title: Python regex - How to look for an arbitrary number of sentences after a digit?
body: <p>Let's say we had the following string: "1. Sentence 1. Sentence 2? Sentence 3!".
  How would I go about looking for ( and returning as a string) a pattern that matches all of the following cases:</p>
  
  <ul>
  <li><code>"1. Sentence 1."</code></li>
  <li><code>"1. Sentence 1. Sentence 2?"</code></li>
  <li><code>"1. Sentence 1. Sentence 2? Sentence 3!"</code></li>
  </ul>
  
  <p>There is always a number in front of the pattern,
  but there could be any number of sentences after it.
  What I've tried thus far is</p>
  
  <pre><code>pattern = re.compile("\d.(\s[A-Ö][^.!?]+[.!?])+?")   
  </code></pre>
  
  <p>and</p>
  
  <pre><code>assignmentText = "".join(pattern.findall(assignment))
  </code></pre>
  
  <p>where the <code>join</code>-method is an ugly hack used to extract the string from the list returned by <code>findall</code>, since <code>list[0]</code> doesn't seem to work ( I know there will only be a single <code>str</code> in the list).
  However, I only ever receive the first sentence, without the digit in front.</p>
  
  <p>How could this be fixed?</p>