函数方法或“看，妈妈，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
或者如果你想有一个“错误输入”的消息从输入提示符中分离出来，就像在其他答案中:
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: "))
This combination of itertools.chain and itertools.repeat will create an iterator
which will yield strings "Enter a number: " once, and "Not a number! Try again: " an infinite number of times:
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) - here map will apply all the prompts strings from the previous step to the input function. E.g.:
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...
We use filter and str.isdigit to filter out those strings that contain only digits:
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...
And to get only the first digits-only string we use next.
其他验证规则:
String methods: Of course you can use other string methods like str.isalpha to get only alphabetic strings, or str.isupper to get only uppercase. See docs for the full list.
Membership testing:
There are several different ways to perform it. One of them is by using __contains__ method:
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
Numbers comparison:
There are useful comparison methods which we can use here. For example, for __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
Or, if you don't like using dunder methods (dunder = double-underscore), you can always define your own function, or use the ones from the operator module.
Path existance:
Here one can use pathlib library and its Path.exists method:
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:
如果您不想通过无数次地询问用户某个问题来折磨用户，您可以在itertools.repeat调用中指定一个限制。这可以与为下一个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，如果用户不小心以大写形式提供输入，或者在字符串的开头或结尾使用空格，我们不想拒绝输入。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，在成员测试的情况下，代码看起来像这样:
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
在有许多函数用于预处理的情况下，使用一个函数执行函数组合可能更容易。例如，使用这里的一个:
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
组合验证规则:
例如，对于一个简单的情况，当程序要求年龄在1到120之间时，可以添加另一个过滤器:
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)
但是在有很多规则的情况下，最好实现一个执行逻辑连接的函数。在下面的例子中，我将使用一个现成的例子:
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定制消息，那么恐怕没有非常实用的方法。或者，至少我找不到。
