给定python中的链接列表，如何在每个网页上抓取街道地址？

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The code below gets the URL for each gym location on this website, beginning with "Albertville, AL."

from urlparse import urljoin
import requests
import urllib3
from bs4 import BeautifulSoup

res = requests.get("https://www.planetfitness.com/sitemap").content
soup = BeautifulSoup(res, 'html.parser')

tds = soup.find_all('td', {'class': 'club-title'})
links = [td.find('a')['href'] for td in tds]
keywords = ['gyms']

for link in links:
    if any(keyword in link for keyword in keywords):
        print urljoin('https://www.planetfitness.com/', link)

前2个链接输出：

https://www.planetfitness.com/gyms/albertville-al
https://www.planetfitness.com/gyms/alexander-city-al

但是，我试图从每个链接输出中抓取以下内容：

  • 街道地址
  • 俱乐部营业时间

以下是我尝试完成街道地址部分的代码。我相信这是行不通的，因为“ ps =”行返回空白，但是我不知道该用什么代替“ p”，“ class”和“ address”。任何有关如何修复代码以使其实际执行此操作的想法，将不胜感激！

res1 = requests.get(urljoin('https://www.planetfitness.com/', link)).content
soup1 = BeautifulSoup(res1, 'html.parser')

ps = soup.find_all('p', {'class': 'address'})
address1 = [p.find('span')['itemprop'] for p in ps]

This image of when you inspect street address may help

感谢您的帮助！