title: Python 2.7.16 - regex lookbehind not working with Findall
body: <p>I have the following piece of code
  Note: Line variable is from a line in a text file I'm reading and pattern variable is saved in a config file which I pickup and apply in the code</p>
  
  <pre><code>line ="[u'INVOICE# SMR/0038 f"', u'', u'', u'']"
  pattern ='(?&lt;=(invoice#)\s)[A-z]{3}/\d{1,5}'
  
  regex = re.compile(r'' + pattern),re.IGNORECASE)
  invNum= re.findall(pattern, str(line),re.IGNORECASE)[0]
        ........
  </code></pre>
  
  <p>I'm expecting to get <code>invNum = SMR/0038</code> but instead I get <code>invoice#</code>. What's the issue?
  if try this pattern on <a href="https://regexr.com/" rel="nofollow noreferrer">https://regexr.com/</a> I see that the lookbehind is working. But transferring it to Python code doesn't work. See image below from <a href="https://regexr.com/" rel="nofollow noreferrer">https://regexr.com/</a></p>
  
  <p><a href="https://i.stack.imgur.com/ojrgl.png" rel="nofollow noreferrer">sample from regexr</a></p>