[Numpy-discussion] how do I speed up this?
Giorgio F. Gilestro
Tue Nov 27 13:41:57 CST 2007
shape of the array is variable but follows the rule (x, y*1440) with
both x an y being values greater than 1 (usually around 10 or 20).
So as result second dimension is always much bigger. (and the bug you
foresaw is actually taken care of)
I figured out that somehow removing the unnecessary *1 multiplication
after the boolean verification will increase speed by almost 100% (wow!).
I guess if I can get rid of the iteration too and use some internal
function I'd probably get even faster.
Could .flat() help me somehow?
Timothy Hochberg wrote:
> On Nov 27, 2007 11:38 AM, Giorgio F. Gilestro <email@example.com
> <mailto:firstname.lastname@example.org>> wrote:
> Hello everyone,
> ma and new_ma are bi-dimensional array with shape (a1, a2) on
> which I am
> performing the following iteration:
> for fd in range(a1):
> new_ma[fd] = [( ma[fd][i-5:i+5].sum() == 0 )*1 for i in range
> This looks like it's probably buggy; when, for example,, a2 == 0, I
> don't think ma[fd][-5:5] is going to return what you want.
> Is there any faster more elegant way to do that with numpy?
> There are faster ways, I'm not sure that they are more elegant. It
> looks like what you want is essentially a moving average, possible
> with periodic boundary conditions. There are a few different tricks
> that can make this fast; which is appropriate depends on what the
> shape of ma is expected to be. If 'a1' if large, then you can
> profitably remove the outer loop; something vaguely like:
> for i in range(a2):
> new_ma[:, i] = ( ma[fd][i-5:i+5].sum() == 0)
> (This is still buggy as above, since I'm not sure what kind of
> boundary conditions you need).
> If that's not applicable, another approach is to loop over the offset,
> in this case -5 to 5, and remove the loop over a2. That method is more
> complicated and I'm going to avoid describing it in detail unless
> needed. One might even be able to do both, but that's probably overkill.
> . __
> . |-\
> . email@example.com <mailto:firstname.lastname@example.org>
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