On Tue, 07 Jun 2005 18:13:01 +0200, "Diez B. Roggisch" <[EMAIL PROTECTED]>
wrote:
>Another optimization im too lazy now would be to do sort of a "tree
>search" of data[i] in rngs - as the ranges are ordered, you could find
>the proper one in log_2(len(rngs)) instead of len(rngs)/2.
I don't see a "break" so why the "/2" ? also IIUC the
ranges are more than just ordered... they're all equal
and computed by
for i in xrange(no_of_bins+1):
rngs[i] = dmin + (rng*i)
so my guess is that instead of searching with
for j in xrange(len(rngs)-1):
if rngs[j] <= data[i] < rngs[j+1]:
one could just do
j = int((data[i] - dmin)/rng)
The code with this change gets roughly about twice faster.
Andrea
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