如何在python中的循環和範圍內生成隨機的非循環數字？

Question

Hi, I'm still a beginner and a bit lost. I'm working on a project for school that requires me to write different small programs that will 'guess' the given password. This is a bruteforce program, and I need it to guess every possible combination of 4 number passwords like those on the old iPhones. My problem is that when I use random.sample it generates the same random numbers multiple times. What function can I use, or what should I change so that the random numbers within the given range don't repeat themselves? I tried doing rand.int but it gave me "TypeError: 'int' object is not iterable"

Additional questions: - How do I get my loop to stop once n == Password4 ? It simply continues, even after the correct password is found. - Is there a way I can count the number of fails(n != Password4) before my success (n == Password4)?

這是我的代碼：

import random 

    Password4 = 1234 
    def crack_password(): 

while True: 
    for n in (random.sample(range(1112, 10000), 1)): 
     while n == Password4: 
      print(n, "is the password") 
      break 
     if n != Password4: 
      print('fail') 
      break 

    crack_password() 

更新：使用代碼現在不產生隨機的非經常性的數字，但適用於我預期目的。請仍然可以自由回答原始問題，並非常感謝您的好意和迅速回復。

新代碼（歸功於@roganjosh）：

import datetime as dt 

    Password4 = 9999 

    def crack_password(): 
     start = dt.datetime.now() 
     for n in range(10000): 
      password_guess = '{0:04d}'.format(n) 
      if password_guess == str(Password4): 
       end = dt.datetime.now() 
       print("Password found: {} in {}".format(password_guess, end - start)) 
       break 
    guesses = crack_password() 

Answer 1

如果你真的想隨機嘗試所有的密碼。這更容易被

import random 
digits = [str(i) for i in range(10)] 
s = [''.join([a,b,c,d]) for a in digits for b in digits for c in digits for d in digits] 
random.shuffle(s) 
real_password = '1234' 
i = 0 
for code in s: 
    if code == real_password: 
     print() 
     print('The password is: ', code) 
     break 
    else: 
     i += 1 
     print(i, ' failures', end='\r') 

Answer 2

你可能要檢查出所有可能的值，在一定規則密碼，e.g「4位」或「8個小寫字符」。考慮這些問題的答案爲出發點：

• Generating 3-letter strings with itertools.product使用字母和產生正長字符串
• itertools.permutations如果你知道的字符集不重複
• 如果你只是想格式化整數零填充在左邊（0000，0001，... 9999）請參閱accomplish that with format strings。

Answer 3

你有兩個while迴路來完成，所以即使你試圖break當你找到密碼，外（第一）while循環剛剛起步而過一遍。

如果你想獨特的猜測，那麼你將不得不考慮排列。但是，由於假定密碼本身是隨機的，因此隨機猜測密碼在破解密碼方面不會比僅僅依次查看整個潛在密碼列表更有效。

嘗試像這樣的：`而真正的原因

import datetime as dt 

Password4 = 5437 

def crack_password(): 
    start = dt.datetime.now() 
    for n in range(9999): 
     password_guess = '{0:04d}'.format(n) 
     if password_guess == str(Password4): 
      end = dt.datetime.now() 
      print "Password found: {} in {}".format(password_guess, end - start) 
      break 
guesses = crack_password()