wim*_*wim 5 python iteration math containers argparse
我有以下代码来创建一个容器,它假装行为像所有素数的集合(实际上隐藏了一个memoised蛮力素数测试)
import math
def is_prime(n):
if n == 2 or n == 3:
return True
if n == 1 or n % 2 == 0:
return False
else:
return all(n % i for i in xrange(3, int(1 + math.sqrt(n)), 2))
class Primes(object):
def __init__(self):
self.memo = {}
def __contains__(self, n):
if n not in self.memo:
self.memo[n] = is_prime(n)
return self.memo[n]
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到目前为止似乎有效:
>>> primes = Primes()
>>> 7 in primes
True
>>> 104729 in primes
True
>>> 100 in primes
False
>>> 100 not in primes
True
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但它并没有很好地与argparse:
>>> import argparse as ap
>>> parser = ap.ArgumentParser()
>>> parser.add_argument('prime', type=int, choices=primes, metavar='p')
_StoreAction(option_strings=[], dest='prime', nargs=None, const=None, default=None, type=<type 'int'>, choices=<__main__.Primes object at 0x7f4e21783f10>, help=None, metavar='p')
>>> parser.parse_args(['7'])
Namespace(prime=7)
>>> parser.parse_args(['11'])
Namespace(prime=11)
>>> parser.parse_args(['12'])
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.7/argparse.py", line 1688, in parse_args
args, argv = self.parse_known_args(args, namespace)
File "/usr/lib/python2.7/argparse.py", line 1720, in parse_known_args
namespace, args = self._parse_known_args(args, namespace)
File "/usr/lib/python2.7/argparse.py", line 1929, in _parse_known_args
stop_index = consume_positionals(start_index)
File "/usr/lib/python2.7/argparse.py", line 1885, in consume_positionals
take_action(action, args)
File "/usr/lib/python2.7/argparse.py", line 1778, in take_action
argument_values = self._get_values(action, argument_strings)
File "/usr/lib/python2.7/argparse.py", line 2219, in _get_values
self._check_value(action, value)
File "/usr/lib/python2.7/argparse.py", line 2267, in _check_value
tup = value, ', '.join(map(repr, action.choices))
TypeError: argument 2 to map() must support iteration
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该文档只是说,
任何支持in运算符的对象都可以作为选项值传递,因此支持dict对象,set对象,自定义容器等.
显然,我不想迭代素数的无限"集合".那么为什么我要argparse尝试map我的素数呢?不只是需要in和not in?
你得到异常的来源非常清楚,请查看:
if action.choices is not None and value not in action.choices:
tup = value, ', '.join(map(repr, action.choices))
msg = _('invalid choice: %r (choose from %s)') % tup
raise ArgumentError(action, msg)
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支票本身没问题.它试图打印出一条有用的错误信息,试图给你所有可能的选择.我想如果你定义迭代器只返回一个repr字符串的东西primes,你可能会破解它做正确的事情.
这是一个文档错误。这是问题 16468 ,已于 2019 年 8 月由PR 15566修复。
编写的库要求choices参数不仅是一个容器,而且是可迭代的,它尝试列出可用的选项,但这不适用于您的情况。您可以尝试通过给它一个__iter__仅返回一些信息字符串的假值来破解它。