从for循环中将结果保存到列表中?

时间:2015-09-17 19:21:44

标签: python list python-2.7 beautifulsoup scrape

url = 'http://www.millercenter.org/president/speeches'

conn = urllib2.urlopen(url)
html = conn.read()


miller_center_soup = BeautifulSoup(html)
links = miller_center_soup.find_all('a')

for tag in links:
    link = tag.get('href',None)
        if link is not None:
            print link

以下是我的一些输出:

/president/washington/speeches/speech-3939
/president/washington/speeches/speech-3939
/president/washington/speeches/speech-3461
https://www.facebook.com/millercenter
https://twitter.com/miller_center
https://www.flickr.com/photos/miller_center
https://www.youtube.com/user/MCamericanpresident
http://forms.hoosonline.virginia.edu/s/1535/16-uva/index.aspx?sid=1535&gid=16&pgid=9982&cid=17637
mailto:mcpa-webmaster@virginia.edu

我正试图在网站millercenter.org/president/speeches上网上搜集所有总统演讲,但我很难保存适当的语音链接,我将从中收集语音数据。更明确地说,我需要乔治华盛顿的演讲,可以在http://www.millercenter.org/president/washington/speeches/speech-3461访问 - 我需要能够访问该网址。我正在考虑将所有演讲的所有网址存储在一个列表中,然后编写一个for循环来抓取并清理所有数据。

2 个答案:

答案 0 :(得分:2)

将其转换为列表理解:

linklist = [tag.get('href') for tag in links if tag.get('href') is not None]

略微优化:

linklist = [href for href in (tag.get('href') for tag in links) if href is not None]

答案 1 :(得分:1)

如果你对列表理解不好或者你不想使用它,你可以创建一个列表并附加到它:

all_links = []
for tag in links:
    link = tag.get('href',None)
        if link is not None:
            all_links.append(link)