对比集合Set | 详解Pandas的DataFrame如何做交集、并集、差集与对称差集

一、简介

"""
@Author   ：叶庭云
@公众号    ：AI庭云君
@CSDN     ：https://yetingyun.blog.csdn.net/
"""

Python的数据类型集合：由不同元素组成的集合，集合中是一组无序排列的可 Hash 的值（不可变类型），可以作为字典的Key

Pandas中的DataFrame：DataFrame是一个表格型的数据结构，可以理解为带有标签的二维数组。

常用的集合操作如下图所示：

二、交集

• pandas的 merge 功能默认为 inner 连接，可以实现取交集
• 集合 set 可以直接用 & 取交集

import pandas as pd
print("CSDN叶庭云：https://yetingyun.blog.csdn.net/")
set1 ={"Python","Go","C++","Java"}
set2 ={"Go","C++","JavaScript","C"}
set1 & set2
df1 = pd.DataFrame([['1','Python'],['2','Go'],['3','C++'],['4','Java'],], columns=['id','name'])
df2 = pd.DataFrame([['2','Go'],['3','C++'],['5','JavaScript'],['6','C'],], columns=['id','name'])
pd.merge(df1, df2, on=['id','name'])

操作如下所示：

三、并集

• Pandas的 merge 方法里参数 how 的取值有 “left”, “right”, “inner”, “outer”，默认是inner。outer外连接可以实现取并集。另一种方法也可以df1.append(df2)后去重，保留第一次出现的也可以实现取并集。
• 集合 set 可以直接用 | 取并集

set1 ={"Python","Go","C++","Java"}
set2 ={"Go","C++","JavaScript","C"}
set1 | set2
print("CSDN叶庭云：https://yetingyun.blog.csdn.net/")
df1 = pd.DataFrame([['1','Python'],['2','Go'],['3','C++'],['4','Java'],], columns=['id','name'])
df2 = pd.DataFrame([['2','Go'],['3','C++'],['5','JavaScript'],['6','C'],], columns=['id','name'])
pd.merge(df1, df2,
         on=['id','name'],
         how='outer')
         
df3 = df1.append(df2)
df3.drop_duplicates(subset=['id'], keep="first")

四、差集

set1 ={"Python","Go","C++","Java"}
set2 ={"Go","C++","JavaScript","C"}
set1 - set2
print("CSDN叶庭云：https://yetingyun.blog.csdn.net/")
set1 ={"Python","Go","C++","Java"}
set2 ={"Go","C++","JavaScript","C"}
set2 - set1
# df1-df2
df1 = pd.DataFrame([['1','Python'],['2','Go'],['3','C++'],['4','Java'],], columns=['id','name'])
df2 = pd.DataFrame([['2','Go'],['3','C++'],['5','JavaScript'],['6','C'],], columns=['id','name'])
df1 = df1.append(df2)
df1 = df1.append(df2)
set_diff_df = df1.drop_duplicates(subset=df1.columns,
                                  keep=False)
set_diff_df
# df2-df1
df1 = pd.DataFrame([['1','Python'],['2','Go'],['3','C++'],['4','Java'],], columns=['id','name'])
df2 = pd.DataFrame([['2','Go'],['3','C++'],['5','JavaScript'],['6','C'],], columns=['id','name'])print("CSDN叶庭云：https://yetingyun.blog.csdn.net/")
df2 = df2.append(df1)
df2 = df2.append(df1)
set_diff_df = df2.drop_duplicates(subset=df2.columns,
                                  keep=False)
set_diff_df
# df1-df2
df1 = pd.DataFrame([['1','Python'],['2','Go'],['3','C++'],['4','Java'],], columns=['id','name'])
df2 = pd.DataFrame([['2','Go'],['3','C++'],['5','JavaScript'],['6','C'],], columns=['id','name'])
pd.concat([df1, df2, df2]).drop_duplicates(keep=False)# df2-df1
df1 = pd.DataFrame([['1','Python'],['2','Go'],['3','C++'],['4','Java'],], columns=['id','name'])
df2 = pd.DataFrame([['2','Go'],['3','C++'],['5','JavaScript'],['6','C'],], columns=['id','name'])
pd.concat([df2, df1, df1]).drop_duplicates(keep=False)

五、对称差集

print("CSDN叶庭云：https://yetingyun.blog.csdn.net/")
set1 ={"Python","Go","C++","Java"}
set2 ={"Go","C++","JavaScript","C"}
set1 ^ set2    # 对称差集# 去重   不保留重复的：即可实现取对称差集
df3 = df1.append(df2)
df3.drop_duplicates(subset=['id'], keep=False)

推荐学习：

• https://www.jianshu.com/p/877e2bc11d93
• https://www.runoob.com/python3/python3-set.html