title: Find third occurring `<p>` tag using with Beautiful Soup
body: <p>As the title suggests, I'm trying to understand how to find the third occurring <code>&lt;p&gt;</code> of a website (as an example, I used the following website: <a href="http://www.musicmeter.nl/album/31759" rel="nofollow noreferrer">http://www.musicmeter.nl/album/31759</a>). </p>
  
  <p>Using the answer to <a href="https://stackoverflow.com/questions/21823229/finding-next-occuring-tag-and-its-enclosed-text-with-beautiful-soup">this question</a>, I tried the following code</p>
  
  <pre><code>from bs4 import BeautifulSoup
  import requests
  html = requests.get("http://www.musicmeter.nl/album/31759").text    # get HTML from http://www.musicmeter.nl/album/31759
  soup = BeautifulSoup(html, 'html5lib')                              # Get data out of HTML
  
  first_paragraph = soup.find('p')    # or just soup.p
  
  print "first paragraph:", first_paragraph
  
  second_paragraph = first_paragraph.find_next_siblings('p')
  
  print "second paragraph:", second_paragraph
  
  third_paragraph = second_paragraph.find_next_siblings('p')
  
  print "third paragraph:", third_paragraph
  </code></pre>
  
  <p>But this code results in the following error for the third_paragraph:</p>
  
  <pre><code>Traceback (most recent call last):
    File "page_109.py", line 21, in &lt;module&gt;
      third_paragraph = second_paragraph.find_next_siblings('p')
  AttributeError: 'ResultSet' object has no attribute 'find_next_siblings'
  </code></pre>
  
  <p>I tried to lookup the error, but I couldn't figure out what is wrong.</p>