使用locals()动态分配变量不适用于递归函数

时间:2014-10-19 17:56:44

标签: python python-3.x recursion exec locals

我有一个递归脚本,正在为汽车抓取一个JSON文件。在每个递归级别,它会添加一个新变量,并将其(以及其他值)传递给递归调用,每次都会在信息中越来越详细。我尝试使用locals()来动态分配变量,但即使在调用之后它仍然是None(我记得有时候locals()是只读的)。

我也尝试使用eval(),它给了我同样的问题(我知道eval并不理想)。理想情况下我想避免使用字典,因为这需要我先用值加载它,这似乎有一些不必要的步骤,但我现在对任何事情持开放态度。

示例:

scraper(manufacturer='Honda')将刮取模型的JSON文件,设置model='Accord'然后递归调用

scraper(manufacturer='Honda, model='Accord')抓取多年的文件,设置为year=2014并递归调用

scraper(manufacturer='Honda', model='Accord', year='2014')这是基本案例

def scraper(self, manufacturers, model=None, year=None):

    if year:
        scrapeurl = '%s&manufacturer=%s&model=%s&year=%s' % (url, manufacturer, model, year)
        return someFinalFunction()

    elif model:
        scrapeurl = '%s&manufacturer=%s&model=%s' % (url, manufacturer, model)

    elif manufacturer:
        scrapeurl = '%s&manufacturer=%s' % (url, manufacturer)

    j = getJSONFromUrl(scrapeurl)
    key, values = j.popitems()

    for value in values:
        locals()[key] = value
        return self.scraper(manufacturer, model, year, color)

我很感激有关如何处理这个的任何意见,我知道Python似乎总是有一些聪明的做事方式,而且我总是在学习更多,所以提前谢谢你!我也在这个例子中使用Python3,如果这改变了什么

2 个答案:

答案 0 :(得分:2)

locals()['key'] = value应为locals()[key] = value


更好的是,使用**kwargs

def scraper(self, manufacturer, model=None, year=None):
    kwargs = dict(manufacturer=manufacturer, model=model, year=year)

    if year:
        scrapeurl = '%s&manufacturer=%s&model=%s&year=%s' % (url, manufacturer, model, year)
        return someFinalFunction()

    elif model:
        scrapeurl = '%s&manufacturer=%s&model=%s' % (url, manufacturer, model)

    elif manufacturer:
        scrapeurl = '%s&manufacturer=%s' % (url, manufacturer)

    j = getJSONFromUrl(scrapeurl)
    key, values = j.popitems()

    for value in values:
        kwargs[key] = value
        return self.scraper(**kwargs)

答案 1 :(得分:1)

目前还不完全清楚你要做什么,但也许这会有所帮助:

def scraper(self, **kwargs):

    if kwargs.get('year') is not None:
        scrapeurl = '{0}&manufacturer={manufacturer}&model={model}&year={year}'
        return someFinalFunction() # not sure why this takes no arguments

    elif kwargs.get('model') is not None:
        scrapeurl = '{0}&manufacturer={manufacturer}&model={model}'

    elif kwargs.get('manufacturer') is not None:
        scrapeurl = '{0}&manufacturer={manufacturer}'

    else:
        raise KeyError

    j = getJSONFromUrl(scrapeurl.format(url, **kwargs))
    key, values = j.popitems()

    for value in values:
        kwargs[key] = value
        return self.scraper(**kwargs)

这使用Python的内置功能将任意关键字参数作为字典处理,以及更现代的str.format字符串格式,以动态处理您正在寻找的参数。唯一的区别是你现在需要调用它:

instance.scraper(manufacturer='...')

而不仅仅是

instance.scraper('...')

字符串格式化的一个示例,混合位置和关键字参数:

>>> '{0}&manufacturer={manufacturer}'.format('foo', **{'manufacturer': 'bar'})
'foo&manufacturer=bar'