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  • 集合的使用

    • 集合:

    特点:   1.无序

        2.没有重复数据

    • 列表转换为集合
    list_1 = [1,3,5,6,7,3,6,9]
    list_1 = set(list_1)#将列表转换为集合
    print(list_1)

    结果:{1, 3, 5, 6, 7, 9}

    • 交集intersection()
    list_1 = [1,3,5,6,7,3,6,9]
    list_1 = set(list_1)#将列表转换为集合
    list_2 = set([2,33,44,5,6,9,77,55])
    #交集
    print(list_1.intersection())

    运算符号显示:

    print(list_1 & list_2)

    结果:{9, 5, 6}

    • 并集union()
    list_1 = [1,3,5,6,7,3,6,9]
    list_1 = set(list_1)#将列表转换为集合
    list_2 = set([2,33,44,5,6,9,77,55])
    #并集
    print(list_1.union(list_2))
    运算符号显示:
    print(list_1. | list_2)
    结果:{1, 33, 3, 2, 5, 6, 7, 9, 44, 77, 55}
    • 差集difference()
    list_1 = [1,3,5,6,7,3,6,9]
    list_1 = set(list_1)#将列表转换为集合
    list_2 = set([2,33,44,5,6,9,77,55])
    #差集
    print(list_1.difference(list_2))
    print(list_2.difference(list_1))
    运算符号显示:
     
    print(list_1 - list_2)
    print(list_2 - list_1)
     
    结果: {1, 3, 7} {33, 2, 44, 77, 55}
    • 子集/父集issubset()/issuperset()
    list_1 = [1,3,5,6,7,3,6,9]
    list_1 = set(list_1)#将列表转换为集合
    list_2 = set([2,33,44,5,6,9,77,55])
    list_3 = set([1,3,5])
    #子集/父集
    print(list_1.issubset(list_2))#list_1是否是list_2的子集,是则True,否则False
    print(list_1.issuperset(list_3))#list_3是list_1的父集,所以返回True
    print(list_3.issubset(list_1))
    print(list_1.issuperset(list_2))#list_1是否是list_2的父集,是则True,否则False

    结果:
     

      False
      True
      True
      False

     
    • 对称差集symmetric_difference()
    list_1 = [1,3,5,6,7,3,6,9]
    list_1 = set(list_1)#将列表转换为集合
    list_2 = set([2,33,44,5,6,9,77,55])
    #对称差集
    print(list_1.symmetric_difference(list_2))
    运算符号显示:
     
    print(list_1 ^ list_2)
    结果: {33, 2, 1, 3, 7, 44, 77, 55}
    • isdisjoint()
    list_1 = [1,3,5,6,7,3,6,9]
    list_1 = set(list_1)#将列表转换为集合
    
    list_3 = set([1,3,5])
    #是否有交集,有则返回True,无则False
    print(list_1.isdisjoint(list_3))
    
    结果:
    False
    • 集合增加值
    list_1 = set([1,3,5])
    #若集合没有这个值,则集合增加一项,否则不变
    list_1.add(999)
    #若集合没有这个值的,则增加到集合,若集合原来就存在的不变
    list_1.update([1,33,44,66])
    print(list_1)
    • 集合删除值 discard()
    list_1 = set([1,3,5])
    #删除
    list_1.remove('2')#remove只能删除集合有的东西,若没有,则报错
    list_1.discard("22")#集合存在该值,则能删除,若没有,则不操作
    print(list_1)
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  • 原文地址:https://www.cnblogs.com/cheng662540/p/7931862.html
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