What's the most Pythonic way to identify consecutive duplicates in a list?

Question

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).

For example, if I have a list like:

[0, 0, 0, 3, 3, 2, 5, 2, 6, 6] 

I want the result to be:

[(0, 3), (3, 2), (2, 1), (5, 1), (2, 1), (6, 2)] 

I have a fairly simple way of doing this with a for-loop, a temp, and a counter:

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)) 

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.

Is there a particularly elegant/pythonic way to do this, especially with generators?

Answer 1

>>> from itertools import groupby >>> L = [0, 0, 0, 3, 3, 2, 5, 2, 6, 6] >>> grouped_L = [(k, sum(1 for i in g)) for k,g in groupby(L)] >>> # Or (k, len(list(g))), but that creates an intermediate list >>> grouped_L [(0, 3), (3, 2), (2, 1), (5, 1), (2, 1), (6, 2)] 

Batteries included, as they say.

Suggestion for using sum and generator expression from JBernardo; see comment.