[Numpy-discussion] Efficient square distance computation

Matthew Brett matthew.brett at gmail.com
Tue Oct 8 17:11:36 EDT 2013


On Tue, Oct 8, 2013 at 12:44 PM, Matthew Brett <matthew.brett at gmail.com> wrote:
> Hi,
> On Tue, Oct 8, 2013 at 4:38 AM, Ke Sun <sunk.cs at gmail.com> wrote:
>> On Tue, Oct 08, 2013 at 01:49:14AM -0700, Matthew Brett wrote:
>>> Hi,
>>> On Tue, Oct 8, 2013 at 1:06 AM, Ke Sun <sunk.cs at gmail.com> wrote:
>>> > Dear all,
>>> >
>>> > I have written the following function to compute the square distances of a large
>>> > matrix (each sample a row). It compute row by row and print the overall progress.
>>> > The progress output is important and I didn't use matrix multiplication.
>>> >
>>> > I give as input a 70,000x800 matrix. The output should be a 70,000x70,000
>>> > matrix. The program runs really slow (16 hours for 1/3 progress). And it eats
>>> > 36G memory (fortunately I have enough).
>>> That is very slow.
>>> As a matter of interest - why didn't you use matrix multiplication?
>> Because it will cost hours and I want to see the progress and
>> know how far it goes. Another concern is to save memory and
>> compute one sample at a time.
>>> On a machine I had access to it took about 20 minutes.
>> How? I am using matrix multiplication (the same code as
>> http://stackoverflow.com/a/4856692) and it runs for around 18 hours.
> I wonder if you are running into disk swap - the code there does
> involve a large temporary array.
> I believe the appended version of the code is correct, and I think it
> is also memory efficient.
> On a fast machine with lots of memory, it ran in about 5 minutes.
> It's using EPD, which might be using multiple cores for the matrix
> multiply.
> Does the code also work for you in reasonable time?
> Another suggestion I saw which only calculates the unique values (say
> lower diagonal) is scipy.spatial.distance
> http://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.distance.pdist.html#scipy.spatial.distance.pdist
> To a first pass that seems to be slower than the matrix multiply.

On a machine that should have been about as fast:

import scipy.spatial.distance as ssd
dists = ssd.pdist(A, 'sqeuclidean')

on the same size matrix took about 40 minutes.



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