python - python 迭代美丽汤结果集

Question

我正在努力从网站( http://sports.yahoo.com/nfl/players/8800/ )上抓取数据，为此我使用 urllib2 和 BeautifulSoup。我现在的代码如下所示:

site=  'http://sports.yahoo.com/nfl/players/8800/'
response = urllib2.urlopen(site)
html = response.read()
soup = BeautifulSoup(html)
rushing=[]
passing=[]
receiving=[]

#here is where my problem arises
for elem in soup.find_all('th', text=re.compile('2008')):
        passing = elem.parent.find_all('td', class_=re.compile('10'))
        rushing = elem.parent.find_all('td', class_=re.compile('20'))
        receiving = elem.parent.find_all('td', class_=re.compile('30'))

此页面上存在 soup.find_all(...'2008')) 部分的三个实例，并且当单独打印该部分时，每个实例都会出现。不过，运行这个 for 循环只会运行一次循环。如何确保循环运行三次？

Answer 1

据我了解，您需要 extend()您在循环之前定义的列表:

rushing = []
passing = []
receiving = []

for elem in soup.find_all('th', text=re.compile('2008')):
    passing.extend([td.text for td in elem.parent.find_all('td', class_=re.compile('10'))])
    rushing.extend([td.text for td in elem.parent.find_all('td', class_=re.compile('20'))])
    receiving.extend([td.text for td in elem.parent.find_all('td', class_=re.compile('30'))])

print passing
print rushing
print receiving

打印:

[u'3']
[u'19', u'58', u'14.5', u'3.1', u'0']
[u'2', u'17', u'4.3', u'8.5', u'11', u'6.5', u'0']