• 3.21 DAY4


    int会默认去除空格

    1.APPEND

    list = [1,'a','b',2,3,'老男孩']
    list.append('alex')
    print(list)

    print(li.append('alex') return:none这个是动作,没有返回值

    2.INSERT

    list = [1,'a','b',2,3,'老男孩']
    list.insert(2,'kate')
    print(list)

    3.EXTEND

    li = [1,'a','b',2,3,'老男孩']
    li.extend('abc')
    print(li)
    ##HR添加新员工
    name_list = ['kate','nero','jake'] while True: name = input('请添加员工姓名:按Q/q退出') if name.upper() == 'Q':break else: name_list.append(name) print('已成功添加新员工%s % name') print(name_list)

    #POP唯一一个有返回值的,按索引删,只能删一个元素

    li = ['kate','nero','jake',1,2,3]
    li.pop()
    print(li)
    空格默认删除最后一个
    li = ['kate','nero','jake',1,2,3]
    li.pop(1)
    print(li)
    ['kate', 'jake', 1, 2, 3]
    li = ['kate','nero','jake',1,2,3]
    print(li.pop(1))
    返回值 nero

    #remove按元素删,默认就删一个

    li = ['kate','nero','jake',1,2,3]
    li.remove('kate')
    print(li)

    # clear 清空内容

    li = ['kate','nero','jake',1,2,3]
    li.clear()
    print(li)

    return[]

    # del 删除列表

    li = ['kate','nero','jake',1,2,3]
    del li[0:2]
    print(li)

    # 跳着删

    li = ['kate','nero','jake',1,2,3]
    del li[0:5:2]
    print(li)

    ['nero', 1, 3]

    1.按照索引改

    li = ['kate','nero','jake',1,2,3]
    li[1]= 'KATE'
    print(li)

    ['kate', 'KATE', 'jake', 1, 2, 3]
    print(li[1]) 只是打印元素 KATE

    2.按切片去改

    li = ['kate','nero','jake',1,2,3]
    li[:3] = '0'
    print(li)
    li = ['kate','nero','jake',1,2,3]
    li[:3] = 'nerosb'
    print(li)

    ['n', 'e', 'r', 'o', 's', 'b', 1, 2, 3]
    li = ['kate','nero','jake',1,2,3]
    li[:3] = [11,22,33,44]
    print(li)

    把原来位置清空,再放入新的元素一个一个添加进去
    [11, 22, 33, 44, 1, 2, 3]

    查:索引,切片步长,查看

    li = ['kate','nero','jake',1,2,3]
    print(li[:3])
    for i in li:
        print(i)

    kate
    nero
    jake
    1
    2
    3

    #sort: 正向排序,从小到大

    li = [2,7,9,0,2,3,4,5,5]
    li.sort()
    print(li)

    [0, 2, 2, 3, 4, 5, 5, 7, 9]

    #反向排序,从大到小

    li = [2,7,9,0,2,3,4,5,5]
    li.sort(reverse=True)
    print(li)

    [9, 7, 5, 5, 4, 3, 2, 2, 0]

    # 翻转 reverse

    li = [2,7,9,0,2,3,4,5,5]
    li.reverse()
    print(li)

    [5, 5, 4, 3, 2, 0, 9, 7, 2]

    # len 长度

    li = [2,7,9,0,2,3,4,5,5]
    print(len(li))

    9

    # count

    li = [2,7,9,0,2,3,4,5,5]
    
    print(li.count(2))

    2

    # index 通过元素找索引

    li = [2,7,9,0,2,3,4,5,5]
    
    print(li.index(2))

    0

    列表的嵌套

    li = [1,2,'alex',['100','wusir',99],22]
    
    li[2]=li[2].capitalize()
    print(li)
    
    li[3][1] = li[3][1].upper()
    print(li)
    
    li[3][2] = li[3][2] + 1
    print(li)

    1.将alex变成Alex

    2.将wusir变成WUSIR

    3.将99变成100

    [1, 2, 'Alex', ['100', 'wusir', 99], 22]
    [1, 2, 'Alex', ['100', 'WUSIR', 99], 22]
    [1, 2, 'Alex', ['100', 'WUSIR', 100], 22]

    #元组:不能增删改,只能查,通过切片,索引,for循环来查

    tu = (1,2,'alex','oldboy')
    print(tu[:2])
    
    print(tu[2])
    
    for i in tu:
        print(i)

    #儿子不能改,孙子可能改
    tul = (1,2,'alex',[1,'taibai'],'1,2,3','oldboy')
    tul[3].append('日天')
    print(tul)

    元组中含列表,只能改列表

    #RANGE:

    1.当成数字列表range(1,10,2)范围加步长

    for i in range(1,100,1):
        print(i)
    打印1-100

    2.还可以倒着取值

    
    
    for i in range(100,0,-1):
    print(i)

    打印100-1倒叙

    3.len

    li = input('请输入内容')
    for i in range(0,len(li)):
        print(i)

    #Join:iterable 对可迭代的对象进行操作,字符串,元组,列表都可用,但是字符串用最方便

    s = '*'.join('老男孩')
    print(s)

    老*男*孩

    # split str转list

    s = 'wusir alex taibai'
    print(s.split())

    ['wusir', 'alex', 'taibai'] 字符串变列表

    # join  list转str

    s =['wusir', 'alex', 'taibai']
    print(''.join(s))

    wusiralextaibai 列表变字符串

    思考题:

    1.把每一个元素依次打印,两种方法

    第一种:
    li = [1,2,3,['alex','wusir','老男孩'],4] for i in li: if type(i) == list: for k in li[3]: print(k) else: print(i)

      第二种:

     
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  • 原文地址:https://www.cnblogs.com/kateli/p/8618875.html
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