Duncan Booth wrote:
> I haven't timed it, but at the very least I'd avoid going through all the
> data twice:
>
> count = {}
> for i in range(256):
> count[chr(i)] = 0
> for x in data:
> count[x] += 1
>
> ordinalSum = sum(ord(c)*count[c] for c in count)
> ordinalSumSquared = sum(ord(c)**2 * count[c] for c in count)
This approach is definitely faster than using a generator in my tests,
and was second fastest overall. I'm actually a bit surprised that it's
as fast as it is; it never would have occurred to me to try it this way.
--
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