阅读目录

• 一、set集合介绍
• 二、集合的方法
• 1、s.add()添加元素
• 3、s.copy()浅拷贝
• 4、s.difference(b)
• 5、s.difference_update(b)
• 6、s.discard(obj)
• 7、s.intersection(b)
• 8、s.intersection_update(b)
• 9、s.isdisjoint(b)
• 10、s.issubset(b)
• 11、s.issuperset(b)
• 12、s.pop()
• 13、s.remove(obj)
• 14、s.symmetric_difference()
• 15、s.symmetric_difference_update(b)
• 16、s.union(b)
• 17、s.update(b)

>>> set1 = {1,2,3,4,1,3}  #定义集合，默认会将重复的去掉

>>> set1

{1, 2, 3, 4}

>>> aa = 'abcde'

>>> set2 = set(aa)  #使用set函数定义集合

>>> set2

{'a', 'b', 'd', 'c', 'e'}

>>> set3 = set('python')

>>> set3

{'y', 'p', 'o', 'h', 'n', 't'}

>>> set2 = set(range(5))  #如果使用数字必须使用迭代序列

>>> set2

{0, 1, 2, 3, 4}

>>> set2

{1, 2, 3, 4, 5}

>>> set2.add(6)

>>> set2

{1, 2, 3, 4, 5, 6}

>>> set2.add('addr')

>>> set2

{1, 2, 3, 4, 5, 6, 'addr'}

2、s.clear()情况所有元素

>>> set2

{1, 2, 3, 4, 5, 6, 'addr'}

>>> set2.clear()

>>> set2

set()

>>> set1

{1, 2, 3, 4, 6, 'addr'}

>>> set3=set1.copy()

>>> set3

{1, 2, 3, 4, 6, 'addr'}

>>> set1 = {1,2,3,4}

>>> set2 = {3,4,5,6}

>>> set1.difference(set2)  #用set2比较set1时，返回set1和set2交集中set1剩余的元素

{1, 2}

>>> set2.difference(set1) #理解同上，返回自身集合中对方集合没有的新集合

{5, 6}

>>> set1 = {1,2,3,4}

>>> set2 = {3,4,5,6}

>>> set1.difference_update(set2)

>>> set1

{1, 2}

>>> set2.difference_update(set1) #没有交集元素，所以不变

>>> set2

{3, 4, 5, 6}

>>> set2

{3, 4, 5, 6}

>>> set2.discard(6)

>>> set2

{3, 4, 5}

>>> set2.discard(7)  #元素不存在，没有报错

>>> set2

{3, 4, 5}

>>> set1 = {1,2,3,4}

>>> set2 = {3,4,5,6}

>>> set1.intersection(set2)  #返回交集值

{3, 4}

>>> set3 = {3,6,8,9}

>>> set1.intersection(set2,set3) #必须三个set都存在才返回

{3}

>>> set1 = {1,2,3,4,5,6}

>>> set2 = {3,4,5,7,8,9}

>>> set1.intersection_update(set2) #取交集值更新到自身set

>>> set1

{3, 4, 5}

>>> set2

{3, 4, 5, 7, 8, 9}

>>> set1 = {1,2,3,4,5}

>>> set2 = {4,5,6,7,8}

>>> set1.isdisjoint(set2)

False

>>> set3 = {22,33}

>>> set1.isdisjoint(set3)

True

>>> s1 = {1,2}

>>> s2 = {1,2,3,4}

>>> s3 = {11,22}

>>> s1.issubset(s2)

True

>>> s1.issubset(s3)

False

>>> s1 = {1,2}

>>> s2 = {1,2,3,4}

>>> s2.issuperset(s1)

True

>>> s1.issuperset(s2)

False

>>> s2 = {'name','age','salary','shool'}

>>> s2

{'salary', 'name', 'shool', 'age'}

>>> s2.pop()

'salary'

>>> s2.pop()

'name'

>>> s2.pop()

'shool'

>>> s1 = {1,2,3,4}

>>> s1.remove(2)

>>> s1

{1, 3, 4}

>>> s1.remove(5)

Traceback (most recent call last):

  File "<stdin>", line 1, in <module>

KeyError: 5

>>> s1 = {1,2,3,4}

>>> s2 = {3,4,5,6}

>>> s1.symmetric_difference(s2)  #返回两个集合不交集的元素

{1, 2, 5, 6}

>>> s2.symmetric_difference(s1)

{1, 2, 5, 6}

>>> s1 = {1,2,3,4}

>>> s2 = {3,4,5,6}

>>> s1.symmetric_difference_update(s2)

>>> s1

{1, 2, 5, 6}

>>> s1 = {1,2,3,4}

>>> s2 = {3,4,5,6}

>>> s1.union(s2)

{1, 2, 3, 4, 5, 6}

>>> s1 = {1,2,3}

>>> s2 = {4,5,6}

>>> s1.update(s2)

>>> s1

{1, 2, 3, 4, 5, 6}