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?

• Aaron Robson: For reference this process is called: en.wikipedia.org/wiki/Run-length_encoding

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.

• JBernardo: +1, maybe you could change len(list(g)) for sum(1 for i in g) to avoid intermediate storage.
• jscs: @JBernardo: Good suggestion, thanks. Creating a list from g has always kind of bothered me when I use groupby for this.
• machine yearning: @JBernardo: Actually I'm gonna go with creating the intermediate list. Although perhaps doing the sum would be more efficient, I think the former is far more readable (really states exactly what we want to happen) and thus more pythonic! I do think that this "adding ones" solution is hinting at something lacking in generators, specifically that there's no way of explicitly, with a built-in function, telling how many elements will be generated. Might this be amended in the future?
• jscs: @machine: It's in principle impossible. Consider: def long_gen(): while True: yield 1 What is the len of this? See: stackoverflow.com/questions/390852/…
• jscs: @machine: You're welcome. I've seen this use of sum in other places but hadn't thought to use it in this case. I think it would be quickly understood by most readers.