给定一系列关键字,例如“Python最佳实践”,我想获得包含关键字的前10个Stack Overflow问题,按相关性(?)排序,比如Python脚本。我的目标是最终得到一个元组列表(标题,URL)。
我怎样才能做到这一点?您会考虑查询Google吗? (你会怎么用Python做的?)
答案 0 :(得分:5)
由于Stackoverflow已具备此功能,您只需获取搜索结果页面的内容并抓取所需信息即可。以下是按相关性搜索的网址:
https://stackoverflow.com/search?q=python+best+practices&sort=relevance
如果您查看来源,您会看到每个问题所需的信息都在这样的一行:
<h3><a href="/questions/5119/what-are-the-best-rss-feeds-for-programmersdevelopers#5150" class="answer-title">What are the best RSS feeds for programmers/developers?</a></h3>
所以你应该能够通过正则表达式搜索该表格的字符串来获得前十个。
答案 1 :(得分:5)
>>> from urllib import urlencode
>>> params = urlencode({'q': 'python best practices', 'sort': 'relevance'})
>>> params
'q=python+best+practices&sort=relevance'
>>> from urllib2 import urlopen
>>> html = urlopen("http://stackoverflow.com/search?%s" % params).read()
>>> import re
>>> links = re.findall(r'<h3><a href="([^"]*)" class="answer-title">([^<]*)</a></h3>', html)
>>> links
[('/questions/5119/what-are-the-best-rss-feeds-for-programmersdevelopers#5150', 'What are the best RSS feeds for programmers/developers?'), ('/questions/3088/best-ways-to-teach-a-beginner-to-program#13185', 'Best ways to teach a beginner to program?'), ('/questions/13678/textual-versus-graphical-programming-languages#13886', 'Textual versus Graphical Programming Languages'), ('/questions/58968/what-defines-pythonian-or-pythonic#59877', 'What defines “pythonian” or “pythonic”?'), ('/questions/592/cxoracle-how-do-i-access-oracle-from-python#62392', 'cx_Oracle - How do I access Oracle from Python? '), ('/questions/7170/recommendation-for-straight-forward-python-frameworks#83608', 'Recommendation for straight-forward python frameworks'), ('/questions/100732/why-is-if-not-someobj-better-than-if-someobj-none-in-python#100903', 'Why is if not someobj: better than if someobj == None: in Python?'), ('/questions/132734/presentations-on-switching-from-perl-to-python#134006', 'Presentations on switching from Perl to Python'), ('/questions/136977/after-c-python-or-java#138442', 'After C++ - Python or Java?')]
>>> from urlparse import urljoin
>>> links = [(urljoin('http://stackoverflow.com/', url), title) for url,title in links]
>>> links
[('http://stackoverflow.com/questions/5119/what-are-the-best-rss-feeds-for-programmersdevelopers#5150', 'What are the best RSS feeds for programmers/developers?'), ('http://stackoverflow.com/questions/3088/best-ways-to-teach-a-beginner-to-program#13185', 'Best ways to teach a beginner to program?'), ('http://stackoverflow.com/questions/13678/textual-versus-graphical-programming-languages#13886', 'Textual versus Graphical Programming Languages'), ('http://stackoverflow.com/questions/58968/what-defines-pythonian-or-pythonic#59877', 'What defines “pythonian” or “pythonic”?'), ('http://stackoverflow.com/questions/592/cxoracle-how-do-i-access-oracle-from-python#62392', 'cx_Oracle - How do I access Oracle from Python? '), ('http://stackoverflow.com/questions/7170/recommendation-for-straight-forward-python-frameworks#83608', 'Recommendation for straight-forward python frameworks'), ('http://stackoverflow.com/questions/100732/why-is-if-not-someobj-better-than-if-someobj-none-in-python#100903', 'Why is if not someobj: better than if someobj == None: in Python?'), ('http://stackoverflow.com/questions/132734/presentations-on-switching-from-perl-to-python#134006', 'Presentations on switching from Perl to Python'), ('http://stackoverflow.com/questions/136977/after-c-python-or-java#138442', 'After C++ - Python or Java?')]
将此转换为函数应该是微不足道的。
编辑:哎呀,我会做的......
def get_stackoverflow(query):
import urllib, urllib2, re, urlparse
params = urllib.urlencode({'q': query, 'sort': 'relevance'})
html = urllib2.urlopen("http://stackoverflow.com/search?%s" % params).read()
links = re.findall(r'<h3><a href="([^"]*)" class="answer-title">([^<]*)</a></h3>', html)
links = [(urlparse.urljoin('http://stackoverflow.com/', url), title) for url,title in links]
return links
答案 2 :(得分:2)
建议在SO中添加REST API。 http://stackoverflow.uservoice.com/
答案 3 :(得分:1)
您可以屏幕从有效的HTTP请求中抓取返回的HTML。但这会导致恶劣的业力,以及失去享受良好睡眠的能力。
答案 4 :(得分:0)
我只是使用Pycurl将搜索词连接到查询uri。