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

# 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)