Python中如何操作集合？很多新手对此不是很清楚，为了帮助大家解决这个难题，下面小编将为大家详细讲解，有这方面需求的人可以来学习下，希望你能有所收获。
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>>> 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'>>>> 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}>>> 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}>>> 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}判断一个集合是否是另一个集合的子集
>>> 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判断一个集合是否是另一个集合的父集
>>> 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把两个集合没有交集的数值取出来
>>> 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}判断两个集合是否有交集，没有交集，则返回True
>>> 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>>> 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}>>> 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}>>> 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}>>> 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}>>> 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>>> 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>>> 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}>>> 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}>>> 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>>> 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>>> 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)>>> name_1 = [1,2,3,4,7,8,7,10]
>>> name_1 = set(name_1)
#结果输出
>>> len(name_1)
7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测试 x 是否不是 s 的成员
>>> name_1 = [1,2,3,4,7,8,7,10]
>>> name_1 = set(name_1)
#输出
>>> 12 not in name_1
True1、云计算，典型应用OpenStack。2、WEB前端开发，众多大型网站均为Python开发。3.人工智能应用，基于大数据分析和深度学习而发展出来的人工智能本质上已经无法离开python。4、系统运维工程项目，自动化运维的标配就是python+Django/flask。5、金融理财分析，量化交易，金融分析。6、大数据分析。
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