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

Posted on

Question :

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

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?

``````>>> from itertools import groupby
Suggestion for using `sum` and generator expression from JBernardo; see comment.