Sun May 4 09:40:40 CDT 2008
On Sat, May 3, 2008 at 5:31 PM, Keith Goodman <email@example.com> wrote:
> On Sat, May 3, 2008 at 5:05 PM, Christopher Barker
> <Chris.Barker@noaa.gov> wrote:
> > Robert Kern wrote:
> > > I can get a ~20% improvement with the following:
> > > In : def mycut(x, i):
> > > ...: A = x[:i,:i]
> > > ...: B = x[:i,i+1:]
> > > ...: C = x[i+1:,:i]
> > > ...: D = x[i+1:,i+1:]
> > > ...: return hstack([vstack([A,C]),vstack([B,D])])
> > Might it be a touch faster to built the final array first, then fill it:
> > def mycut(x, i):
> > r,c = x.shape
> > out = np.empty((r-1, c-1), dtype=x.dtype)
> > out[:i,:i] = x[:i,:i]
> > out[:i,i:] = x[:i,i+1:]
> > out[i:,:i] = x[i+1:,:i]
> > out[i:,i+1:] = x[i+1:,i+1:]
> > return out
> > totally untested.
> > That should save the creation of two temporaries.
> Initializing the array makes sense. And it is super fast:
> >> timeit mycut(x, 6)
> 100 loops, best of 3: 7.48 ms per loop
> >> timeit mycut2(x, 6)
> 1000 loops, best of 3: 1.5 ms per loop
If you don't need the old array after the cut, I think that you could use
the input array as the output array and then take a slice, saving a
temporary and one-quarter of your assignments (on average). Something like.
def destructive_cut(x, i): # Untested
out = x[:-1,:-1]
out[:i,i:] = x[:i,i+1:]
out[i:,:i] = x[i+1:,:i]
out[i:,i:] = x[i+1:,i+1:]
If you were really clever, you could take different initial slices based on
i so that you always skipped at least one-quarter of the assignments, but
that's probably not worth the effort.
> The time it takes to cluster went from about 1.9 seconds to 0.7
> seconds! Thank you.
> When I run the single linkage clustering on my data I get one big
> cluster and a bunch of tiny clusters. So I need to try a different
> linkage method. Average linkage sounds good, but it sounds hard to
> Numpy-discussion mailing list
-------------- next part --------------
An HTML attachment was scrubbed...
More information about the Numpy-discussion