zoukankan      html  css  js  c++  java
  • day3-课堂代码

    # a = ('哈哈', 'xixi', 'hehe')
    # print(a[0])
    # print(a[0:2])
    #
    # # 列表
    # a = ['哈哈', 'xixi', 'hehe', 1, True, (1, 2, 3)]
    # print(a[0])
    # print(a[0:2])
    # a[1] = '嘻嘻'  # 修改,列表:可变数据类型
    # a[-1] = '呵呵'
    # print(a)
    # a.append('旺旺')  # 追加元素
    # print(a)
    # b = [1, 2, 3]
    # a.append(b)
    # print(a)
    # a.extend(b)
    # print(a)
    # a += b
    # print(a)
    # a[1:1] = b
    # print(a)
    # a[1:] = [1, 2, 3]  # 分片赋值
    # print(a)
    # a[:] = [1, 2, 3]
    # print(a)
    # a[0:1] = []  # 删除
    # print(a)
    # del (a[0:1])  # 删除
    # print(a)
    #
    # str1 = 'hello huice'
    # a = list(str1)
    # print(a)
    # print(a.count('e'))
    # print(a.index('l'))
    # print(a.append('mm'))  # 打印函数,但a.append()没有返回值
    # print(a)  # 打印操作的对象有返回值,说明a.append()做的是原地操作,改变对象本身
    #
    # a.pop()  # 弹出最后一个元素
    # a.pop(1)
    # print(a.pop(1))  # 既有返回值,又修改了对象
    # print(a)
    # a.remove('h')  # 没有返回值,pop可以不传参数,传的是索引,remove传的是元素
    # print(a)
    #
    # a.reverse()  # 没有返回值,原地反向
    # print(a)
    
    a = ['Java', 'python', 'C', 'Ruby', 'PHP', 'JS']
    
    # a.sort()
    # print(a)
    # a.sort(reverse=True)
    # print(a)
    # a.sort(reverse=True, key=len)
    # print(a)
    
    # def my_sort(a):
    #     #return len(a)
    #     return a[-1]
    #
    # a.sort(key=my_sort)
    # print(a)
    
    
    # 匿名函数
    # a.sort(key=lambda a: a[-1])
    # print(a)
    #
    # table = [(1, 'zhangsan', 3000), (2, 'lisi', 2500), (3, 'tiantian', 20000)]
    #
    # table.sort(reverse=True, key=lambda a: a[-1])
    # print(table)
    #
    # # sort原地排序,sorted有返回值,原来列表不变
    # table = [(1, 'zhangsan', 3000), (2, 'lisi', 2500), (3, 'tiantian', 20000)]
    # print(sorted(table, key=lambda x: x[-1], reverse=True)[0][1])
    # print(table)
    
    # 遍历列表
    # a = ['Java', 'python', 'C', 'Ruby', 'PHP', 'JS']
    # for ele in a:
    #     print('*')
    #
    # for i in range(len(a)):  # '0,'省略
    #     print(a[i])
    #
    # for i, j in enumerate(a):
    #     print(i, j)
    #
    # a = [1, 2, 3, 4, 5]
    # print([ele ** 2 for ele in a])
    
    # a = [1, 2, 3]
    # b = [2, 3, 4]
    # print(a + b)
    # print(a * 3)
    
    # a = [1, 2, 3]
    # b = ['zhangsan', 'lisi', 'tiantian']
    # c = ['2000', '3000', '20000']
    # print(tuple(zip(a, b, c)))
    # for i in zip(a, b, c):
    #     print(i)
    
    # str0 = 'xiaoming&xiaohua'
    # print(str0.split('i'))
    # print(''.join(str0))
    
    # eval使用的前提是,这个字符串长的像列表或元祖
    # str = '[1,"a",23]'
    # print(type(eval(str)))
    #
    # str = '(1,"a",3)'
    # print(type(eval(str)))
    
    # Set集合 无序不重复元素集 不支持索引 只能用for in 循环,不能用 for in range
    # s1 = {1, 5, 2, 3, 5, 2}  # 集合
    # d1 = {'name': 'zhangsan', 'age': '18'}  # 字典 key value
    # print(type(s1))
    # print(type(d1))
    # print(s1)
    #
    # for i in s1:
    #     print(i)
    #
    # s1 = [8, 1, 5, 2, 3, 5, 2]
    # print(list(set(s1)))  # 用来给列表去重
    # lst0 = list(set(s1))
    # lst0.sort(key=lambda x: s1.index(x))
    # print(lst0)
    #
    # new = []
    # for ele in s1:
    #     if ele not in new:
    #         new.append(ele)
    # print(new)
    
    # 字典 {k1:v2,k2:v2}
    a = {'zhangsan': {'语文': '90', '数学': '100'}, 'lisi': {'语文': '90', '数学': '100'}}
    
    dic = {'name': 'zhangsan', 'age': '17'}
    print(dic['age'])
    dic['age'] = 18  # 修改
    print(dic)
    
    print(dic.get('name'))
    print(dic['name'])
    
    # print(dic.get('six')) #当取不到时会报错返回None
    # print(dic['six'])     #当取不到时会报错
    
    dic['six'] = '女'  # 赋值
    print(dic)
    
    # dic{}
    # dic[key]=value
    
    # dic = {}
    # str = 'k1:1|k2:2|k3:3'
    # str = str.split('|')
    # for i in str:
    #     i = i.split(':')
    #     n = i[0]
    #     m = i[1]
    #     dic[n] = int(m)
    # print(dic)
    
    # dic = {}
    # str = 'k1:1|k2:2|k3:3'
    # str = str.split('|')
    # for i in str:
    #     key = i.split(':')[0]
    #     value = i.split(':')[-1]
    #     dic[key] = int(value)
    # print(dic)
    #
    # print('k1' in dic)
    # print(dic.keys())
    # print(dic.values())
    # print(dic.items())
    #
    # dic.pop('k1')
    # print(dic)
    #
    # # dic = {}
    # # a = ['name', 'age', 'six']
    # # print(dic.fromkeys(a))
    #
    # a = {'k2': 10, 'k4': 5}
    # dic.update(a)
    # print(dic)
    
    # 字典的遍历
    dic = {'k1': 1, 'k2': 2}
    for key in dic.keys():
        print(key)
    for value in dic.values():
        print(value)
    for item in dic.items():
        print(item)
    for k, v in dic.items():
        print(k, v)
    
    # k1:1|k2:2|k3:3
    #
    # dic ={'k3': 3, 'k2': 2, 'k1': 1}
    
    print(sorted(dic.keys(), reverse=True))
    print(sorted(dic.values(), reverse=True))
    print(sorted(dic.items(), reverse=True))
    
    s = {'zhagnsan': 98, 'lily': 99}
    print(sorted(s.items(), key=lambda x: x[-1], reverse=True))
    
    
    # 函数
    def add_two_number(x, y):  # x、y形式参数 形参
        '求加法的函数'
        return x + y
    
    
    a = 3
    b = 4
    # add_two_number(a, b)  # a、b 实际参数 实参
    # add_two_number(3, 4)  # 3、4 实际参数 实参
    print(add_two_number.__doc__)
    print(add_two_number(y=4, x=3))
    
  • 相关阅读:
    洛谷 P2756 飞行员配对方案问题 (二分图匹配)
    HDU 1879 继续畅通工程 (最小生成树)
    POJ 3090 Visible Lattice Points (欧拉函数)
    SPOJ VFMUL
    洛谷 P3803 【模板】多项式乘法(FFT)
    JAVA MyBatis 逆向工程 遇到的坑
    RabbitMQ遇到的坑
    .net webapi action拦截器 统计action耗时
    CEFCharp下载问题
    【进击.NET高级程序员之路】【二】
  • 原文地址:https://www.cnblogs.com/lp475177107/p/9908203.html
Copyright © 2011-2022 走看看