我有一些html,这种排序看起来像字典:
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(所以findAll,div类名(。
是否有一种优雅且简单的方法将此类代码提取到字典中?或者必须遍历每个div,找到两个文本项,并假设第一个文本项是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.
想要的输出是这个
expected result result = {'manufacturer website':"ford.com", 'Headquarters': 'Dearborn, Mi', 'Model':'Mustang'}
附言:在这一点上,我已经用这种不优雅的方式做过几次了,只是想知道我是否错过了什么,以及是否有更好的方法来做到这一点。提前谢谢!
或者,为了使事情或多或少地通用和简单,您可以将字段的处理与标签和制造商网站链接分开:
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'}
当前的HTML结构非常通用,它包含多个具有子内容的infoEntity
div,这些子内容可以通过多种方式格式化。为了处理这个问题,您可以迭代infoEntity
div并应用格式化对象,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'}