集合是无序的,天生不重复的数据组合,它的作用如下:

  • 去重,即:把一个列表变成集合,就去重了
  • 关系测试,即:测试两组集合的交集、并集和差集等

一、集合常用方法总结

Python集合操作方法详解

二、定义

1、语法

> name_1 = [1,2,3,4,7,8,7,10]
#把列表转换为集合
> name_1 = set(name_1)
#转换后,去重
> print(name_1,type(name_1))
{1, 2, 3, 4, 7, 8, 10} <class 'set'>

"htmlcode">
> name_1 = [1,2,3,4,7,8,7,10]
> name_2 = [1,3,5,8,10]
> name_1 = set(name_1)
> name_2 = set(name_2)
#输出结果
> name_1.intersection(name_2)
{8, 1, 10, 3}

2、并集(union())

> name_1 = [1,2,3,4,7,8,7,10]
> name_2 = [1,3,5,8,10]
> name_1 = set(name_1)
> name_2 = set(name_2)
#输出结果
> name_1.union(name_2)
{1, 2, 3, 4, 5, 7, 8, 10}

"htmlcode">
> name_1 = [1,2,3,4,7,8,7,10]
> name_2 = [1,3,5,8,10]
> name_1 = set(name_1)
> name_2 = set(name_2)
#输出结果
> name_1.difference(name_2)
{2, 4, 7}

"htmlcode">
> name_1 = [1,2,3,4,7,8,7,10]
> name_3 = [1,2,3,4]
> name_1 = set(name_1)
> name_3 = set(name_3)
#输出结果
> name_3.issubset(name_1)
True

"htmlcode">
> name_1 = [1,2,3,4,7,8,7,10]
> name_3 = [1,2,3,4]
> name_1 = set(name_1)
> name_3 = set(name_3)
#输出结果
> name_1.issuperset(name_3)
True

"htmlcode">
> name_1 = [1,2,3,4,7,8,7,10]
> name_2 = [1,3,5,8,10]
> name_1 = set(name_1)
> name_2 = set(name_2)
#输出结果
> name_1.symmetric_difference(name_2)
{2, 4, 5, 7}

"htmlcode">
> name_1 = [1,2,3,4,7,8,7,10]
> name_2 = [1,3,5,8,10]
> name_3 = [11]
> name_1 = set(name_1)
> name_2 = set(name_2)
> name_3 = set(name_3)
#有交集
> name_1.isdisjoint(name_2)
False
#无交集
> name_1.isdisjoint(name_3)
True

"htmlcode">
> name_1 = [1,2,3,4,7,8,7,10]
> name_2 = [1,3,5,8,10]
> name_1 = set(name_1)
> name_2 = set(name_2)
#结果输出
> name_1 & name_2
{8, 1, 10, 3}

2、并集(|)

> name_1 = [1,2,3,4,7,8,7,10]
> name_2 = [1,3,5,8,10]
> name_1 = set(name_1)
> name_2 = set(name_2)
#结果输出
> name_1 | name_2
{1, 2, 3, 4, 5, 7, 8, 10}

"htmlcode">
> name_1 = [1,2,3,4,7,8,7,10]
> name_2 = [1,3,5,8,10]
> name_1 = set(name_1)
> name_2 = set(name_2)
#结果输出
> name_1 - name_2
{2, 4, 7}

4、对称差集(^)

> name_1 = [1,2,3,4,7,8,7,10]
> name_2 = [1,3,5,8,10]
> name_1 = set(name_1)
> name_2 = set(name_2)
#输出
> name_1 ^ name_2
{2, 4, 5, 7}

5、是否是子集(<=)

> name_1 = [1,2,3,4,7,8,10]
> name_3 = [1,2,3,4]
> name_1 = set(name_1)
> name_3 = set(name_3)
#输出
> name_3 <= name_1
True

6、是否是父集(>=)

> name_1 = [1,2,3,4,7,8,10]
> name_3 = [1,2,3,4]
> name_1 = set(name_1)
> name_3 = set(name_3)
#输出
> name_1 >= name_3
True

"htmlcode">
> name_2 = [1,3,5,8,10]
> name_2 = set(name_2)
#添加已存在,不报错
> name_2.add(1)
> name_2
{8, 1, 10, 3, 5}
#添加不存在,添加一个新的数值
> name_2.add(11)
> name_2
{1, 3, 5, 8, 10, 11}

"htmlcode">
> name_2 = [1,3,5,8,10]
> name_2 = set(name_2)
> name_2.update([12,13,14])
#输出结果
> name_2
{1, 3, 5, 8, 10, 12, 13, 14}

"htmlcode">
> name_2 = [1,3,5,8,10]
> name_2 = set(name_2)
> name_2
{8, 1, 10, 3, 5}
> name_2.remove(1)
#输出
> name_2
{8, 10, 3, 5}
#删除不存在的元素,会报错
> name_2.remove(1)
Traceback (most recent call last):
 File "<input>", line 1, in <module>
KeyError: 1

"htmlcode">
> name_2 = [1,3,5,8,10]
> name_2 = set(name_2)
> name_2
{8, 1, 10, 3, 5}
#输出
> name_2.pop()
8

"htmlcode">
> name_2 = [1,3,5,8,10]
> name_2 = set(name_2)
> name_2.discard(10)
#输出结果
> name_2
{8, 1, 3, 5}
#删除不存在元素,不报错
> name_2.discard(10)

特别提示:用discard删除不存在的元素,不会出现报错

4、长度(len())

> name_1 = [1,2,3,4,7,8,7,10]
> name_1 = set(name_1)
#结果输出
> len(name_1)
7

"htmlcode">
> name_1 = [1,2,3,4,7,8,7,10]
> name_1 = set(name_1)
#结果输出
> 1 in name_1
True

"htmlcode">
> name_1 = [1,2,3,4,7,8,7,10]
> name_1 = set(name_1)
#输出
> 12 not in name_1
True

更多关于Python集合操作方法可查看下面的相关链接

标签:
Python集合操作方法,Python集合函数,Python集合交集,Python集合并集,Python集合差集

免责声明:本站文章均来自网站采集或用户投稿,网站不提供任何软件下载或自行开发的软件! 如有用户或公司发现本站内容信息存在侵权行为,请邮件告知! 858582#qq.com
白云城资源网 Copyright www.dyhadc.com

评论“Python集合操作方法详解”

暂无“Python集合操作方法详解”评论...

稳了!魔兽国服回归的3条重磅消息!官宣时间再确认!

昨天有一位朋友在大神群里分享,自己亚服账号被封号之后居然弹出了国服的封号信息对话框。

这里面让他访问的是一个国服的战网网址,com.cn和后面的zh都非常明白地表明这就是国服战网。

而他在复制这个网址并且进行登录之后,确实是网易的网址,也就是我们熟悉的停服之后国服发布的暴雪游戏产品运营到期开放退款的说明。这是一件比较奇怪的事情,因为以前都没有出现这样的情况,现在突然提示跳转到国服战网的网址,是不是说明了简体中文客户端已经开始进行更新了呢?