• 集合类型内置方法


    1.作用

    是存储一大堆元素的东西,容器数据类型。用于关系运算的集合体,由于集合内的元素无序且集合元素不可重复,因此集合可以去重,但是去重后的集合会打乱原来元素的顺序。

    2.定义方式

    以{}用逗号隔开不可变数据类型

    空大括号是字典,不是集合,定义空集合必须得用set()

    s = {1,2,1,'a','a','c'}
    print(s)
    

    3.内置方法

    ## 优先掌握
    # 1.长度len
    pythoners = {'jason','nick','tank','sean'}
    print(len(pythoners))
    
    # 2.成员运算in和not in
    print('nick' in pythoners)
    
    pythoners = {'jason','nick','tank','sean'}
    linuxers = {'nick','egon','kevin'}
    # 3.并集|
    print(pythoners | linuxers)
    print(pythoners.union(linuxers))
    
    # 4.交集&
    print(pythoners & linuxers)
    print(pythoners.intersection(linuxers))
    
    # 5.差集
    print(pythoners - linuxers)
    print(pythoners.difference(linuxers))
    
    # 6.对称差集^
    print(pythoners ^ linuxers)
    print(pythoners.symmetric_difference(linuxers))
    
    # 7.==
    print(pythoners == linuxers)
    
    # 8.父集:>,>=
    print(pythoners > linuxers)
    print(pythoners >= linuxers)
    
    # 9.子集:<,<=
    print(pythoners < linuxers)
    print(pythoners <= linuxers)
    
    ## 了解
    pythoners = {'jason','nick','tank','sean'}
    linuxers = {'nick','egon','kevin'}
    
    # 随机删除
    pythoners.pop()
    print(pythoners)
    
    pythoners.update(linuxers)
    print(pythoners)
    
    # 清空
    pythoners.clear()
    print(pythoners)
    
    # 复制
    print(pythoners.copy())
    
    # 移除
    pythoners.remove('nickk')  # 没有报错
    print(pythoners)
    
    pythoners.discard('nickk')  # 没有不报错
    print(pythoners)
    
    pythoners = {'jason','nick','tank','sean'}
    pythoners2 = {'jason','nick','tank','sean','nick2'}
    # 子集
    print(pythoners.issubset(pythoners2))   
    # 父集
    print(pythoners.issuperset(pythoners2))  
    
    pythoners = {'jason','nick','tank','sean'}
    linuxers = {'nick','egon','kevin'}
    pythoners.difference_update(linuxers)
    print(pythoners)
    
    pythoners.symmetric_difference_update(linuxers)
    print(pythoners)
    
    # isdisjoint() 集合没有共同的部分返回True,否则返回False
    pythoners = {'jason','nick','tank','sean'}
    linuxers = {'nick','egon','kevin'}
    res = pythoners.isdisjoint(linuxers)
    print(res)
    

    4.存一个值还是多个值

    多个值

    5.有序 or 无序

    无序

    6.可变 or 不可变(重点)

    可变

    s = {1,2}
    print(id(s))
    s.add(3)
    print(id(s))
    
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  • 原文地址:https://www.cnblogs.com/yushan1/p/11305151.html
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