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

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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?

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.

Answered By: jscs

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