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Home/ Questions/Q 8617961
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T05:59:47+00:00 2026-06-12T05:59:47+00:00

I’ve seen other Python programmers use defaultdict from the collections module for the following

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I’ve seen other Python programmers use defaultdict from the collections module for the following use case:

from collections import defaultdict

s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)]

def main():
    d = defaultdict(list)
    for k, v in s:
        d[k].append(v)

I’ve typically approached this problem by using setdefault instead:

def main():
    d = {}
    for k, v in s:
        d.setdefault(k, []).append(v)

The docs do in fact claim that using defaultdict is faster, but I’ve seen the opposite to be true when testing myself:

$ python -mtimeit -s "from withsetdefault import main; s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)];" "main()"
100000 loops, best of 3: 4.51 usec per loop
$ python -mtimeit -s "from withdefaultdict import main; s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)];" "main()"
100000 loops, best of 3: 5.38 usec per loop

Is there something wrong with how I’ve set up the tests?

For reference, I’m using Python 2.7.3 [GCC 4.2.1 (Apple Inc. build 5666)

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  1. Editorial Team
    Editorial Team
    2026-06-12T05:59:48+00:00Added an answer on June 12, 2026 at 5:59 am

    Yes, there is something “wrong”:

    You have put the creation of the (default)dict into the statement instead of the setup. Constructing a new defaultdict is more expensive than a normal dict, and usually that’s not the bottleneck you should be profiling in a program – after all, you build your data structures once but you use them many times.

    If you do your tests like below, you see that defaultdict operations are indeed faster:

    >>> import timeit
    >>> setup1 = """from collections import defaultdict
    ... s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)]
    ... d = defaultdict(list)"""
    >>> stmt1 = """for k, v in s:
    ...     d[k].append(v)"""
    >>> setup2 = """s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)]
    ... d = {}"""
    >>> stmt2 = """for k, v in s:
    ...     d.setdefault(k, []).append(v)"""
    >>> timeit.timeit(setup=setup1, stmt=stmt1)
    1.0283400125194078
    >>> timeit.timeit(setup=setup2, stmt=stmt2)
    1.7767367580925395
    

    Python 2.7.3 on Win7 x64.

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