Question

我需要根据给定的字符串创建自定义对象的实例（下例中的Bar）。如果我不将类型更改为Bar并运行以下代码：

import argparse

VALID_BAR_NAMES = ['alfa', 'beta', 'gamma', 'delta']


class Bar:

    def __init__(self, name):
        if not name in VALID_BAR_NAMES:
            raise RuntimeError('Bar can not be {n}, '
                               'it must be one of {m}'.format(
                                    n=name, m=', '.join(VALID_BAR_NAMES)))
        self.name = name


if __name__ == '__main__':
    parser = argparse.ArgumentParser()

    parser.add_argument('foo', help='Specify the foo!')

    parser.add_argument('-b', '--bar', nargs='*',
                        choices=VALID_BAR_NAMES,
                        type=str,  # SELECTED TYPE
                        help=('Specify one or many valid bar(s)'))
    parsed_arguments = parser.parse_args()

在将无效的段hello传递给-b时，我得到了相当不错的输出：

usage: Example.py [-h]
                  [-b [{alfa,beta,gamma,delta} [{alfa,beta,gamma,delta} ...]]]
                  foo
Example.py: error: argument -b/--bar: invalid choice: 'hello' (choose from 'alfa', 'beta', 'gamma', 'delta')

但是，如果我将type=str更改为type=Bar并再次运行该示例，我会得到此输出：

Traceback (most recent call last):
  File "C:\PyTest\Example.py", line 25, in <module>
    parsed_arguments = parser.parse_args()
  File "C:\Python27\lib\argparse.py", line 1688, in parse_args
    args, argv = self.parse_known_args(args, namespace)
  File "C:\Python27\lib\argparse.py", line 1720, in parse_known_args
    namespace, args = self._parse_known_args(args, namespace)
  File "C:\Python27\lib\argparse.py", line 1926, in _parse_known_args
    start_index = consume_optional(start_index)
  File "C:\Python27\lib\argparse.py", line 1866, in consume_optional
    take_action(action, args, option_string)
  File "C:\Python27\lib\argparse.py", line 1778, in take_action
    argument_values = self._get_values(action, argument_strings)
  File "C:\Python27\lib\argparse.py", line 2218, in _get_values
    value = [self._get_value(action, v) for v in arg_strings]
  File "C:\Python27\lib\argparse.py", line 2233, in _get_value
    result = type_func(arg_string)
  File "C:\PyTest\Example.py", line 12, in __init__
    n=name, m=', '.join(VALID_BAR_NAMES)))
RuntimeError: Bar can not be hello, it must be one of alfa, beta, gamma, delta

哪个看起来很糟糕。我知道这是由于在检查之前发生的类型转换，对可用选项进行了检查。处理这个问题的最佳方法是什么？

Answer 1

您需要创建自定义操作（未经测试）：

class Bar: 
    ...


class BarAction(argparse.Action):
    def __call__(self,parser,namespace,values,option_string=None):
        try:  #Catch the runtime error if it occures.
           l=[Bar(v) for v in values] #Create Bars, raise RuntimeError if bad arg passed.
        except RuntimeError as E:
           #Optional:  Print some other error here.  for example: `print E; exit(1)`  
           parser.error()

        setattr(namespace,self.dest,l) #add the list to the namespace

...
parser.add_argument('-b', '--bar', nargs='*',
                    choices=VALID_BAR_NAMES,
                    action=BarAction,  # SELECTED TYPE -- The action does all the type conversion instead of the type keyword.
                    help=('Specify one or many valid bar(s)'))

...

Answer 2

将参数保留为字符串，直到解析完成。解析完成后首先将它们转换为域对象。

Answer 3

不知道是否有帮助，但我将argparse.FileType视为参考：

class Bar:
    def __call__(self, strting):
        if not name in VALID_BAR_NAMES:
            raise argparse.ArgumentError(None,'Bar can not be {n}, '
                               'it must be one of {m}'.format(
                                    n=name, m=', '.join(VALID_BAR_NAMES)))
        self.name = string

    def __init__(self):
        pass

    def __repr__(self):
        return name

parser.add_argument('-b', '--bar', nargs='*',
                    choices=VALID_BAR_NAMES,
                    type=Bar(),  # SELECTED TYPE
                    help=('Specify one or many valid bar(s)'))

使用带有argparse的自定义类型时的正常参数错误

3 个答案: