cached_function.clear_cache is inefficent

[dkrenn]

@cached_function
def noop():
    return None

R = Zmod(42)
D = dict((x^k, R(k)^k) for k in srange(1000000))

E = copy(D)
timeit("E.clear()", number=1)

noop.cache = copy(D)
timeit("noop.clear_cache()", number=1)

gives

1 loops, best of 3: 954 ns per loop
1 loops, best of 3: 9.56 ms per loop

This is because .clear_cache does

        cdef object cache = self.cache
        for key in cache.keys():
            del cache[key]

which is very inefficient for large caches.

[dkrenn]

Is there a reason why not using self.cache.clear()?

[nbruin]

Your point is probably correct, but your code timing doesn't really corroborate it. Timeit, as used, it still repeating the command thrice. It's a mutating command, so the second time around, the original length of the dict doesn't come into play at all.

sage: %time E.clear()
CPU times: user 11 ms, sys: 6 ms, total: 17 ms
sage: %time noop.clear_cache()
CPU times: user 59 ms, sys: 5 ms, total: 64 ms
Wall time: 64 ms

probably gives a fairer picture. Your original numbers would have meant that E.clear() would have visited al 1000000 entries (they all have to be decreffed!) in less than 1 microsecond, meaning your computer would be performing more than 10¹² operations per second. Terahertz performance is usually not so easily attained.