• python列表补充、循环


    优先掌握部分
     
        切片
    l=['a','b','c','d','e','f']

    print(l[1:5])
    print(l[1:5:2])
    print(l[2:5])
    print(l[-1])


    了解
    print(l[-1:-4])
    print(l[-4:])
    l=['a','b','c','d','e','f']
    print(l[-2:])

    追加
    hobbies=['play','eat','sleep','study']
    hobbies.append('girls')
    print(hobbies)

        删除
    hobbies=['play','eat','sleep','study']
    x=hobbies.pop(1) #不是单纯的删除,是删除并且把删除的元素返回,我们可以用一个变量名去接收该返回值
    print(x)
    print(hobbies)

    x=hobbies.pop(0)
    print(x)

    x=hobbies.pop(0)
    print(x)

    队列:先进先出
    queue_l=[]
    入队
    queue_l.append('first')
    queue_l.append('second')
    queue_l.append('third')
    print(queue_l)
    出队
    print(queue_l.pop(0))
    print(queue_l.pop(0))
    print(queue_l.pop(0))


    堆栈:先进后出,后进先出
    l=[]
    入栈
    l.append('first')
    l.append('second')
    l.append('third')
    出栈
    print(l)
    print(l.pop())
    print(l.pop())
    print(l.pop())

    了解
    del hobbies[1] 单纯的删除
    hobbies.remove('eat') 单纯的删除,并且是指定元素去删除


        长度
    hobbies=['play','eat','sleep','study']
    print(len(hobbies))

        包含in
    hobbies=['play','eat','sleep','study']
    print('sleep' in hobbies)

    msg='hello world egon'
    print('egon' in msg)


    掌握部分
    hobbies=['play','eat','sleep','study','eat','eat']
    hobbies.insert(1,'walk')
    hobbies.insert(1,['walk1','walk2','walk3'])
    print(hobbies)

    print(hobbies.count('eat'))
    print(hobbies)
    hobbies.extend(['walk1','walk2','walk3'])
    print(hobbies)

    hobbies=['play','eat','sleep','study','eat','eat']
    print(hobbies.index('eat'))


    了解部分
    hobbies=['play','eat','sleep','study','eat','eat']
    hobbies.clear()
    print(hobbies)

    l=hobbies.copy()
    print(l)

    l=[1,2,3,4,5]
    l.reverse()
    print(l)

    l=[100,9,-2,11,32]
    l.sort(reverse=True)
    print(l)

    循环

    msg_dic={
    'apple':10,
    'tesla':1000000,
    'mac':3000,
    'lenovo':30000,
    'chicken':10,
    }

    index=0
    while index < len(msg):
    print(msg[index])
    index+=1



    不依赖索引的取值
    msg_dic={
    'apple':10,
    'tesla':1000000,
    'mac':3000,
    'lenovo':30000,
    'chicken':10,
    }
    for item in msg_dic:
    print(item,msg_dic[item])
     
    补充
    range:顾头不顾尾,默认从0开始
    print(type(range(1,2)))
    for i in range(10):
    print(i,type(i))


    break
    continue


    for i in range(10):
    if i == 4:
    # break
    continue
    print(i)


    for+else


    for i in range(10):
    print(i)
    break
    else:
    print('===>')
     
     


  • 相关阅读:
    移动端解决fixed和input弹出虚拟键盘时样式错位
    JS的面向对象
    js计算两个时间范围间的间隔秒数
    使用js过滤字符串前后的空格
    C#时间格式-摘自http://www.cnblogs.com/xiaogongzhu/p/3825600.html
    [dp/贪心]435. 无重叠区间-----经典问题
    【dp】Leetcode面试题 17.16. 按摩师
    [dp]Leetcode.376.摆动序列
    Leetcode 945 使数组唯一的最小增量
    LeetCode 365.水壶问题
  • 原文地址:https://www.cnblogs.com/mengqingjian/p/7217906.html
Copyright © 2020-2023  润新知