# 四 声明式编程练习题
# 1、将names=['egon','alex_sb','wupeiqi','yuanhao']中的名字全部变大写
names = ['egon', 'alex_sb', 'wupeiqi', 'yuanhao']
# # 方式一:手动实现
# new_names=[]
# for line in names:
# new_names.append(line.swapcase())
# print(new_names)
#
# # 方式二:列表生成式
# new_names=[line.swapcase() for line in names]
# print(new_names)
#
# # 方式三:map+匿名函数
# res=map(lambda line :line.swapcase() ,names)
# print(list(res))
# names=[name.upper() for name in names]
# print(names)
# 2、将names=['egon','alex_sb','wupeiqi','yuanhao']中以sb结尾的名字过滤掉，然后保存剩下的名字长度
names = ['egon', 'alex_sb', 'wupeiqi', 'yuanhao']
# new_names=[]
# for line in names:
# if not line.endswith('sb'):
# new_names.append(line)
# print(new_names)
# new_names=[len(line) for line in names if not line.endswith('sb') ]
# print(new_names)
# res=filter(lambda line:not line.endswith('sb'),names)
# print(list(res))
# 3、求文件a.txt中最长的行的长度（长度按字符个数算，需要使用max函数）
# file=[]
# with open('a.txt','rt',encoding='utf-8')as f:
# for line in f:
# res=f.read()
# # print(res)
# res=res.split('\n')
# file.append(res)
# print(file)
# def funcs(x):
# return len(x)
# res=max(file,key=funcs)
# print(res)
# res2=max(res,key=funcs)
# print(len(res2))
# print(max(len(line) for line in f))
# 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','rt',encoding='utf-8')as f:
# print(sum(len(line) for line in f))
# print(sum(len(line) for line in f)) # 求包换换行符在内的文件所有的字符数，为何得到的值为0?
# 5、思考题
# with open('a.txt') as f:
# g=(len(line) for line in f)
# print(sum(g)) #为何报错？
# with open('a.txt') as f:
# print(sum(len(line) for line in f))
# print(sum(g)) #为何报错？(len(line) for line in f)时循环多个值
#
# 6、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
# with open('shopping.txt','rt',encoding='utf-8') as f:
# info=[line.strip('\n').split(',') for line in f]
# print(info)
# cost=sum(float(unit_price) *int(count) for _,unit_price,count in info)
# print(cost)
# with open('shopping.txt', 'rt', encoding='utf-8') as f:
# # for line in f:
# # print(line)
# info=[{'name':line.strip('\n').split(',')[0],
# 'price':int(line.strip('\n').split(',')[1]),
# 'count':int(line.strip('\n').split(',')[2])} for line in f]
# print(info)
# with open('shopping.txt',encoding='utf-8') as f:
# info = [{'name': line.strip('\n').split(',')[0],
# 'price': int(line.strip('\n').split(',')[1]),
# 'count': int(line.strip('\n').split(',')[2])} for line in f if int(line.strip('\n').split(',')[1]) > 10000]
# print(info)
# ====================
# 函数递归
# 四 二分法
#
# 想从一个按照从小到大排列的数字列表中找到指定的数字，
# 遍历的效率太低，用二分法（算法的一种，算法是解决问题的方法）可以极大低缩小问题规模
# 实现类似于in的效果
# l = [1, 2, 10, 30, 33, 99, 101, 200, 301, 311, 402, 403, 500, 900, 1000] # 从小到大排列的数字列表
# # find_index = 900
# def search(find_index, l):
# print(l)
# if len(l)==0:
# print('not find!')
# return
# mid_num = len(l) // 2
# mid_index = l[mid_num]
# # print(mid_index)
# if find_index > mid_index:
# print('right')
# l = l[mid_num + 1:]
# search(find_index, l)
# elif find_index < mid_index:
# print('left')
# l = l[0:mid_num]
# search(find_index, l)
# else:
# print('find it！')
#
# search(901, l)
# 实现类似于l.index(30)的效果
# 查找一个值是否存在于列表中并返回索引
# l = [1, 2, 10, 30, 33, 99, 101, 200, 301, 311, 402, 403, 500, 900, 1000] # 从小到大排列的数字列表
# def search(num,l,start=0,stop=len(l)-1):
# if start <= stop:
# mid=start+(stop-start)//2
# print('start:[%s] stop:[%s] mid:[%s] mid_val:[%s]' %(start,stop,mid,l[mid]))
# if num > l[mid]:
# start=mid+1
# elif num < l[mid]:
# stop=mid-1
# else:
# print('find it',mid)
# return
# search(num,l,start,stop)
# else: #如果stop > start则意味着列表实际上已经全部切完，即切为空
# print('not exists')
# return
#
# search(301,l)