来自metavar`[[USER @] HOST:] FILE的Python argparse AssertionError [英] Python argparse AssertionError from metavar `[[USER@]HOST:]FILE`
问题描述
Argparse 1.1或1.4失败,并显示AssertionError
-读取metavar值的怪异正则表达式似乎炸毁了argparse-可能与当参数数量超过阈值时,Python argparse AssertionError ?
Argparse 1.1 or 1.4 fails with AssertionError
- the weird regex that reads the metavar value seems to blow up argparse - Possibly related to Python argparse AssertionError when number of arguments exceeds threshold?
是否有其他替代方法可以创建或使用元变量[[USER@]HOST:]FILE
?
Is there any alternative way to create or use the metavar [[USER@]HOST:]FILE
?
测试设置:
$ virtualenv-3.5 --always-copy test2
$ ./test2/bin/pip install argparse
Python代码test.py:
Python code test.py:
#!/Users/[username]/Development/test2/bin/python3.5
import os
import sys
print('sys.prefix', sys.prefix)
sys.path.insert(
0, os.path.join(sys.prefix, 'lib/python3.5/site-packages'))
import argparse
print('argparse', argparse.__version__)
parser = argparse.ArgumentParser()
parser.add_argument(
'files',
metavar='[[USER@]HOST:]FILE',
nargs=argparse.PARSER,
)
parser.add_argument('-a', '-A', metavar='PTRN', dest='patterns', default=[])
parser.add_argument('-b', '-B', metavar='PTRN', dest='patterns', default=[])
parser.add_argument('-c', '-C', metavar='PTRN', dest='patterns', default=[])
parser.add_argument('-d', '-D', metavar='PTRN', dest='patterns', default=[])
parser.add_argument('-e', '-E', metavar='PTRN', dest='patterns', default=[])
parser.add_argument('-f', '-F', metavar='PTRN', dest='patterns', default=[])
print(parser.parse_args())
Shell输出:
$ ./test.py
sys.prefix /Users/[username]/Development/test2/bin/..
argparse 1.4.0
Traceback (most recent call last):
File "./test.py", line 24, in <module>
print(parser.parse_args())
File "/Users/[username]/Development/test2/bin/../lib/python3.5/site-packages/argparse.py", line 1725, in parse_args
args, argv = self.parse_known_args(args, namespace)
File "/Users/[username]/Development/test2/bin/../lib/python3.5/site-packages/argparse.py", line 1754, in parse_known_args
namespace, args = self._parse_known_args(args, namespace)
File "/Users/[username]/Development/test2/bin/../lib/python3.5/site-packages/argparse.py", line 1971, in _parse_known_args
self.error(_('too few arguments'))
File "/Users/[username]/Development/test2/bin/../lib/python3.5/site-packages/argparse.py", line 2391, in error
self.print_usage(_sys.stderr)
File "/Users/[username]/Development/test2/bin/../lib/python3.5/site-packages/argparse.py", line 2353, in print_usage
self._print_message(self.format_usage(), file)
File "/Users/[username]/Development/test2/bin/../lib/python3.5/site-packages/argparse.py", line 2309, in format_usage
return formatter.format_help()
File "/Users/[username]/Development/test2/bin/../lib/python3.5/site-packages/argparse.py", line 306, in format_help
help = self._root_section.format_help()
File "/Users/[username]/Development/test2/bin/../lib/python3.5/site-packages/argparse.py", line 236, in format_help
func(*args)
File "/Users/[username]/Development/test2/bin/../lib/python3.5/site-packages/argparse.py", line 358, in _format_usage
assert ' '.join(pos_parts) == pos_usage
AssertionError
推荐答案
解决方法是创建自定义帮助格式化程序,并使用清理后的值进行后处理.这样做的一种方法可能是这样的:
The workaround is to create a custom help formatter and do post-processing using a sanitized value. One way to do so can be like so:
class MyHelpFormatter(argparse.HelpFormatter):
def _format_usage(self, usage, actions, groups, prefix):
result = super(MyHelpFormatter, self)._format_usage(
usage, actions, groups, prefix)
return result.format(user_host_file='[[USER@]HOST]:FILE')
然后使用自定义格式器构造解析器
Then construct the parser using the custom formattter
parser = argparse.ArgumentParser(formatter_class=MyHelpFormatter)
parser.add_argument(
'files',
metavar='{user_host_file}',
nargs=argparse.PARSER,
)
将其余的放入所提供的代码中,将产生类似这样的结果
Plugging the rest into the code supplied, something like this is produced
$ python demo.py
('argparse', '1.1')
usage: demo.py [-h] [-a PTRN] [-b PTRN] [-c PTRN] [-d PTRN] [-e PTRN]
[-f PTRN]
[[USER@]HOST]:FILE ...
demo.py: error: too few arguments
自然,确切的策略可以像在格式化程序中使用str.replace
方法一样简单,也可以比str.format
方法更复杂.
Naturally, the exact strategy can be as simple as using the str.replace
method in the formatter, or more complicated than the str.format
method.
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