我的数据样本:
<table id = "history"> <tr class = "printCol"> <td class="name">Google</td><td class="date">07/11/2001</td><td class="state"> <span>CA</span> </td> </tr> <tr class = "printCol"> <td class="name">Apple</td><td class="date">27/08/2001</td> </tr> <tr class = "printCol"> <td class="name">Microsoft</td><td class="date">01/11/1991</td> </tr> </table>
Beautifulsoup代码:
table = soup.find("table", id = "history") rows = table.findAll('tr') for tr in rows: cols = tr.findAll('td') for td in cols: print td.find(text=True)
MySQL存储的所需输出(列表):
['Google|07/11/2001|CA', 'Apple|27/08/2001', 'Microsoft|01/11/1991']
我的输出(难以将正确的日期与正确的公司相关联):
Google 07/11/2001 Apple 27/08/2001 Microsoft 01/11/1991
我编写了一个从每个 tr中 提取元素的函数,但我认为在原始的 for 循环中有一种更有效的方法来完成所有操作。我想将它们作为数据对存储在列表中。有什么想法吗?
列表理解将使它更容易:
table = soup.find("table", id = "history") rows = table.findAll('tr') data = [[td.findChildren(text=True) for td in tr.findAll("td")] for tr in rows] # data now contains: [[u'Google', u'07/11/2001'], [u'Apple', u'27/08/2001'], [u'Microsoft', u'01/11/1991']] # If the data may contain extraneous whitespace you can clean it up # Additional processing could also be done - but once you hit much more # complex than this later maintainers, yourself included, will thank you # for using a series of for loops that call clearly named functions to perform # the work. data = [[u"".join(d).strip() for d in l] for l in data] # If you want to store it joined as name | company # then simply follow that up with: data = [u"|".join(d) for d in data]
列表理解基本上是一个for带有聚合的反向循环:
for
[[td.findNext(text=True) for td in tr.findAll("td")] for tr in rows]
转换为 * :
final_list = [] intermediate_list = [] for tr in rows: for td in tr.findAll("td") intermediate_list.append(td.findNext(text=True)) final_list.append(intermediate_list) intermediate_list = [] data = final_list
* 粗略地讲-我们不考虑生成器 而不 建立中间列表的麻烦,因为我现在不能在不弄乱示例的情况下添加生成器。