将抓取的结果并排分类成一行

Classify scraped results side by side into a row

所以我正在使用 python/scrapy 从网页中抓取数据。网页基本上由包含各种信息的 15 个块组成。我的蜘蛛重复遍历每个块以抓取一些特定内容。我对结果的内容很满意,但对数据的呈现方式不满意。我希望属于一个块的所有抓取信息都显示在一行中。你会从下面的截图中看到,同一个区块的结果并没有并排呈现,这就是我想要的。

def parse(self, response):
    for i in response.css('span.dir'):
        yield {'address': i.css('b::text').extract()}
    for l in response.css('div.datos'):
        yield {'area': l.css('i::text').extract()}
    for x in response.css('div.opciones'):
        yield {'price stable': x.css('span.eur::text').extract()}
    for o in response.css('div.opciones'):
        yield {'price drop': o.css('div.mp_pvpant.baja::text').extract()}
    for y in response.css('div.opciones'):
        yield {'price decreased': y.css('span.eur_m::text').extract()}
    for u in response.css('div.datos'):
        yield {'link': u.css('a::attr(href)').extract_first()}

您可以将提取的值附加到列表中,然后生成相同的值,就像这样

def parse(self, response):
    # create more lists for the remaining fields
    address = []
    area = []
    for i in response.css('span.dir'):
        address.append(i.css('b::text').extract())
    yield {'address':address}
    for l in response.css('div.datos'):
        area.append(l.css('i::text').extract())
    yield {'area':area}

如果每行的结果数量相同,您可以这样做:

def parse(self, response):
    addresses = []
    areas = []
    prices_stable = []
    prices_drop = []
    prices_decreased = []
    links = []
    for i in response.css('span.dir'):
        addresses.append(i.css('b::text').extract())
    for l in response.css('div.datos'):
        areas.append(l.css('i::text').extract())
    for x in response.css('div.opciones'):
        prices_stable.append(x.css('span.eur::text').extract())
    for o in response.css('div.opciones'):
        prices_drop.append(o.css('div.mp_pvpant.baja::text').extract())
    for y in response.css('div.opciones'):
        prices_decreased.append(y.css('span.eur_m::text').extract())
    for u in response.css('div.datos'):
        links.append(u.css('a::attr(href)').extract_first())

    for address, area, price_stable, price_drop, price_decreased, link in zip(addresses, areas, prices_stable, prices_drop, prices_decreased, links):
        yield {
            'address': address,
            'area': area,
            'price_stable': price_stable,
            'price_drop': price_drop,
            'price_decreased': price_decreased,
            'link': link,
        }