Python简洁性: Comparing string to multiple sets with return value

Question

我从未找到一种简洁的方法，就是将一个字符串与多个集合进行比较，或者可能是一个能够提供一些有意义的输出的长字符串。这是我当前的代码:

def input():  
    text = input("Choose a number: ").lower()  
    if "1" in text or "one" in text: return 1  
    elif "2" in text or "two" in text: return 2  
    ... and so on...  
    elif "9" in text or "nine" in text: return 9
    else: return 0  

考虑到它现在有多难看，我的目标是让它更加简洁。

另外，还有一点问题:这样做更快，还是:

    if any(x in text for x in (["1", "one"])): return 1  

哪一种被认为更主流？

抱歉；我的Python确实是入门级的。

--- 编辑 ---

对造成的困惑表示歉意；我实际上指的是更普遍的东西，例如:

def input():  
        text = input("Give an input: ").lower()  
        if any(x in text for x in (["eggplant", "emoji", "purple"])): return 1  
        elif any(x in text for x in (["tomato", "red", "tomatina"])): return 2  
        ... and so on...  
        elif any(x in text for x in (["lettuce", "green", "avatar"])): return 9
        else: return 0  

Answer 1

根据查询的规律性，您可以使用某种循环:

for i, s in enumerate('one two three four ... nine'.split(), 1):
  if str(i) in text or s in text:
    return i
return 0