Python的argparse选择限制了打印 [英] Python's argparse choices constrained printing

问题描述

当前，我希望Python的argparse模块仅输出'1-65535'而不是{1,2,3，... 65535}，但是文档似乎没有为此提供任何方法.有什么建议吗?

Currently I want Python's argparse module to only print out '1 - 65535' rather than {1, 2, 3, ... 65535}, but the documentation doesn't seem to provide any method for this. Any suggestions?

推荐答案

您可以通过设置

You can alter the way defaults are formatted by setting the formatter_class option.

我将 HelpFormatter类子类化为更改格式化choices值的方式.此类正式是实现细节"，但我怀疑它会在较新的python版本中发生很大变化.

I'd subclass the HelpFormatter class to alter the way it formats your choices values. This class is officially an "implementation detail" but I doubt it'll change much with newer python versions.

_metavar_formatter方法格式化{1, 2, ..., 65535}字符串，您的子类可以覆盖它:

The _metavar_formatter method formats the {1, 2, ..., 65535} string and your subclass could override that:

class RangeChoiceHelpFormatter(HelpFormatter):
    def _metavar_formatter(self, action, default_metavar):
         if action.metavar is not None:
             result = action.metavar
         elif action.choices is not None:
             result = '{%s .. %s}' % (min(action.choices), max(action.choices])
         else:
             result = default_metavar

          def format(tuple_size):
              if isinstance(result, tuple):
                  return result
              else:
                  return (result, ) * tuple_size
          return format

另一种选择是不在如此大的范围内使用choices参数，而是定义一个新的

Another option is to not use the choices argument for such a large range, and instead define a new argument type.

这只是一个可调用的，传递的字符串，如果无法将字符串转换为目标类型，则引发argparse.ArgumentTypeError，TypeError或ValueError，否则将转换为值:

This is just a callable, passed a string, that raises argparse.ArgumentTypeError, TypeError or ValueError if the string cannot be converted to the target type, or the converted value otherwise:

class IntRange(object):
    def __init__(self, start, stop=None):
        if stop is None:
            start, stop = 0, start
        self.start, self.stop = start, stop

    def __call__(self, value):
        value = int(value)
        if value < self.start or value >= self.stop:
            raise argparse.ArgumentTypeError('value outside of range')
        return value

您可以像这样使用它作为参数类型:

You can use this as the argument type like this:

parser.add_argument('foo', type=IntRange(1, 65536))

并调整您的帮助消息以指示可以接受的值.

and adjust your help message to indicate what values are acceptable.

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