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 = 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 = 1234random = str(random)#guess = str(input('enter your guess'))guess = '1222'correct = 0for i in range(len(guess)):    if guess[i] == random[i]:        correct += 1print(correct)ouput: 2或者简单地：correct = sum(guess[i] == random[i] for i in range(len(guess)))