day03-集合

一、概述

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

二、语法

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

三、关系测试

1、交集(intersection())

>>> 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}

交集(&)

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

2、并集(union())

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

并集(|)

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

3、差集(difference())

差集取的是数值在第一个集合中,但是不在第二个集合中(在我不在你)

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

差集(-)

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

4、issubset()

判断一个集合是否是另一个集合的子集

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

是否是子集(<=)

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

 5、issuperset()

判断一个集合是否是另一个集合的父集

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

是否是父集(>=)

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

 6、对称差集(symmetric_difference())

把两个集合没有交集的数值取出来

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

对称差集(^)

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

 7、isdisjoint()

判断两个集合是否有交集,没有交集,则返回True

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

四、基本操作

1、添加(add())

 1 >>> name_2 = [1,3,5,8,10]
 2 >>> name_2 = set(name_2)
 3 #添加已存在,不报错
 4 >>> name_2.add(1)
 5 >>> name_2
 6 {8, 1, 10, 3, 5}
 7 #添加不存在,添加一个新的数值
 8 >>> name_2.add(11)
 9 >>> name_2
10 {1, 3, 5, 8, 10, 11}

 2、添加多项(update())

1 >>> name_2 = [1,3,5,8,10]
2 >>> name_2 = set(name_2)
3 >>> name_2.update([12,13,14])
4 #输出结果
5 >>> name_2
6 {1, 3, 5, 8, 10, 12, 13, 14}

 3、删除(remove(),pop(),discard())

①remove()

 1 >>> name_2 = [1,3,5,8,10]
 2 >>> name_2 = set(name_2)
 3 >>> name_2
 4 {8, 1, 10, 3, 5}
 5 >>> name_2.remove(1)
 6 #输出
 7 >>> name_2
 8 {8, 10, 3, 5}
 9 #删除不存在的元素,会报错
10 >>> name_2.remove(1)
11 Traceback (most recent call last):
12   File "<input>", line 1, in <module>
13 KeyError: 1

 注:用remove删除时,当元素不存在,会报错

②pop()

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

 注:pop是随机删除集合中的某个元素,并且打印

③discard()

1 >>> name_2 = [1,3,5,8,10]
2 >>> name_2 = set(name_2)
3 >>> name_2.discard(10)
4 #输出结果
5 >>> name_2
6 {8, 1, 3, 5}
7 #删除不存在元素,不报错
8 >>> name_2.discard(10)

注:用discard删除不存在的元素,不会出现报错

4、长度(len())

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

 5、x in s

测试 x 是否是 s 的成员

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

 6、x not in s

测试 x 是否不是 s 的成员

1 >>> name_1 = [1,2,3,4,7,8,7,10]
2 >>> name_1 = set(name_1)
3 #输出
4 >>> 12 not in name_1
5 True
原文地址:https://www.cnblogs.com/Study-Blog/p/6580759.html