Wrote a memoize function: I do not mind feedback

Ian Kelly ian.g.kelly at gmail.com
Wed Apr 29 03:02:39 EDT 2015


On Wed, Apr 29, 2015 at 12:06 AM, Cecil Westerhof <Cecil at decebal.nl> wrote:
> Op Monday 27 Apr 2015 22:35 CEST schreef Albert-Jan Roskam:
>> def some_func(arg, _memoize={}):
>> try:
>> return _memoize[arg]
>> except KeyError:
>> result = some_expensive_operation(arg)
>> _memoize[arg] = result
>> return result
>
> That is in a way the same as what I do, except I do not use an
> exception. Iunderstand it is not as expensive as it was anymore, but I
> do not like to use an exception (when not necessary).

It's useful to keep in mind which case it is that you're trying to
optimize. The expensive case for exceptions is when one actually gets
raised. A try that doesn't raise an exception is pretty cheap,
probably cheaper than looking up the key in the dict twice as the code
you linked does. Compare:

>>> from timeit import timeit
>>> timeit("if x in y: y[x]", setup="x = 1; y = {1: 2}")
0.2224265851985905
>>> timeit("""
... try:
...     y[x]
... except KeyError:
...     pass
... """, setup="x = 1; y = {1: 2}")
0.15237198271520924

On the other hand, if the KeyError does get raised, then it will be
more expensive, but that would already be the expensive case for
memoized function, and if that computation isn't significantly more
expensive than the cost of raising an exception, then it probably
isn't worth memoizing in the first place.

Regardless of which way you write it, this sort of micro-optimization
hardly ever matters, and in the situations where it does matter,
you'll gain much more performance by rewriting in C than by
obsessively tuning your Python code.



More information about the Python-list mailing list