python-code-15

作业：
   1 使用递归打印斐波那契数列(前两个数的和得到第三个数，如：0 1 1 2 3 4 7...)

#递归
def fib(a,b,stop):
    if  a > stop:
        return
    print(a,end=' ')
    fib(b,a+b,stop)

fib(0,1,100)

#非递归
def fib(n):
    a,b=0,1
    while a < n:
        print(a,end=' ')
        a,b=b,a+b
    print()

fib(10)

 

   2 一个嵌套很多层的列表，如l=［1,2,[3,[4,5,6,[7,8,[9,10,[11,12,13,[14,15]]]]]]］，用递归取出所有的值

l = [1,2,[3,[4,5,6,[7,8,[9,10,[11,12,13,[14,15]]]]]]]
def tell(l):
    for item in l:
        if type(item) is list:
            tell(item)
        else:
            print(item)

tell(l)

 

   3 编写用户登录装饰器，在登录成功后无需重新登录，同一账号重复输错三次密码则锁定5分钟

import time
import sys
current_user = {'user':None}
def auth(func):
    def wrapper(*args,**kwargs):
        if current_user['user']:
            res = func(*args,**kwargs)
            return res
        tag = True
        count = 0
        while tag:
            if count > 2:
                lock_time = time.time()
                with open('db_lock','at',encoding='utf-8') as f:
                    f.write('%s:%s\n' %(name_inp,lock_time))  #db_lock egon:时间戳
                print('超过3次，锁定5分钟')
                sys.exit()

            # while True:
            name_inp = input('username>>: ').strip() #用户名是否存在不验证，输入存在的用户名
            current_time = time.time()
            #判断是否在锁定中 1 在 判断是否超过5分钟 超过重新输入 不超过退出   2 不在 锁定文件
            with open('db_lock',encoding='utf-8') as f2:
                for line in f2:  #是否锁定 遍历完都没有匹配到 则没锁定
                    line = line.strip('\n').split(':')
                    if name_inp == line[0]:   #是否超过5分钟
                        if current_time - float(line[1]) > 300:
                            # print('锁定解除')
                            break
                        else:
                            print('锁定中剩余时间:%s秒' %(300 - (current_time - float(line[1]))))
                            sys.exit()
                else:
                    pass

            pwd_inp = input('password>>: ')
            with open('db',encoding='utf-8') as f1:  #db  egon:123
                for line in f1:
                    line = line.strip('\n').split(':')
                    if name_inp == line[0] and pwd_inp == line[1]:
                        current_user['user'] = name_inp
                        print('验证成功')
                        res = func(*args,**kwargs)
                        return res
                else:
                    print('用户名或密码错误')
                    count += 1
    return wrapper

@auth
def index():
    print('hello world')

@auth
def home(name):
    print('welcome %s to home page' %name)

index()
home('alex')

 

   
   4、求文件a.txt中总共包含的字符个数？思考为何在第一次之后的n次sum求和得到的结果为0？（需要使用sum函数）

with open('a.txt',encoding='utf-8') as f:
    g=(len(line) for line in f)  #生成器
    print(sum(g))
    print(sum(g))
    print(sum(g))

生成器的next方法是从头到尾一次性的

 

   5、文件shopping.txt内容如下

      mac,20000,3
      lenovo,3000,10
      tesla,1000000,10
      chicken,200,1
      求总共花了多少钱？

   打印出所有商品的信息，格式为[{'name':'xxx','price':333,'count':3},...]

   求单价大于10000的商品信息,格式同上

常规法
total = 0
with open('shopping.txt',encoding='utf-8') as f:
    for line in f:
        line = line.strip('\n').split(',')
        line_monny = int(line[1]) * int(line[2])
        total += line_monny
print('total:%s' %total)

生成器方法
with open('shopping.txt',encoding='utf-8') as f:
    g=(int(line.strip('\n').split(',')[1]) * int(line.strip('\n').split(',')[2]) for line in f)
    print(sum(g))

shopping_info = []
with open('shopping.txt',encoding='utf-8') as f:
    for line in f:
        line = line.strip('\n').split(',')
        keys = ['name','price','count']
        d = {k:v for k,v in zip(keys,line)}  #字典生成式
        shopping_info.append(d)
print(shopping_info)

shopping_info_10000 = []
with open('shopping.txt',encoding='utf-8') as f:
    for line in f:
        line = line.strip('\n').split(',')
        if int(line[1]) > 10000:
            keys = ['name','name','count']
            d = {k:v for k,v in zip(keys,line)}
            shopping_info_10000.append(d)
print(shopping_info_10000)

 

明日默写：
   二分查找

nums=[1,3,7,11,22,34,55,78,111,115,137,149,246,371]
def search(search_num,nums):
    print(nums)
    if len(nums) == 0:
        print('not exists')
        return
    mid_index = len(nums) // 2

    if search_num > nums[mid_index]:
        nums = nums[mid_index+1:]
        search(search_num,nums)
    elif search_num < nums[mid_index]:
        nums = nums[:mid_index-1]
        search(search_num,nums)
    else:
        print('exist')

search(149,nums)

 

预习：模块与包的使用http://www.cnblogs.com/linhaifeng/articles/6379069.html