设计猜测 python 中的数字列表的游戏时遇到问题

Question

今天我想设计一个游戏来猜我设置的数字列表。玩家应该打印出所有 5 个数字才能赢得游戏。并且不允许打印重复的数字。代码如下：

def guess(): 

   print "Please input a number smaller than 10, let's see if it is in my plan."

   print "You should figure out all the 5 numbers."

   n = int(raw_input('>'))
   my_list = [1, 3, 5, 7, 9]
    your_list = []
    count = 0
    while count < 5:
        if n in my_list:
            if n not in your_list:
                your_list.append(n) 
                count = count + 1
                print "Good job! You have got %d numbers!" % count
                n = int(raw_input('>'))
            else:
                print "You have already typed that. Input again!"
                n = int(raw_input('>'))
        else:
            print "That is not what I want."
            n = int(raw_input('>'))
    print "Here you can see my plan:", my_list
    print "You are so smart to  guess out, you win!"

guess()

当我尝试运行时，看到结果我很困惑：

 Please input a number smaller than 10, let's see if it is in my plan
 You should figure out all the 5 numbers
>1
Good job! You have got 1 numbers
>2
That is not what I want
>3
Good job! You have got 2 numbers
>5
Good job! You have got 3 numbers
>7
Good job! You have got 4 numbers
>4
Here you can see my plan: [1, 3, 5, 7, 9]
You are so smart to  guess out, you win!

当我输入 4 时，它应该打印 "That is not what I want" 而不是显示“win”。为什么会出错？

Answer 1

您刚刚在 raw_input.

的订单中出现了一些逻辑错误

你在代码的开头多了一个。

而且每个条件的末尾都有一个 raw_input。最好在 while 循环的开头只有一个 raw_input，并根据该输入显示一条消息。

按照你的逻辑，即使你找到了所有 5 个猜测，仍然有一个输入等待，无论输入是什么，它都会显示获胜消息。因此，即使您在最后键入 4，您也会收到结束消息，因为 count<5 条件已完成，但 raw_input 的定位意味着它仍然会请求输入，即使您赢了。

def guess(): 
    print "Please input a number smaller than 10, let's see if it is in my plan."
    print "You should figure out all the 5 numbers."
    my_list = [1, 3, 5, 7, 9]
    your_list = []
    count = 0

    while count < 5:
        n = int(raw_input('>'))
        if n in my_list:
            if n not in your_list:
                your_list.append(n) 
                count = count + 1
                print "Good job! You have got %d numbers!" % count
            else:
                print "You have already typed that. Input again!"
        else:
            print "That is not what I want."

    print "Here you can see my plan:", my_list
    print "You are so smart to  guess out, you win!"

guess()

这是我的固定代码的示例输出：

Please input a number smaller than 10, let's see if it is in my plan.
You should figure out all the 5 numbers.
>1
Good job! You have got 1 numbers!
>2
That is not what I want.
>3
Good job! You have got 2 numbers!
>5
Good job! You have got 3 numbers!
>7
Good job! You have got 4 numbers!
>4
That is not what I want.
>9
Good job! You have got 5 numbers!
Here you can see my plan: [1, 3, 5, 7, 9]
You are so smart to  guess out, you win!

Answer 2

在我这边，它正在工作。唯一一点不起作用，我需要在退出 while 循环之前输入第 6 个元素。我只是要向您概述可能的改进和不同的实施方式。

此外，您正在使用 python 2。您应该考虑转向 python 3，特别是如果您刚刚开始学习它。

改进：

while count < 5: 可以简单地变成 while sorted(my_list) != sorted(your_list):.
可以简化嵌套语句。
检查输入就好了。

实施：

def guess_2():
    print ("Please input a number smaller than 10, let's see if it is in my plan.")
    print ("You should figure out all the 5 numbers.")

    # Parameters
    my_list = [1, 3, 5, 7, 9]
    your_list = []
    count = 0

    while sorted(my_list) != (sorted(your_list)):

        # Input:
        not_valid = True
        while not_valid:
            try:
                n = int(input('>'))
                not_valid = False
            except:
                print ("Please input a number.")

        if n in my_list and not n in your_list:
            your_list.append(n) 
            count = count + 1
            print ("Good job! You have got %d numbers!" % count)

        elif n in my_list and n in your_list:
            print ("You have already typed that. Input again!")

        else:
            print ("That is not what I want.")

    print ("Here you can see my plan:", my_list)
    print ("You are so smart to  guess out, you win!")

Answer 3

截至 2018 年 7 月 24 日，相关代码出现缩进错误。第 3、5、7 和 8 行缩进了 3 个空格而不是 4 个空格。

更正后代码运行（在 Python 2.7 中），但 "raw_input()" 位于错误的位置。在您的代码中：

while count < 5:
    # do some stuff
    # If previous guess correct, increment count
    n = int(raw_input('>'))  # This prompts the user again after the 5th success

raw_input() 应该只有一次：

while count < 5:
    n = int(raw_input('>'))
    # If previous guess correct, increment count

正如 Mathieu 所写，用户输入应该始终得到验证。在这种情况下，如果输入了一个非数字字符，程序会因 ValueError 异常而崩溃。这可以很容易地用 Try: Except: 成语来捕捉：

guess = raw_input('>')
try:
    value = int(guess)
except ValueError:
    # Handle the error

还有一个问题是，如果"my_list"有一个或多个重复的数字，游戏是不可能完成的。

一个可能的解决方案是；无需为 "your_list" 构建第二个列表，您可以简单地从 "my_list" 中删除元素，直到没有元素为止：

def guess(): 
    print "Please input a number smaller than 10, let's see if it is in my plan."
    print "You should figure out all the 5 numbers."
    my_list = [1, 3, 5, 3, 9]
    my_plan = str(my_list)

    while len(my_list) > 0:
        guess = int(raw_input('>'))
        try:
            int(guess)
        except ValueError:
            print "'%s' is not a number." % str(guess)
            continue

        if guess < 0 or guess > 9:
            print "ONE digit 0 to 9 please."
        else:
            if guess in my_list:
                my_list.remove(guess)  # Remove ONE instance of 'guess'
                # or remove ALL instances of 'guess'
                # my_list = [num for num in my_list if num != guess]
                print "Good job! You have got %d numbers!" % (5 - len(my_list))
            else:
                print "That is not what I want."
    print "Here you can see my plan:", my_plan
    print "You are so smart to  guess out, you win!"

guess()

(Python 2.x is legacy, Python 3.x is the present and future of the language)

设计猜测 python 中的数字列表的游戏时遇到问题

Trouble in designing a game of guessing a list of numbers in python

python

python-2.x