python – 有没有更好的方法来评估用户输入没有’for’循环并将检查器从False更改为True？

2019年7月28日 166次阅读

对
python来说很新,并且作为练习,已经做了一个基于文本的棋盘游戏版本(我是一个狂热的棋盘游戏玩家).

显然有很多地方供用户输入内容,当然我需要评估这些输入并确保它们有效.

我学会了使用True / False检查和来自这个网站的循环(谢谢！),现在我已经做了很多,我想知道是否有更好的方法 – 如果它们是嵌套的,它们会变得有点疯狂我想尽我所能坚持那些像Zen一样的Python规则.

这就是我的意思,在一个简单的例子中,使用通用代码使它有意义(并非所有人都知道这个棋盘游戏的规则！)

目标是确保代码循环,直到给出颜色或“无”.我知道这完成了,我只是想知道是否还有一种我尚未学习的更简化的方法.

colors = ["red", "green", "white", "blue", "yellow"]

for color in colors:
    print(f"You can choose {color}.")

choice = input("Which color would you choose? (input a color or 'none' : ")
checker = False

while not checker:
    if choice not in colors:
        choice = input("That's not one of your options. Choose again: ")

    elif choice == "none":
        print("You don't want a color. That's fine.")
        checker = True

    else:
        print("You chose {color}! Have fun!")
        checker = True

最佳答案 如果你发现自己重复相同的事情,你可以定义一个泛型函数

def prompt_for_valid_input(options):
    while True:
        print(f"You can choose one of {options}.")
        choice = input("Which would you choose? (or 'none' : ")
        if choice == "none" and "none" not in options:
            print("You didn't pick anything. That's fine.")
            return None
        elif choice in options:
            return choice
        else:
            print("That's not one of your options. Choose again. ")

colors = ["red", "green", "white", "blue", "yellow"]
color = prompt_for_valid_input(colors)
if color is not None:
    print(f"You chose {color}! Have fun!")

numbers = [1, 10, 100]
num = prompt_for_valid_input(numbers)
if num is not None:
    print(f"You chose {num}! Have fun!")

所以“没有一会儿/ for循环”,不是真的.没有sentinel变量,是的,如果条件足够简单.

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