使用HTML标记递归73033个元素的列表，并从i获取上下文

Question

我有一个长73033的元素列表。我想从中了解情况。在列表中，每个元素都有相同的结构（如果下面的代码块有帮助的话），它看起来像这样<div align="center" class="photocaption"> Author/Designer Carleton Varney with Jim Druckman </div>。我感兴趣的是文本Author/Designer Carleton Varney with Jim Druckman.

主代码

NewSoups = [BeautifulSoup(NewR) for NewR in NewRs]. 
captions = [soup.find_all("div", class_ = "photocaption") for soup in NewSoups]
flattened_captions = []
for x in captions:
    for y in x:
        flattened_captions.append(y)

print(len(flattened_captions)) #73033

import re
results = [re.sub('<[^>]*>', '', y) for y in flattened_captions] #where the error comes from

错误

Traceback (most recent call last):
File "picked.py", line 22, in <module>
results = [re.sub('<[^>]*>', '', y) for y in flattened_captions]
File "/opt/conda/lib/python2.7/re.py", line 155, in sub
return _compile(pattern, flags).sub(repl, string, count)
TypeError: expected string or buffer

我想知道是否有一种方便的方法来循环浏览<div ></div>的长列表。如果没有，那么提取我想要的所有文本的最佳方法是什么？非常感谢你。你知道吗

Answer 1

我要发布的不是处理发布的问题的最优雅或最有效的方法。正如Welbog所指出的，BeautifulSoup本身提供了提取上下文的功能。然而，当我在发布最初的问题时收到了错误，我只是好奇这个错误是从哪里来的。原来，从压扁的标题返回的东西不是字符串。这很容易解决。方法如下。你知道吗

str_flattened_captions = [str(flattened_captions[i]) for i in range(len(flattened_captions))]

gains = [re.sub('<[^>]*>', '', item) for item in str_flattened_captions]

测试

print(gains[:5])
r Barbara Schorr ', ' Architect Joan Dineen with Alyson Liss ', ' Author/Designer Carleton Varney with Jim Druckman ', ' Designers Richard Cerrone, Lisa Hyman and Rhonda Eleish (front) in their room called "Holiday Nod To Nature" ']