我有一个像这样的站点地图:http://www.site.co.uk/sitemap.xml其结构如下:
<sitemapindex>
<sitemap>
<loc>
http://www.site.co.uk/drag_it/dragitsitemap_static_0.xml
</loc>
<lastmod>2015-07-07</lastmod>
</sitemap>
<sitemap>
<loc>
http://www.site.co.uk/drag_it/dragitsitemap_alpha_0.xml
</loc>
<lastmod>2015-07-07</lastmod>
</sitemap>
...
我想从中提取数据。首先,我需要计算 xml 中有多少个
<sitemap>
,然后对于每个,提取 <loc>
和 <lastmod>
数据。有没有一种简单的方法可以在Python中做到这一点?
我见过类似的其他问题,但它们都提取了例如 xml 中的每个
<loc>
元素,我需要从每个元素中单独提取数据。
我尝试将
lxml
与此代码一起使用:
import urllib2
from lxml import etree
u = urllib2.urlopen('http://www.site.co.uk/sitemap.xml')
doc = etree.parse(u)
element_list = doc.findall('sitemap')
for element in element_list:
url = store.findtext('loc')
print url
但是
element_list
是空的。
我选择使用 Requests 和 BeautifulSoup 库。我创建了一个字典,其中键是 url,值是最后修改日期。
from bs4 import BeautifulSoup
import requests
xml_dict = {}
r = requests.get("http://www.site.co.uk/sitemap.xml")
xml = r.text
soup = BeautifulSoup(xml, "lxml")
sitemap_tags = soup.find_all("sitemap")
print(f"The number of sitemaps are {len(sitemapTags)}")
for sitemap in sitemap_tags:
xml_dict[sitemap.findNext("loc").text] = sitemap.findNext("lastmod").text
print(xml_dict)
或者使用 lxml:
from lxml import etree
import requests
xml_dict = {}
r = requests.get("http://www.site.co.uk/sitemap.xml")
root = etree.fromstring(r.content)
print(f"The number of sitemap tags are {len(root)}")
for sitemap in root:
children = sitemap.getchildren()
xml_dict[children[0].text] = children[1].text
print(xml_dict)
使用 Python 3、请求、Pandas 和列表理解:
import requests
import pandas as pd
import xmltodict
url = "https://www.gov.uk/sitemap.xml"
res = requests.get(url)
raw = xmltodict.parse(res.text)
data = [[r["loc"], r["lastmod"]] for r in raw["sitemapindex"]["sitemap"]]
print("Number of sitemaps:", len(data))
df = pd.DataFrame(data, columns=["links", "lastmod"])
输出:
links lastmod
0 https://www.gov.uk/sitemaps/sitemap_1.xml 2018-11-06T01:10:02+00:00
1 https://www.gov.uk/sitemaps/sitemap_2.xml 2018-11-06T01:10:02+00:00
2 https://www.gov.uk/sitemaps/sitemap_3.xml 2018-11-06T01:10:02+00:00
3 https://www.gov.uk/sitemaps/sitemap_4.xml 2018-11-06T01:10:02+00:00
4 https://www.gov.uk/sitemaps/sitemap_5.xml 2018-11-06T01:10:02+00:00
此函数将从 xml 中提取所有 url
from bs4 import BeautifulSoup
import requests
def get_urls_of_xml(xml_url):
r = requests.get(xml_url)
xml = r.text
soup = BeautifulSoup(xml)
links_arr = []
for link in soup.findAll('loc'):
linkstr = link.getText('', True)
links_arr.append(linkstr)
return links_arr
links_data_arr = get_urls_of_xml("https://www.gov.uk/sitemap.xml")
print(links_data_arr)
这里使用
BeautifulSoup
来获取 sitemap
计数并提取文本:
from bs4 import BeautifulSoup as bs
html = """
<sitemap>
<loc>
http://www.site.co.uk/drag_it/dragitsitemap_static_0.xml
</loc>
<lastmod>2015-07-07</lastmod>
</sitemap>
<sitemap>
<loc>
http://www.site.co.uk/drag_it/dragitsitemap_alpha_0.xml
</loc>
<lastmod>2015-07-07</lastmod>
</sitemap>
"""
soup = bs(html, "html.parser")
sitemap_count = len(soup.find_all('sitemap'))
print("sitemap count: %d" % sitemap)
print(soup.get_text())
输出:
sitemap count: 2
http://www.site.co.uk/drag_it/dragitsitemap_static_0.xml
2015-07-07
http://www.site.co.uk/drag_it/dragitsitemap_alpha_0.xml
2015-07-07
您可以使用
advertools
,它具有用于 解析 XML 站点地图的特殊功能。默认情况下,它还可以解析压缩的站点地图 (.xml.gz)。
如果您有站点地图索引文件,它还会递归地将它们全部放入一个 DataFrame 中。
import advertools as adv
economist = adv.sitemap_to_df('https://www.economist.com/sitemap-2022-Q1.xml')
economist.head()
洛克 | 最后修改 | 更改频率 | 优先 | 网站地图 | 电子标签 | sitemap_last_modified | 站点地图大小_mb | 下载日期 | |
---|---|---|---|---|---|---|---|---|---|
0 | https://www.economist.com/printedition/2022-01-22 | 2022-01-20 15:57:17+00:00 | 每日 | 0.6 | https://www.economist.com/sitemap-2022-Q1.xml | e2637d17284eefef7d1eafb9ef4ebe3a | 2022-01-22 04:00:54+00:00 | 0.0865097 | 2022-01-23 00:01:41.026416+00:00 |
1 | https://www.economist.com/the-world-this-week/2022/01/22/kals-cartoon | 2022-01-20 16:53:34+00:00 | 每日 | 0.6 | https://www.economist.com/sitemap-2022-Q1.xml | e2637d17284eefef7d1eafb9ef4ebe3a | 2022-01-22 04:00:54+00:00 | 0.0865097 | 2022-01-23 00:01:41.026416+00:00 |
2 | https://www.economist.com/united-states/2022/01/22/a-new-barbie-doll-commemorates-a-19th- century-suffragist | 2022-01-20 16:10:36+00:00 | 每日 | 0.6 | https://www.economist.com/sitemap-2022-Q1.xml | e2637d17284eefef7d1eafb9ef4ebe3a | 2022-01-22 04:00:54+00:00 | 0.0865097 | 2022-01-23 00:01:41.026416+00:00 |
3 | https://www.economist.com/britain/2022/01/22/tory-mps-love-to-hate-the-bbc-but-tory-voters-love-to-watch-it | 2022-01-20 17:09:59+00:00 | 每日 | 0.6 | https://www.economist.com/sitemap-2022-Q1.xml | e2637d17284eefef7d1eafb9ef4ebe3a | 2022-01-22 04:00:54+00:00 | 0.0865097 | 2022-01-23 00:01:41.026416+00:00 |
4 | https://www.economist.com/china/2022/01/22/the-communist-party-revisits-its-egalarian-roots | 2022-01-20 16:48:14+00:00 | 每日 | 0.6 | https://www.economist.com/sitemap-2022-Q1.xml | e2637d17284eefef7d1eafb9ef4ebe3a | 2022-01-22 04:00:54+00:00 | 0.0865097 | 2022-01-23 00:01:41.026416+00:00 |
这是一个很好的库:https://github.com/mediacloud/ultimate-sitemap-parser。
Python 3.5+ 的网站站点地图解析器。
安装:
pip install ultimate-sitemap-parser
从站点地图中提取 nytimes.com 网站的所有页面的示例:
from usp.tree import sitemap_tree_for_homepage
tree = sitemap_tree_for_homepage("https://www.nytimes.com/")
for page in tree.all_pages():
print(page)
在现代 Python3 中使用正确的库:
requests
和 lxml
,即使使用来自 utf8
声明的 XML
编码:
import requests
from lxml import etree
from pprint import pprint
session = requests.session()
headers = {
'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/111.0.0.0 Safari/537.36'
}
res = session.get('https://example.org/sitemap-xml', headers=headers)
xml_bytes = res.text.encode('utf-8')
# Parse the XML bytes
root = etree.fromstring(xml_bytes)
# Define the namespace
ns = {'sitemap': 'http://www.sitemaps.org/schemas/sitemap/0.9'}
urls = root.xpath('//sitemap:url[./sitemap:loc[contains(., "/en-us/")]]', namespaces=ns)
# List comprehension
urls = [u.xpath('./sitemap:loc/text()', namespaces=ns)[0] for u in urls]
pprint(urls)
我今天刚接到任务。我使用了 requests 和 re (正则表达式) 导入请求 进口重新
sitemap_url = "https://www.gov.uk/sitemap.xml"
#if you need to send some headers
headers = {'user-agent': 'myApp'}
response = requests.get(sitemap_url,headers = headers)
xml = response.text
list_of_urls = []
for address in re.findall(r"https://.*(?=/</)", xml):
list_of_urls.append(address+'/')#i add trailing slash, you might want to skip it