数控制的是外连接还是内连接,默认外连接,保留两个表中的所有信息;如果设置成内连接,拼接结果只保留两个表共有的信息

● 返回两个表的交集 (join = ‘inner’)
● 返回两个表的并集 (join = ‘outer’ 默认)
● 指定返回的集合(如, join_axes = [df2.columns] 注意:新版本的Pandas 已经不支持该选项了)

举例:

join=inner:
行拼接,上下拼接的时候,保留了共有的列信息!
列拼接,左右拼接的时候,保留了共有的行信息!

完整代码:

import pandas as pd

idnumber = [1,2,5]
fname = ['Kate','John','Eli']
fname_2 = ['Kate 2','John 2','Eli 2']
age = [10,20,50]
grade = ['A','B','C']

df1=pd.Dataframe({'id':idnumber,'fname':fname}, index = [3, 5, 7])
df2= pd.Dataframe({'age':age,'grade':grade, 'fname':fname_2}, index = [3, 6, 9])
print("df1:n{}".format(df1))
print("df2:n{}".format(df2))

# Intersection: join = inner
df_intesection_axis_0 = pd.concat ([df1, df2], join = 'inner', axis=0, keys=["df1", "df2"], names=["From"])   
df_intesection_axis_1 = pd.concat ([df1, df2], join = 'inner', axis=1,keys=["df1", "df2"], names=["From"])

# Union: join = outer
df_union_axis_0 = pd.concat([df1, df2], join = 'outer', axis=0,keys=["df1", "df2"], names=["From"])
df_union_axis_1 = pd.concat([df1, df2], join = 'outer', axis=1,keys=["df1", "df2"], names=["From"])

# Specification: which columns to keep, discard in pandas new version
# df_specification = pd.concat ([df1, df2], join_axes = [df2.columns])

print("df_intesection_axis_0:n{}".format(df_intesection_axis_0))
print("df_intesection_axis_1:n{}".format(df_intesection_axis_1))
print("df_union_axis_0:n{}".format(df_union_axis_0))
print("df_union_axis_1:n{}".format(df_union_axis_1))

输出:

df1:       
   id fname
3   1  Kate
5   2  John
7   5   Eli
df2:
   age grade   fname
3   10     A  Kate 2
6   20     B  John 2
9   50     C   Eli 2
df_intesection_axis_0:
         fname        
From
df1  3    Kate
     5    John
     7     Eli
df2  3  Kate 2
     6  John 2
     9   Eli 2
df_intesection_axis_1:
From df1       df2
      id fname age grade   fname
3      1  Kate  10     A  Kate 2
df_union_axis_0:
         id   fname   age grade
From
df1  3  1.0    Kate   NaN   NaN
     5  2.0    John   NaN   NaN
     7  5.0     Eli   NaN   NaN
df2  3  NaN  Kate 2  10.0     A
     6  NaN  John 2  20.0     B
     9  NaN   Eli 2  50.0     C
df_union_axis_1:
From  df1         df2
       id fname   age grade   fname
3     1.0  Kate  10.0     A  Kate 2
5     2.0  John   NaN   NaN     NaN
6     NaN   NaN  20.0     B  John 2
7     5.0   Eli   NaN   NaN     NaN
9     NaN   NaN  50.0     C   Eli 2

欢迎分享,转载请注明来源:内存溢出

原文地址: https://outofmemory.cn/zaji/5689505.html

(0)
打赏 微信扫一扫 微信扫一扫 支付宝扫一扫 支付宝扫一扫
上一篇 2022-12-17
下一篇 2022-12-17

发表评论

登录后才能评论

评论列表(0条)

保存