路径存在：
这里可以使用pathlib库及其Path.exists方法：
from itertools import chain, repeat
from pathlib import Path
prompts = chain(["Enter a path: "], repeat("This path doesn't exist! Try again: "))
replies = map(input, prompts)
paths = map(Path, replies)
valid_response = next(filter(Path.exists, paths))
print(valid_response)
Enter a path: a b c
This path doesn't exist! Try again: 1
This path doesn't exist! Try again: existing_file.txt
existing_file.txt
限制尝试次数：
如果您不想通过无限次地询问用户来折磨用户，您可以在itertools.repeat 的调用中指定一个限制。这可以与为 next 函数提供默认值相结合：
from itertools import chain, repeat
prompts = chain(["Enter a number:"], repeat("Not a number! Try again:", 2))
replies = map(input, prompts)
valid_response = next(filter(str.isdigit, replies), None)
print("You've failed miserably!" if valid_response is None else 'Well done!')
Enter a number: a
Not a number! Try again: b
Not a number! Try again: c
You've failed miserably!
预处理输入数据：
有时我们不想拒绝用户不小心提供的输入大写或者在字符串的开头或结尾有一个空格。为了考虑到这些简单的错误，我们可以通过应用str.lower和str.strip方法来预处理输入数据。例如，对于成员资格测试，代码将如下所示：
from itertools import chain, repeat
fruits = {'apple', 'orange', 'peach'}
prompts = chain(["Enter a fruit: "], repeat("I don't know this one! Try again: "))
replies = map(input, prompts)
lowercased_replies = map(str.lower, replies)
stripped_replies = map(str.strip, lowercased_replies)
valid_response = next(filter(fruits.__contains__, stripped_replies))
print(valid_response)
Enter a fruit: duck
I don't know this one! Try again: Orange
orange
如果您有许多函数可用于预处理，使用执行 function composition 的函数可能更容易。例如，使用来自here 的那个：
from itertools import chain, repeat
from lz.functional import compose
fruits = {'apple', 'orange', 'peach'}
prompts = chain(["Enter a fruit: "], repeat("I don't know this one! Try again: "))
replies = map(input, prompts)
process = compose(str.strip, str.lower) # you can add more functions here
processed_replies = map(process, replies)
valid_response = next(filter(fruits.__contains__, processed_replies))
print(valid_response)
Enter a fruit: potato
I don't know this one! Try again: PEACH
peach
结合验证规则：
举个简单的例子，比如程序要求年龄在1到120之间，就可以再加一个filter：
from itertools import chain, repeat
prompt_msg = "Enter your age (1-120): "
bad_input_msg = "Wrong input."
prompts = chain([prompt_msg], repeat('
'.join([bad_input_msg, prompt_msg])))
replies = map(input, prompts)
numeric_replies = filter(str.isdigit, replies)
ages = map(int, numeric_replies)
positive_ages = filter((0).__lt__, ages)
not_too_big_ages = filter((120).__ge__, positive_ages)
valid_response = next(not_too_big_ages)
print(valid_response)
但是在有很多规则的情况下，最好实现一个执行logical conjunction的函数。在下面的示例中，我将使用来自here 的现成的：
from functools import partial
from itertools import chain, repeat
from lz.logical import conjoin
def is_one_letter(string: str) -> bool:
return len(string) == 1
rules = [str.isalpha, str.isupper, is_one_letter, 'C'.__le__, 'P'.__ge__]
prompt_msg = "Enter a letter (C-P): "
bad_input_msg = "Wrong input."
prompts = chain([prompt_msg], repeat('
'.join([bad_input_msg, prompt_msg])))
replies = map(input, prompts)
valid_response = next(filter(conjoin(*rules), replies))
print(valid_response)
Enter a letter (C-P): 5
Wrong input.
Enter a letter (C-P): f
Wrong input.
Enter a letter (C-P): CDE
Wrong input.
Enter a letter (C-P): Q
Wrong input.
Enter a letter (C-P): N
N
不幸的是，如果有人需要为每个失败的案例定制一条消息，那么恐怕没有漂亮的功能方式。或者，至少，我找不到。