title: What's the most Pythonic way to identify consecutive duplicates in a list?
body: <p>I've got a list of integers and I want to be able to identify contiguous blocks of duplicates: that is, I want to produce an order-preserving list of duples where each duples contains (int_in_question, number of occurrences).</p>
  
  <p>For example, if I have a list like:</p>
  
  <pre><code>[0, 0, 0, 3, 3, 2, 5, 2, 6, 6]
  </code></pre>
  
  <p>I want the result to be:</p>
  
  <pre><code>[(0, 3), (3, 2), (2, 1), (5, 1), (2, 1), (6, 2)]
  </code></pre>
  
  <p>I have a <em>fairly</em> simple way of doing this with a for-loop, a temp, and a counter:</p>
  
  <pre><code>result_list = []
  current = source_list[0]
  count = 0
  for value in source_list:
      if value == current:
          count += 1
      else:
          result_list.append((current, count))
          current = value
          count = 1
  result_list.append((current, count))
  </code></pre>
  
  <p>But I really like python's functional programming idioms, and I'd like to be able to do this with a simple generator expression. However I find it difficult to keep sub-counts when working with generators. I have a feeling a two-step process might get me there, but for now I'm stumped.</p>
  
  <p>Is there a particularly elegant/pythonic way to do this, especially with generators?</p>