Question

我正在学习Python，尝试做的其中一项练习是制作“猜数字”游戏。我做了一个非常简单的程序，但是现在我想进一步介绍一下，并为输入设置一些边界，以便程序可以防止错误。这是我的代码：

# Guess the number game.

import random

print('Hello. What is your name?')

yourName = input() # collects user's name

solution = random.randint(1,20)

print('Well, ' + str(yourName) + ', I am thinking of a number between 1 and 20.')

acceptable = ['1','2','3','4','5','6','7','8','9','10','11','12','13','14','15','16','17','18','19','20'] # acceptable answers

def game():

    for attempts in range(1,6):

                    print('Take a guess. You have ' + str(6 - attempts) + ' attempt(s) remaining')

                    # try:

                    guess = (input())

                    while guess:

                                if guess != acceptable:

                                    print("That is not a valid answer!")

                                    guess = (input())

                                else:

                                    moveon()

def moveon():

                    while guess != solution:

                                if guess < solution:

                                    print('Your guess is too low. Try again.')

                                elif guess > solution:

                                    print('Your guess is too high. Try again.')

                                else:

                                    endofgame()

'''
                    except ValueError:

                            print('Please enter a guess that is a number.')

                            attempts + 1 == attempts
'''

def endofgame():

    if guess == solution:

            print('You guessed it! It took you ' + str(attempts) + ' attempt(s)')

            playAgain()

    else:

            print('Game over! The number I was thinking of was ' + str(solution))

            playAgain()



# Option to play again with a new number

def playAgain():

    global solution

    print('Play again? Y/N')

    playAgain = input()

    if playAgain == 'Y':

            print('Okay, ' + str(yourName) + ', I have another number between 1 and 20.')

            solution = random.randint(1,20)

            game()

    else: print('Thanks for playing!')



# Start game
game()

所以我要做的是确保提示用户输入1到20之间的数字时，输入“ 22”，“咖啡”或“ 14.5”之类的答案将是无效的，并提示他们再试一次并输入有效答案。但是，当我现在运行该程序时，输入的任何答案将被视为无效。我该如何做，以便仅接受某些答案，而其他答案则不被接受？我怀疑除了使用我不知道的列表之外，还有其他方法。预先感谢！

Answer 1

您需要检查列表中是否包含，这是我们在Python中的处理方式：

if guess not in acceptable:

Answer 2

您想使用if guess != acceptable而不是使用if guess not in acceptable。

使用python，您可以使用in命令检查数组中是否存在元素。

Python-是否可以将列表用作if语句的条件？

2 个答案: