python | 密码生成器，保证有至少一个数字和符号

题目要求：

定义一个函数，第一传参指的是密码的长度；如果第二个传参是True，生成的密码中至少含有一个数字；如果第三个传参是True，生成的密码中至少含有这些字符（!?=+-()#）中的一个。

我的代码虽然能够成功实现题目要求的功能，但是非常冗余，找不到缩减之法。让我们来与标准答案进行对比学习：

1. 我的代码

   import string
   from random import sample
   
   def generate_strong_password(length, num, characters):
       pool = string.ascii_lowercase
       if num:
           for number in range(0, 10):
               pool += str(number)
   
       if characters:
           char = "!?=+-()"
           pool += char
   # pool has all ascii_lowercase, puntuations and chars.
   
       while True:
           count_punc = 0
           count_num = 0
           pw = sample(pool,length)  # 获得个数为length的随机取样字符，return as a list.
           line = ""
           for item in pw:
               if item in string.punctuation:
                   count_punc += 1  # The number of punctuation recorded
               elif item.isdigit():
                   count_num +=1  # The number of numbers recorded
           if num == True and characters == True:
               if count_num >= 1 and count_punc >=1:  # ensure at least one num and punctuation included
                   for pwl in pw:
                       line += pwl
                   return line
           elif num == False and characters == False:
               for pwl in pw:
                   line += pwl
               return line
           elif num == False and characters == True:
               if count_punc >= 1:  # # ensure at least one punctuation included
                   for pwl in pw:
                       line += pwl
                   return line
           else:
               if count_num>=1:  # ensure at least one num included
                   for pwl in pw:
                       line += pwl
                   return line
   
   if __name__ == "__main__":
       print(generate_strong_password(8, True, True))
   

2. 标准答案

   from random import choice, randint
   from string import ascii_lowercase, digits
    
   def generate_strong_password(length: int, numbers: bool, special_characters: bool):
       special_chars = "!?=+-()#"
       # One letter at beginning of the password
       passwd = choice(ascii_lowercase)
       choice_set = ascii_lowercase
    
       # If numbers is wanted, add at least one number
       if numbers:
           passwd = add_character(passwd, digits)  # string的digits功能提供了0-9这10个数字
           choice_set += digits
    
       # same for special characters
       if special_characters:
           passwd = add_character(passwd, special_chars)
           choice_set += special_chars
    
       # build the rest of the password from the whole set
       while (len(passwd) < length):
           passwd = add_character(passwd, choice_set)
    
       return passwd
    
   # Add a random character from the given set either
   # at the beginning or end of the string
   def add_character(passwd: str, characters):
       character = choice(characters)
       if randint(1,2) == 1:
           return character + passwd
       else:
           return passwd + character
   

   边看标准答案变思考，为什么答案这么写能够保证获得的密码是无序的呢？（特别是看到One letter at beginning of the password这一行时）

   看到最后，明白了是add_character函数使之变得无序：两个字符排序时，只有前和后两种顺序，利用randint(1,2)保证了前后的任意性。

   以下是用对表核准标准答案的简单具体分析。
   反观我的代码，是先随机生成length长度的密码，再用if-else和while组合判断其中字母和符号的个数再输出。随机性只用到了sample。另外还有一个繁琐之处在于sample的使用使密码变成了一个数组，需要for循环生成字符串，又是一个浪费时间的点。

题源：Helsinki, programming, Part 7