检查用户输入是否匹配（随机）整数的任何数字

Question

random = random.randint(1000, 9999)
guess = int(input("Enter your guess: "))
while guess != random:
    guess = int(input("That was incorrect! Enter your guess: "))

这是一个非常简单的猜谜游戏，但我想包括一些内容，在每次尝试不成功后，它会说出四位数中有多少个数字是正确的。

我没有尝试过，主要是因为我不确定如何做到这一点。

例如

random = 1234
Enter your guess: 1111
You guessed 1 number correct
Enter your guess: 1222
You guessed 2 numbers correct
...... and so on

Answer 1

如果您真的想将两个数字都保留为 int，您可以使用 %（模运算符）除以其以 10 为底的值后的余数。

例如

    correctnum = 0
    random = 1234
    guess = 1111
    if ((((random-(random % 1000 )) /1000) == (((guess-(guess % 1000 )) /1000)):
      correctnum++

为 100/10/1 重复的公式应该都比较该“点”中的数字，而无需转换数据类型。接着输出 correctnum 的值，你应该有你需要的东西。

Answer 2

random = 1234
random = str(random)

#guess = str(input('enter your guess'))
guess = '1222'
correct = 0
for i in range(len(guess)):
    if guess[i] == random[i]:
        correct += 1

print(correct)
ouput: 2

或者简单地说：

correct = sum(guess[i] == random[i] for i in range(len(guess)))

Answer 3

如果你指的是位置

random = random.randint(1000, 9999)
guess = int(input("Enter your guess: "))
while guess != random:
    right = 0
    index = 0
    for char in str(guess):
        if char == str(random)[index]:
            index += 1
            right += 1
    print(f'{right} were right')
    guess = int(input("That was incorrect! Enter your guess: "))

检查用户输入是否匹配（随机）整数的任何数字

Check if userinput matches any digits of a (random) integer

python

random

integer

selection