DDR爱好者之家 Design By 杰米

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

DDR爱好者之家 Design By 杰米
广告合作:本站广告合作请联系QQ:858582 申请时备注:广告合作(否则不回)
免责声明:本站资源来自互联网收集,仅供用于学习和交流,请遵循相关法律法规,本站一切资源不代表本站立场,如有侵权、后门、不妥请联系本站删除!
DDR爱好者之家 Design By 杰米