我有一些html,看起来像字典:
制造商网站:网站,
总部:地点等 ..
每个部分都包含在自己的div中(因此,findAll,div类名称)。
是否存在优雅的简单方法将此类代码提取到字典中?还是必须遍历每个div,找到两个文本项,并假定第一个文本项是Dictionary的键,第二个值是同一dict元素的值。
示例站点代码:
car = '''
<div class="info flexbox">
<div class="infoEntity">
<span class="manufacturer website">
<a class="link" href="http://www.ford.com" rel="nofollow noreferrer" target="_blank">
www.ford.com
</a>
</span>
</div>
<div class="infoEntity">
<label>
Headquarters
</label>
<span class="value">
Dearbord, MI
</span>
</div>
<div class="infoEntity">
<label>
Model
</label>
<span class="value">
Mustang
</span>
</div>
'''
car_soup = BeautifulSoup(car, 'lxml')
print(car_soup.prettify())
elements = car_soup.findAll('div', class_ = 'infoEntity')
for x in elements:
print(x) ###and then we start iterating over x, with beautiful soup, to find value of each element.
期望的输出是这个
expected result result = {'manufacturer website':"ford.com", 'Headquarters': 'Dearborn, Mi', 'Model':'Mustang'}
P.S。在这一点上,我已经完成了几次非优雅的尝试,只是想知道我是否缺少某些东西,以及是否有更好的方法可以做到这一点。预先谢谢你!
答案 0 :(得分:2)
当前的HTML结构非常通用,它包含多个infoEntity
div,这些div的子内容可以采用多种格式设置。要解决此问题,您可以遍历infoEntity
div并应用格式设置对象,如下所示:
from bs4 import BeautifulSoup as soup
result, label = {}, None
for i in soup(car, 'html.parser').find_all('div', {'class':'infoEntity'}):
for b in i.find_all(['span', 'label']):
if b.name == 'label':
label = b.get_text(strip=True)
elif b.name == 'span' and label is not None:
result[label] = b.get_text(strip=True)
label = None
else:
result[' '.join(b['class'])] = b.get_text(strip=True)
输出:
{'manufacturer website': 'www.ford.com', 'Headquarters': 'Dearbord, MI', 'Model': 'Mustang'}
答案 1 :(得分:2)
或者,为了使事情或多或少保持通用和简单,您可以使用标签和制造商网站链接来分隔字段的处理:
soup = BeautifulSoup(car, 'lxml')
car_info = soup.select_one('.info')
data = {
label.get_text(strip=True): label.find_next_sibling().get_text(strip=True)
for label in car_info.select('.infoEntity label')
}
data['manufacturer website'] = car_info.select_one('.infoEntity a').get_text(strip=True)
print(data)
打印:
{'Headquarters': 'Dearbord, MI',
'Model': 'Mustang',
'manufacturer website': 'www.ford.com'}