Can't avoid BeautifulSoup warning
<p>In a python 2.7 script, I'm using BeautifulSoup:</p>

<pre><code>from bs4 import BeautifulSoup

url_page_accueil_pluzz = "http://pluzz.francetv.fr/"
f = urllib.urlopen(url_page_accueil_pluzz)
page_accueil_pluzz = f.read()
f.close()
soup = str(BeautifulSoup(page_accueil_pluzz, "html.parser"))
root = fromstring(soup)
new_soup = (tostring(root, pretty_print=True).strip())
new_soup = BeautifulSoup(new_soup, "html.parser")
</code></pre>

<p>I can not avoid this message:</p>

<pre><code>/usr/lib/python2.7/dist-packages/bs4/__init__.py:166: UserWarning: No parser was explicitly specified, so I'm using the best available HTML parser for this system ("html.parser"). This usually isn't a problem, but if you run this code on another system, or in a different virtual environment, it may use a different parser and behave differently.

To get rid of this warning, change this:

 BeautifulSoup([your markup])

to this:

 BeautifulSoup([your markup], "html.parser")

  markup_type=markup_type))
</code></pre>

<p>I'm on Kubuntu 16.04 (x64)</p>

<p>python-bs4 version 4.4.1-1</p>

<p>Python v. 2.7.12</p>

<p>Adding "html.parser" seems no effect for my script. Someone has an idea?</p>

<pre><code>$ cat dPluzz.py | grep "BeautifulSoup("
soup = BeautifulSoup(page_video, "html.parser")
soup = BeautifulSoup(self.page_pluzz, "html.parser")
soup = BeautifulSoup(self.feuille_canal, "html.parser")
soup = BeautifulSoup(self.sous_feuille_canal, "html.parser")
soup = str(BeautifulSoup(page_accueil_pluzz, "html.parser"))
new_soup = BeautifulSoup(new_soup, "html.parser")
</code></pre>