功能性方法或“ 看起来没有循环！ ”：
from itertools import chain, repeat
prompts = chain(["Enter a number: "], repeat("Not a number! Try again: "))
replies = map(input, prompts)
valid_response = next(filter(str.isdigit, replies))
print(valid_response)
Enter a number: a
Not a number! Try again: b
Not a number! Try again: 1
1
或者，如果您想将“错误输入”消息与输入提示分开，如其他答案所示：
prompt_msg = "Enter a number: "
bad_input_msg = "Sorry, I didn't understand that."
prompts = chain([prompt_msg], repeat('\n'.join([bad_input_msg, prompt_msg])))
replies = map(input, prompts)
valid_response = next(filter(str.isdigit, replies))
print(valid_response)
Enter a number: a
Sorry, I didn't understand that.
Enter a number: b
Sorry, I didn't understand that.
Enter a number: 1
1
它是如何工作的？
prompts = chain(["Enter a number: "], repeat("Not a number! Try again: "))
的组合itertools.chain和itertools.repeat将创建一个迭代器，这将产生串"Enter a number: "一次，"Not a number! Try again: "中无数次：
for prompt in prompts:
print(prompt)
Enter a number:
Not a number! Try again:
Not a number! Try again:
Not a number! Try again:
# ... and so on
replies = map(input, prompts)-这里map会将prompts上一步中的所有字符串应用于input函数。例如：
for reply in replies:
print(reply)
Enter a number: a
a
Not a number! Try again: 1
1
Not a number! Try again: it doesn't care now
it doesn't care now
# and so on...
- 我们使用
filter和str.isdigit过滤掉那些只包含数字的字符串：
only_digits = filter(str.isdigit, replies)
for reply in only_digits:
print(reply)
Enter a number: a
Not a number! Try again: 1
1
Not a number! Try again: 2
2
Not a number! Try again: b
Not a number! Try again: # and so on...
并且仅使用第一个数字字符串next。
其他验证规则：
字符串方法：当然，您可以使用其他字符串方法，例如str.isalpha仅获取字母字符串或str.isupper仅获取大写字母。请参阅文档以获取完整列表。
成员资格测试：
有几种不同的执行方式。其中之一是通过使用__contains__方法：
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)
valid_response = next(filter(fruits.__contains__, replies))
print(valid_response)
Enter a fruit: 1
I don't know this one! Try again: foo
I don't know this one! Try again: apple
apple
数字比较：
这里有一些有用的比较方法。例如，对于__lt__（<）：
from itertools import chain, repeat
prompts = chain(["Enter a positive number:"], repeat("I need a positive number! Try again:"))
replies = map(input, prompts)
numeric_strings = filter(str.isnumeric, replies)
numbers = map(float, numeric_strings)
is_positive = (0.).__lt__
valid_response = next(filter(is_positive, numbers))
print(valid_response)
Enter a positive number: a
I need a positive number! Try again: -5
I need a positive number! Try again: 0
I need a positive number! Try again: 5
5.0
或者，如果您不喜欢使用dunder方法（dunder =双下划线），则始终可以定义自己的函数，也可以使用 operator模块中。
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：
如果您不想无限次地问某人来折磨他，可以在呼叫中指定一个限制itertools.repeat。这可以与为next函数提供默认值结合使用：
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，如果用户不小心以大写形式提供了输入，或者在字符串的开头或结尾有空格，我们就不想拒绝输入。为了考虑这些简单的错误，我们可以通过应用str.lower和str.strip方法对输入数据进行预处理。例如，对于成员资格测试，代码如下所示：
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
如果要使用许多函数进行预处理，则使用执行函数合成的函数可能会更容易。例如，使用此处的一个：
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
合并验证规则：
例如，在一个简单的情况下，当程序要求输入1到120岁之间的年龄时，可以添加另一个filter：
from itertools import chain, repeat
prompt_msg = "Enter your age (1-120): "
bad_input_msg = "Wrong input."
prompts = chain([prompt_msg], repeat('\n'.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，在规则很多的情况下，最好实现执行逻辑合取的函数。在下面的例子中我将使用一个现成的一个位置：
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('\n'.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情况下，自定义消息，然后，我很害怕，也没有漂亮的功能性的方式。或者，至少，我找不到一个。