Question

import random

options = ['Rock','Paper', 'Scissor']
npc = random.choice(options)

print('Hello')
print('We are about to play Rock,Paper,Scissors.')

while True:
  npc = random.choice(options)
  player = str(input('Please declare your weapon: ')).capitalize()
  if (player == npc):
    print('Your choice: ', player)
    print('npc choice: ', npc)
    print('Oopsie looks like we have a tie!')
    print('Lets Try again!')
    continue
  if (player != 'Rock') or (player != 'Paper') or (player != 'Scissor'):
    print('Poo Poo, that is not a valid option! Please try again!')
    continue
  if ((player == 'Rock') and (npc == 'Scissor')) or ((player == 'Scissor') and (npc == 'Paper')) or ((player == 'Paper') and (npc == 'Rock')):
    print('Your choice: ', player)
    print('npc choice: ', npc)
    print('You win!')
    break
  if ((player == 'Rock') and (npc == 'Paper')) or ((player == 'Scissor') and (npc == 'Rock')) or ((player == 'Paper') and (npc == 'Scissor')):
    print('Your choice: ', player)
    print('npc choice: ', npc)
    print('You lose!')
    break

它不断打印出它是一个领带，不会显示任何其他结果。我刚开始编程。任何意见都将非常感谢！

编辑：循环已经解决。

这是根据请求的示例输出：

   Output: Hello
           We are about to play Rock,Paper,Scissors.
           Please declare your weapon: rock
           Your choice:  Rock
           npc choice:  Paper
           You lose!

Answer 1

这一行

http://localhost:4000

如果没有平局，

将始终为if (player != 'Rock') or (player != 'Paper') or (player != 'Scissor'):。将其更改为

True

改善代码的一些建议

您可以删除所有if player not in options:中的()。此

if

与

相同

if (player == npc):

您还应该使用if player == npc:而不是if/elif/else。这样就可以使用if unncessary。

修改：改进版本：

continue

Answer 2

您的计划中存在逻辑错误。

特别是这一行：

if (player != 'Rock') or (player != 'Paper') or (player != 'Scissor'):

如果至少有一个它所链接的语句为True，则'或'运算符返回True。

例如，假设玩家选择了“Rock”。现在第一个语句player != 'Rock'为False，但第二个语句player != 'Paper'为True，player != 'Scissor'也是如此。

因此，整个语句变为False or True or True，这是True，程序最终告诉用户他们的选择无效。

您可以使用'和'代替'或'来轻松解决此问题：

if (player != 'Rock') and (player != 'Paper') and (player != 'Scissor'):

此处，语句变为False and True and True，这是假的。仅当玩家按预期输入的选项不是Rock, Paper, Scissor选项时，此语句才返回True。

更多Pythonic的做法是用以下内容替换整个语句，如另一个答案所述：

if player not in options:

Answer 3

我建议您进行一些改进，例如使用if-elif，而不是continue。同时使用.format(...)表示回复。

对于循环问题，将第二个if语句中的逻辑运算符更改为and运算符以进行包含迭代。

最终格式化的代码如下所示：

import random

options = ['Rock','Paper', 'Scissor']
npc = random.choice(options)

print('Hello')
print('We are about to play Rock,Paper,Scissors.')

while True:
    npc = random.choice(options)
    player = str(input('Please declare your weapon: ')).capitalize()
    if (player == npc):
        print('Your choice: {}'.format(player))
        print('npc choice: {}'.format(npc))
        print('Oopsie looks like we have a tie!')
        print('Lets Try again!')
    elif (player != 'Rock') and (player != 'Paper') and (player != 'Scissor'):
        print('Poo Poo, that is not a valid option! Please try again!')
    elif ((player == 'Rock') and (npc == 'Scissor')) or ((player == 'Scissor') and (npc == 'Paper')) or ((player == 'Paper') and (npc == 'Rock')):
        print('Your choice: {}'.format(player))
        print('npc choice: {}'.format(npc))
        print('You win!')
        break
    elif ((player == 'Rock') and (npc == 'Paper')) or ((player == 'Scissor') and (npc == 'Rock')) or ((player == 'Paper') and (npc == 'Scissor')):
        print('Your choice: {}'.format(player))
        print('npc choice: {}'.format(npc))
        print('You lose!')
        break

摇滚纸剪刀游戏，无限循环

3 个答案: