检查用户输入是否与（随机）整数的任何数字匹配

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，但是我想添加一些内容，在每次不成功的尝试之后，它会显示四位数中有多少个数字是正确的。我还没有尝试过这一点，主要是因为我不确定如何做到这一点。例如random = 1234Enter your guess: 1111You guessed 1 number correctEnter your guess: 1222You guessed 2 numbers correct...... and so on

Answer 1

如果你的意思是职位

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: "))

Answer 2

如果您确实想将这两个数字保留为 int，则可以使用 %（模运算符）来查找整数除以其以 10 为基数的值后的余数。

例如。

    correctnum = 0

    random = 1234

    guess = 1111

    if ((((random-(random % 1000 )) /1000) == (((guess-(guess % 1000 )) /1000)):

      correctnum++

对 100/10/1 重复的公式应该全部比较该“位置”中的数字，而无需转换数据类型。接下来输出 Correctnum 的值，您应该得到您需要的内容。

Answer 3

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：

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