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A_PYTHONESS

#1 Created at ... [Delete] [Delete and Lock User]

第一题

def normalize(name):
    return str.capitalize(name)

第二题

return reduce(lambda x,y:x*y,L)

第三题

 sn=s.split('.')[0]+s.split('.')[-1]
    def char2num(sn):
        return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[sn]
    def str2int(sn):
        return reduce(lambda x, y: x * 10 + y, map(char2num, sn))   
    return str2int(sn)/pow(10,len(s)-s.index('.')-1)

吴童舒

#2 Created at ... [Delete] [Delete and Lock User]

赞，简洁！

毒遍一条街

#3 Created at ... [Delete] [Delete and Lock User]

第一题是不是这样更好？

def normalize(name):
    return name.capitalize()

A_PYTHONESS

#4 Created at ... [Delete] [Delete and Lock User]

谢谢，受教了^^

宇智波星哥_pearl

#5 Created at ... [Delete] [Delete and Lock User]

涉及到好多没学过的函数，不是很懂

逆风的向日葵ZJUESTC

#6 Created at ... [Delete] [Delete and Lock User]

from functools import reduce


def str2float(s):
    sn = s.split('.')[0]+s.split('.')[-1]  
    # split()通过指定分隔符（'.'）对字符串进行切片，分成'123'和'456'，再通过+连接字符串，执行该语句后sn='123456'

    def char2num(sn):
        return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[sn]

    def str2int(sn):
        return reduce(lambda x, y: x * 10 + y, map(char2num, sn))  
        # 结果是123456

    return str2int(sn)/pow(10,len(s)-s.index('.')-1)  
    # pow(10,len(s)-s.index('.')-1的结果是pow（10,3）=1000，123456/1000=123.456

print('str2float(\'123.456\') =', str2float('123.456'))