[SOLVED] What is the fastest (to access) struct-like object in Python?

Issue

I’m optimizing some code whose main bottleneck is running through and accessing a very large list of struct-like objects. Currently I’m using namedtuples, for readability. But some quick benchmarking using ‘timeit’ shows that this is really the wrong way to go where performance is a factor:

Named tuple with a, b, c:

>>> timeit("z = a.c", "from __main__ import a")
0.38655471766332994

Class using __slots__, with a, b, c:

>>> timeit("z = b.c", "from __main__ import b")
0.14527461047146062

Dictionary with keys a, b, c:

>>> timeit("z = c['c']", "from __main__ import c")
0.11588272541098377

Tuple with three values, using a constant key:

>>> timeit("z = d[2]", "from __main__ import d")
0.11106188992948773

List with three values, using a constant key:

>>> timeit("z = e[2]", "from __main__ import e")
0.086038238242508669

Tuple with three values, using a local key:

>>> timeit("z = d[key]", "from __main__ import d, key")
0.11187358437882722

List with three values, using a local key:

>>> timeit("z = e[key]", "from __main__ import e, key")
0.088604143037173344

First of all, is there anything about these little timeit tests that would render them invalid? I ran each several times, to make sure no random system event had thrown them off, and the results were almost identical.

It would appear that dictionaries offer the best balance between performance and readability, with classes coming in second. This is unfortunate, since, for my purposes, I also need the object to be sequence-like; hence my choice of namedtuple.

Lists are substantially faster, but constant keys are unmaintainable; I’d have to create a bunch of index-constants, i.e. KEY_1 = 1, KEY_2 = 2, etc. which is also not ideal.

Am I stuck with these choices, or is there an alternative that I’ve missed?

Solution

One thing to bear in mind is that namedtuples are optimised for access as tuples. If you change your accessor to be a[2] instead of a.c, you’ll see similar performance to the tuples. The reason is that the name accessors are effectively translating into calls to self[idx], so pay both the indexing and the name lookup price.

If your usage pattern is such that access by name is common, but access as tuple isn’t, you could write a quick equivalent to namedtuple that does things the opposite way: defers index lookups to access by-name. However, you’ll pay the price on the index lookups then. Eg here’s a quick implementation:

def makestruct(name, fields):
    fields = fields.split()
    import textwrap
    template = textwrap.dedent("""\
    class {name}(object):
        __slots__ = {fields!r}
        def __init__(self, {args}):
            {self_fields} = {args}
        def __getitem__(self, idx): 
            return getattr(self, fields[idx])
    """).format(
        name=name,
        fields=fields,
        args=','.join(fields), 
        self_fields=','.join('self.' + f for f in fields))
    d = {'fields': fields}
    exec template in d
    return d[name]

But the timings are very bad when __getitem__ must be called:

namedtuple.a  :  0.473686933517 
namedtuple[0] :  0.180409193039
struct.a      :  0.180846214294
struct[0]     :  1.32191514969

ie, the same performance as a __slots__ class for attribute access (unsurprisingly – that’s what it is), but huge penalties due to the double lookup in index-based accesses. (Noteworthy is that __slots__ doesn’t actually help much speed-wise. It saves memory, but the access time is about the same without them.)

One third option would be to duplicate the data, eg. subclass from list and store the values both in the attributes and listdata. However you don’t actually get list-equivalent performance. There’s a big speed hit just in having subclassed (bringing in checks for pure-python overloads). Thus struct[0] still takes around 0.5s (compared with 0.18 for raw list) in this case, and you do double the memory usage, so this may not be worth it.

Answered By – Brian

Answer Checked By – Marilyn (BugsFixing Volunteer)

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