Python我如何让石头剪纸运行多轮？

Question

编辑：谢谢你的帮助

所以我在一个初级班，没有太多经验，我遇到了这个问题： 我试图让比赛重复五次，我想发生的是

Player inputs choice

Game spits out result

Move onto round 2

Ask for choice again

Game spits out result... etc

但最终发生的是

Player inputs choice

Game spits out result

Move onto round 2

Program asks for choice (I think)

The if/elif is completely disregarded after round 1 as in whatever I input spits no result

Can hit enter five times before program ends

我在这里的挑战是在尽可能简洁的同时做到这一点，因为我可以做到，但最终的程序会非常长，非常难看

无论如何，要问一个特定的问题，是否有可能像这样重复使用同一个变量五次？为什么在第一轮之后完全忽略了if/else

import random

cpu_choice = ["rock","paper","scissors","dynamite"]

for i in range(5):

    choice = input().lower().strip("!.?")
    cpu_choice = random.choice(cpu_choice)

    if "rock" in choice:
        if cpu_choice == "rock":
            print("It's a tie!")
        elif cpu_choice == "scissors":
            print("Dang, I lost!")
        elif cpu_choice == "paper" or cpu_choice == "dynamite":
            print("Haha! I won!")

    elif "paper" in choice:
        if cpu_choice == "rock":
            print("Dang, I lost!")
        elif cpu_choice == "paper":
            print("It's a tie!")
        elif cpu_choice == "scissors" or cpu_choice == "dynamite":
            print("Haha! I won!")

    elif "scissors" in choice:
        if cpu_choice == "rock":
            print("Haha! I won!")
        elif cpu_choice == "scissors":
            print("It's a tie!")
        elif cpu_choice == "paper" or cpu_choice == "dynamite":
            print("Dang, I lost!")

    elif "dynamite" in choice:
        if cpu_choice == "scissors":
            print("Dang, I lost!")
        elif cpu_choice == "dynamite":
            print("It's a tie!")
        elif cpu_choice == "paper" or cpu_choice == "rock":
            print("Haha! I won!")
    else:
        print("Please use rock, paper, scissors, or dynamite!")
    
    ```

Answer 1

您将cpu_选项用于两种不同的事情。第一次，它是cpu拥有的潜在选择列表。但随后，您将其重新定义为cpu实际选择的结果。这意味着（假设cpu在第一次迭代中选择了rock），在第二次迭代中，您有“cpu_choice=random.choice（“rock”）”，因为python将字符串视为列表，所以它将输出“r”、“o”、“c”或“k”。您不会因此收到错误消息，因为您的内部if语句中没有else语句

只需将第二行重命名为潜在的_选项，就可以了