Python：随机猜测4位密码，无法解决1个问题

Question

我是Python新手。我有一个程序，让你猜一个随机的4位密码。例如，如果密码是5530，如果我猜里面有一个5（例如5111），它说我猜对了两个数字。如果密码是5535，它说我正确地猜到了3个数字

但实际上，它解释了我答案中的单个5对于其他5是正确的。我只想说，如果有两个或三个5，我猜对了一个数字；如果我为5530键入了两个5，我猜对了两个；如果我为5535键入了三个5，我猜对了三个

import random

no1 = random.randint(0, 9)
no2 = random.randint(0, 9)
no3 = random.randint(0, 9)
no4 = random.randint(0, 9)

password = str(no1) + str(no2) + str(no3) + str(no4)
count = 0

if no1 % 2 == 0:
    count += 1

if no2 % 2 == 0:
    count += 1

if no3 % 2 == 0:
    count += 1

if no4 % 2 == 0:
    count += 1

print(password)
print(f"Hint: The password consists of {count} even number(s)")

guess = input("Guess the 4 numbers: ")
present = 0

if str(no1) in guess:
    present += 1

if str(no2) in guess:
    present += 1

if str(no3) in guess:
    present += 1

if str(no4) in guess:
    present += 1

if present == 4:

    if guess == password:
        print("Congrats! You have guessed the password correctly.")

    else:
        print("All the numbers are present but not in the correct sequence.")
        print(f"You did not make it! The password is {password}.")

else:
    print(f"{present} number(s) of your guess are present in the password.")
    print(f"You did not make it! The password is {password}.")

p.S.我希望我下面的代码可能会被range之类的东西缩短，但我没有信心这样做，所以有人能帮我吗

Answer 1

更改此项：

if str(no1) in guess:
    present += 1
if str(no2) in guess:
    present += 1
if str(no3) in guess:
    present += 1
if str(no4) in guess:
    present += 1

致：

present += guess[0] == str(no1)
present += guess[1] == str(no2)
present += guess[2] == str(no3)
present += guess[3] == str(no4)

另一种方式：

rand_nums = [no1, no2, no3, no4]
for i in range(4)
    if (int(guess[i]) == rand_nums[i]):
        present += 1