BeautifulSoup: Strip specified attributes, but preserve the tag and its contents

Question

BeautifulSoup: Strip specified attributes, but preserve the tag and its contents

19.2k views Asked by bgibson At 28 January 2012 at 09:03

I'm trying to 'defrontpagify' the html of a MS FrontPage generated website, and I'm writing a BeautifulSoup script to do it.

However, I've gotten stuck on the part where I try to strip a particular attribute (or list attributes) from every tag in the document that contains them. The code snippet:

REMOVE_ATTRIBUTES = ['lang','language','onmouseover','onmouseout','script','style','font',
                        'dir','face','size','color','style','class','width','height','hspace',
                        'border','valign','align','background','bgcolor','text','link','vlink',
                        'alink','cellpadding','cellspacing']

# remove all attributes in REMOVE_ATTRIBUTES from all tags, 
# but preserve the tag and its content. 
for attribute in REMOVE_ATTRIBUTES:
    for tag in soup.findAll(attribute=True):
        del(tag[attribute])

It runs without error, but doesn't actually strip any of the attributes. When I run it without the outer loop, just hard coding a single attribute (soup.findAll('style'=True), it works.

Anyone see know the problem here?

PS - I don't much like the nested loops either. If anyone knows a more functional, map/filter-ish style, I'd love to see it.

Original Q&A

There are 5 answers

oxidworks On 16 November 2018 at 15:03

I use this one:

if "align" in div.attrs:
    del div.attrs["align"]

or

if "align" in div.attrs:
    div.attrs.pop("align")

Thanks to https://stackoverflow.com/a/22497855/1907997

Cyril On 16 May 2020 at 17:58

I use this method to remove a list of attributes, very compact :

attributes_to_del = ["style", "border", "rowspan", "colspan", "width", "height", 
                     "align", "valign", "color", "bgcolor", "cellspacing", 
                     "cellpadding", "onclick", "alt", "title"]
for attr_del in attributes_to_del: 
    [s.attrs.pop(attr_del) for s in soup.find_all() if attr_del in s.attrs]

dtk On 13 July 2018 at 12:04

Just ftr: the problem here is that if you pass HTML attributes as keyword arguments, the keyword is the name of the attribute. So your code is searching for tags with an attribute of name attribute, as the variable does not get expanded.

This is why

hard-coding your attribute name worked[0]
the code does not fail. The search just doesn't match any tags

To fix the problem, pass the attribute you are looking for as a dict:

for attribute in REMOVE_ATTRIBUTES:
    for tag in soup.find_all(attrs={attribute: True}):
        del tag[attribute]

Hth someone in the future, dtk

[0]: Although it needs to be find_all(style=True) in your example, without the quotes, because SyntaxError: keyword can't be an expression

Nóra On 11 October 2016 at 11:16

Here's a Python 2 version of unutbu's answer:

REMOVE_ATTRIBUTES = ['lang','language','onmouseover']

doc = '''<html><head><title>Page title</title></head><body><p id="firstpara" align="center">This is <i>paragraph</i> <a onmouseout="">one</a>.<p id="secondpara" align="blah">This is <i>paragraph</i> <b>two</b>.</html>'''

soup = BeautifulSoup.BeautifulSoup(doc)

for tag in soup.recursiveChildGenerator():
    if hasattr(tag, 'attrs'):
        tag.attrs = {key:value for key,value in tag.attrs.iteritems()
                    if key not in REMOVE_ATTRIBUTES}

**unutbu** · Accepted Answer · 2012-01-28 13:48:57

The line

for tag in soup.findAll(attribute=True):

does not find any tags. There might be a way to use findAll, I'm not sure.

However, this works (as of beautifulsoup 4.8.1):

import bs4
REMOVE_ATTRIBUTES = [
    'lang','language','onmouseover','onmouseout','script','style','font',
    'dir','face','size','color','style','class','width','height','hspace',
    'border','valign','align','background','bgcolor','text','link','vlink',
    'alink','cellpadding','cellspacing']

doc = '''<html><head><title>Page title</title></head><body><p id="firstpara" align="center">This is <i>paragraph</i> <a onmouseout="">one</a>.<p id="secondpara" align="blah">This is <i>paragraph</i> <b>two</b>.</html>'''
soup = bs4.BeautifulSoup.BeautifulSoup(doc)
for tag in soup.descendants:
    if isinstance(tag, bs4.element.Tag):
        tag.attrs = {key: value for key, value in tag.attrs
                     if key not in REMOVE_ATTRIBUTES}
print(soup.prettify())

This is previous code that may have worked with an older version of beautifulsoup:

import BeautifulSoup
REMOVE_ATTRIBUTES = [
    'lang','language','onmouseover','onmouseout','script','style','font',
    'dir','face','size','color','style','class','width','height','hspace',
    'border','valign','align','background','bgcolor','text','link','vlink',
    'alink','cellpadding','cellspacing']

doc = '''<html><head><title>Page title</title></head><body><p id="firstpara" align="center">This is <i>paragraph</i> <a onmouseout="">one</a>.<p id="secondpara" align="blah">This is <i>paragraph</i> <b>two</b>.</html>'''
soup = BeautifulSoup.BeautifulSoup(doc)
for tag in soup.recursiveChildGenerator():
    try:
        tag.attrs = [(key,value) for key,value in tag.attrs
                     if key not in REMOVE_ATTRIBUTES]
    except AttributeError: 
        # 'NavigableString' object has no attribute 'attrs'
        pass
print(soup.prettify())

Note this this code will only work in Python 3. If you need it to work in Python 2, see Nóra's answer below.

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BeautifulSoup: Strip specified attributes, but preserve the tag and its contents

There are 5 answers

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