在一个范围内迭代超过生成所需输出所需的次数是否有任何不利之处？

Question

我正在学习 100 天的 Python 代码，我正在尝试通过接收用户输入的密码来创建 Python 密码生成器。

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下面的程序运行并生成所需的输出，但我知道肯定有比在范围内迭代任意次数来生成固定长度密码更好的方法。

import random
letters = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']
numbers = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
symbols = ['!', '#', '$', '%', '&', '(', ')', '*', '+']

print("Welcome to the PyPassword Generator!")
nr_letters= int(input('How many letters would you like in your password?: ')) 
nr_symbols = int(input('How many symbols would you like?: '))
nr_numbers = int(input('How many numbers would you like?: '))

password = ""

# Generate unshuffled password

# for i in range(1, (nr_letters + 1)):
#   password += random.choice(letters)
# for i in range(1, (nr_symbols + 1)):
#   password += random.choice(numbers)
# for i in range(1, (nr_numbers +1)):
#   password += random.choice(symbols)
# print(password)

letter_counter = 0
symbol_counter = 0
number_counter = 0

# NOTE: This seems dumb but it works so...

for i in range(0, 100):
  random_int = random.randint(0, 2)
  if random_int == 0 and letter_counter < nr_letters:
    password += random.choice(letters)
    letter_counter += 1
  elif random_int == 1 and symbol_counter < nr_symbols:
    password += random.choice(symbols)
    symbol_counter += 1
  elif random_int == 2 and number_counter < nr_numbers:
    password += random.choice(numbers)
    number_counter += 1

print(password)

有没有一种更简洁的方法可以通过 Python for 循环创建一个随机的、固定长度的字符串？

对于未来，循环遍历的次数多于生成所需输出的次数是否有一个主要缺点？

Answer 1

我想这就是你要找的东西？：

import random
letters = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']
numbers = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
symbols = ['!', '#', '$', '%', '&', '(', ')', '*', '+']

print("Welcome to the PyPassword Generator!")
nr_letters= int(input('How many letters would you like in your password?: ')) 
nr_symbols = int(input('How many symbols would you like?: '))
nr_numbers = int(input('How many numbers would you like?: '))

password = ""

for i in range(1, sum([nr_letters,nr_numbers,nr_symbols])+1):
  if i<= nr_letters:
    password += random.choice(letters)
  elif i > nr_letters and i<=nr_symbols+nr_letters:
    password += random.choice(symbols)
  elif i > nr_symbols+nr_letters:
    password += random.choice(numbers)

password=[x for x in password]
random.shuffle(password)
password="".join(password)
print(password)

基本上，它不是遍历任意数字，而是将所需的字符数相加，然后遍历范围，并在满足要求时添加数字/符号/字母。

要求是基本的。当少于或等于所需字母数时，添加一个字母。然后，您将字母和符号的数量加在一起，并说它必须同时满足字母数量的上方和字母+符号的下方。然后，如果i 在字母+符号上方，则添加数字。差不多就是这样。

缺点是之后你必须像我上面所做的那样洗牌。 password=[x for x in password] 基本上将password 变成了一个列表。然后你用random.shuffle(password)洗牌，然后加入password="".join(password)。

这基本上是您第一次注释掉捆绑在 1 中的迭代。