python argparse，如何通过其名称引用args [英] python argparse, how to refer args by their name

问题描述

这是关于python中的argparse的问题，可能很容易

this is a question about argparse in python, it is probably very easy

import argparse

parser=argparse.ArgumentParser()

parser.add_argument('--lib')

args = parser.parse_known_args()

if args.lib == 'lib':
    print 'aa'

这可以工作，但是我只想说'lib'(我不想输入更多)，而不是调用args.lib，有没有办法将所有args变量导出到模块之外(即更改范围) ).这样我就可以直接检查lib的值，而不必在前面指定模块的名称

this would work, but instead of calling args.lib, i only want to say 'lib' (i dont want to type more), is there a way to export all the args variable out of the module (ie changing scope). so that i can directly check the value of lib not by specifying name of the module at the front

PS:我有很多变量，我不想重新分配每个变量

PS: i have a lot of variables, i do not want to reassign every single one

推荐答案

首先，我将建议使用args说明符.很清楚lib的来源.就是说，如果您发现自己经常引用某个参数，则可以将其分配给一个较短的名称:

First, I'm going to recommend using the args specifier. It makes it very clear where lib is coming from. That said, if you find you're referring to an argument a lot, you can assign it to a shorter name:

lib = args.lib

有一种方法可以将所有属性立即转储到全局名称空间中，但不适用于函数的本地名称空间，并且在没有充分理由的情况下使用globals是个坏主意.我不会考虑保存几个args.实例是足够好的理由.就是说，这里是:

There's a way to dump all the attributes into the global namespace at once, but it won't work for a function's local namespace, and using globals without a very good reason is a bad idea. I wouldn't consider saving a few instances of args. to be a good enough reason. That said, here it is:

globals().update(args.__dict__)

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