PYTHON-匿名函数，递归与二分法，面向过程编程-练习

# 四 声明式编程练习题# 一、将names=['egon','alex_sb','wupeiqi','yuanhao']中的名字所有变大写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)# 二、将names=['egon','alex_sb','wupeiqi','yuanhao']中以sb结尾的名字过滤掉，而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))# 三、求文件a.txt中最长的行的长度（长度按字符个数算，须要使用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))# 四、求文件a.txt中总共包含的字符个数？思考为什么在第一次以后的n次sum求和获得的结果为0？（须要使用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全部的字符数，为什么获得的值为0?# 五、思考题# with open('a.txt') as f:#     g=(len(line) for line in f)# print(sum(g)) #为什么报错？# with open('a.txt') as f:#     print(sum(len(line) for line in f))# print(sum(g)) #为什么报错？(len(line) for line in f)时循环多个值## 六、文件shopping.txt内容以下# mac,20000,3# lenovo,3000,10# tesla,1000000,10# chicken,200,1# 求总共花了多少钱？# 打印出全部商品的信息，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，# 遍历的效率过低，用二分法（算法的一种，算法是解决问题的方法）能够极大低缩小问题规模#  实现相似于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)# 实现相似于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: #若是stop > start则意味着列表实际上已经所有切完，即切为空#         print('not exists')#         return## search(301,l)