geckon geckon - 5 months ago 15
Python Question

How to get "subsoups" and concatenate/join them?

I have a HTML document I need to process. I'm using 'beautifoulsoup' for that. Now I would like to retrieve a few "subsoups" from that document and join them into one soup so I can later use it as a parameter for a function that expects a soup object.

If it's not clear, I'll give you an example...

from bs4 import BeautifulSoup

my_document = """
<html>
<body>

<h1>Some Heading</h1>

<div id="first">
<p>A paragraph.</p>
<a href="another_doc.html">A link</a>
<p>A paragraph.</p>
</div>

<div id="second">
<p>A paragraph.</p>
<p>A paragraph.</p>
</div>

<div id="third">
<p>A paragraph.</p>
<a href="another_doc.html">A link</a>
<a href="yet_another_doc.html">A link</a>
</div>

<p id="loner">A paragraph.</p>

</body>
</html>
"""

soup = BeautifulSoup(my_document)

# find the needed parts
first = soup.find("div", {"id": "first"})
third = soup.find("div", {"id": "third"})
loner = soup.find("p", {"id": "loner"})
subsoups = [first, third, loner]

# create a new (sub)soup
resulting_soup = do_some_magic(subsoups)

# use it in a function that expects a soup object and calls its methods
function_expecting_a_soup(resulting_soup)


The goal is to have an object in
resulting_soup
that is/behaves like a soup with the following content:

<div id="first">
<p>A paragraph.</p>
<a href="another_doc.html">A link</a>
<p>A paragraph.</p>
</div>

<div id="third">
<p>A paragraph.</p>
<a href="another_doc.html">A link</a>
<a href="yet_another_doc.html">A link</a>
</div>

<p id="loner">A paragraph.</p>


Is there a convenient way to do that? If there is a better way to retrieve the "subsoups" than
find()
, I can use it instead. Thanks.

Update

There is a solution advised by Wondercricket that concatenates strings containing the found tags and parses them again into a new BeautifulSoup Object. While it's a possible way to solve the problem, the re-parsing may take longer than I'd like especially when I want to retrieve the most of them and there are many such documents I need to process.
find()
returns a
bs4.element.Tag
. Isn't there a way how to concatenate several
Tag
s into one soup without converting the
Tag
s to a string and parsing the string?

Answer

SoupStrainer would do exactly what you are asking about and, as a bonus, you'll get a performance boost since it would parse exactly what you want it to parse - not the complete document tree:

from bs4 import BeautifulSoup, SoupStrainer

parse_only = SoupStrainer(id=["first", "third", "loner"])
soup = BeautifulSoup(my_document, "html.parser", parse_only=parse_only)

Now, the soup object would contain only the desired elements:

<div id="first">
 <p>
  A paragraph.
 </p>
 <a href="another_doc.html">
  A link
 </a>
 <p>
  A paragraph.
 </p>
</div>
<div id="third">
 <p>
  A paragraph.
 </p>
 <a href="another_doc.html">
  A link
 </a>
 <a href="yet_another_doc.html">
  A link
 </a>
</div>
<p id="loner">
 A paragraph.
</p>

Is it also possible to specify not only ids but also tags? For example if I want to filter all paragraphs with class="someclass but not divs with the same class?

In this case, you can make a search function to join multiple criteria for the SoupStrainer:

from bs4 import BeautifulSoup, SoupStrainer, ResultSet

my_document = """
<html>
<body>

    <h1>Some Heading</h1>

    <div id="first">
    <p>A paragraph.</p>
    <a href="another_doc.html">A link</a>
    <p>A paragraph.</p>
    </div>

    <div id="second">
    <p>A paragraph.</p>
    <p>A paragraph.</p>
    </div>

    <div id="third">
    <p>A paragraph.</p>
    <a href="another_doc.html">A link</a>
    <a href="yet_another_doc.html">A link</a>
    </div>

    <p id="loner">A paragraph.</p>

    <p class="myclass">test</p>
</body>
</html>
"""

def search(tag, attrs):
    if tag == "p" and "myclass" in attrs.get("class", []):
        return tag

    if attrs.get("id") in ["first", "third", "loner"]:
        return tag


parse_only = SoupStrainer(search)

soup = BeautifulSoup(my_document, "html.parser", parse_only=parse_only)

print(soup.prettify())
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