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  • python入门_老男孩_集合_元祖

     集合  可变数据类型,内置元素必须是不可变类型,无序,不重复

      增

        set.add

        set.update

      

        del

        pop

        remove

        clear

      查

        for

        切片

    # 集合
    # 创建
    set1 = set({1,2,3})
    # 集合内置元素必须是不可变类型
    # set2 = set({1,[2,3],{'name':larry}}) false
    
    #
    set1.add('小赵女神')
    print(set1)
    
    set1.update('love')
    print(set1)
    
    #
    set1.pop() # 随机删除
    print(set1.pop()) # 有返回值
    print(set1)
    
    set1.remove('l') # 按元素删除
    print(set1)
    
    set1.clear() # 清空集合
    print(set1) # 返回set()
    
    # del set1    # 删除集合
    # print(set1)
    
    #
    set1 = set({1,2,3})
    for i in set1:
        print(i,type(i))
    
    # print(set1[0]) # 不支持索引
    # print(set1[0:3] # 不支持切片
    View Code

      运算

        intersection

        union

        difference

        issubset

        issuperset

    # 运算操作
    # 交集
    set2 = {1,2,3,4,5}
    set3= {3,4,5,6,7,8}
    
    set4 = set2 & set3 
    set5 = set2.intersection(set3)
    
    print(set4, set5)
    # 并集
    set6 = set2 | set3
    set7 = set2.union(set3)
    
    print(set6, set7)
    # 差集
    set8 = set2 - set3
    set9 = set2.difference(set3)
    
    print(set8, set9)
    # 子集
    set3 = {1,2,3}
    set4 = {1,2,3,4,5,6}
    
    print(set3 < set4)
    print(set3.issubset(set4))
    
    print(set4 > set3)
    View Code

      去重小案例

    # 列表去重的两种方法
    # 方法一,转换成集合
    li = [1,3,3,44,5,33,44]
    temp = set(li)
    li = list(temp)
    print(li)
    
    # 方法二,for循环
    li = ['a', 'b', 'c', 'a']
    l = []
    for i in li:
        for i not in l:
            l.append(i)
    print(l)
    View Code

    元祖

      创建

    >>>tup1 = ('Google', 'Runoob', 1997, 2000);
    >>> tup2 = (1, 2, 3, 4, 5 );
    >>> tup3 = "a", "b", "c", "d";   #  不需要括号也可以
    >>> type(tup3)
    <class 'tuple'>
    View Code

      类型

    >>>tup1 = (50)
    >>> type(tup1)     # 不加逗号,类型为整型
    <class 'int'>
     
    >>> tup1 = (50,)
    >>> type(tup1)     # 加上逗号,类型为元组
    <class 'tuple'>
    View Code

      查

    tup1 = ('Google', 'Runoob', 1997, 2000)
    tup2 = (1, 2, 3, 4, 5, 6, 7 )
     
    print ("tup1[0]: ", tup1[0])
    print ("tup2[1:5]: ", tup2[1:5])
    View Code

      拼接

    tup1 = (12, 34.56);
    tup2 = ('abc', 'xyz')
     
    # 以下修改元组元素操作是非法的。
    # tup1[0] = 100
     
    # 创建一个新的元组
    tup3 = tup1 + tup2;
    print (tup3)
    View Code

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  • 原文地址:https://www.cnblogs.com/dignity/p/9754737.html
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