[Numpy-discussion] Faster

Keith Goodman kwgoodman@gmail....
Sun May 4 10:46:11 CDT 2008


On Sun, May 4, 2008 at 8:14 AM, Hoyt Koepke <hoytak@gmail.com> wrote:
> >  and then update the others only if the row you're updating has that
>  >  minimum value in it.  Then, when scanning for the min dist, you only
>  >  need to scan O(n) rows.
>
>  Sorry, let me clarify -- Update the entries corresponding to entries
>  in the row you're updating if they are the same as the minimum
>  distance; in that case you need to rescan the row.  Also, ySorry, ou
>  only need to scan O(n) entries in your cached array.

If I understand your improvement, I can speed up dist.min() in

        i, j = np.where(dist == dist.min())
        return i[0], j[0]

since it is faster to find the min in a n-element array than in a nxn
array. But the new code, thanks to Robert, is

        ij = x.argmin()
        i, j = divmod(ij, N)

Would a 1d array of the column minimums still help?

On a separate note, the distance matrix is symmetric so I could fill
the lower half of the distance matrix with large values and, after
updating a column skip the step of keeping the matrix symmetric by
copying the column into the row (d[i,:] = d[:,i]).

Does d[i,:] = d[:,i] make a copy?

Here's what I have so far. It's fast for my needs.


import time

import numpy as np

class Cluster:
    "Single linkage hierarchical clustering"

    def __init__(self, dist, label=None, linkage='single', debug=False):
        """
        dist   Distance matrix, NxN numpy array
        label  Labels of each row of the distance matrix, list of N items,
               default is range(N)
        """
        assert dist.shape[0] == dist.shape[1], 'dist must be square (nxn)'
        assert (np.abs(dist - dist.T) < 1e-8).all(), 'dist must be symmetric'
        if label is None:
            label = range(dist.shape[0])
        assert dist.shape[0] == len(label), 'dist and label must match in size'
        msg = 'linkage must be single or complete'
        assert linkage in ('single', 'complete'), msg
        self.c = [[[z] for z in label]]
        self.label = label
        self.linkage = linkage
        self.dist = dist
        self.debug = debug

    def run(self):
        for level in xrange(len(self.label) - 1):
            i, j = self.min_dist()
            self.join(i, j)

    def join(self, i, j):

        assert i != j, 'Cannot combine a cluster with itself'

        # Join labels
        new = list(self.c[-1])
        new[i] = new[i] + new[j]
        new.pop(j)
        self.c.append(new)

        # Join distance matrix
        if self.linkage == 'single':
            self.dist[:,i] = self.dist[:,[i,j]].min(1)
        elif self.linkage == 'complete':
            self.dist[:,i] =  self.dist[:,[i,j]].max(1)
        else:
            raise NotImplementedError, 'Unknown linkage method'
        self.dist[i,:] = self.dist[:,i]
        # A faster verion of this code...
        # idx = range(self.dist.shape[1])
        # idx.remove(j)
        # self.dist = self.dist[:,idx]
        # self.dist = self.dist[idx,:]
        # ...is this...
        out = self.dist[:-1,:-1]
        out[:i,i:] = self.dist[:i,i+1:]
        out[i:,:i] = self.dist[i+1:,:i]
        out[i:,i:] = self.dist[i+1:,i+1:]
        self.dist = out

        # Debug output
        if self.debug:
            print
            print len(self.c) - 1
            print 'Clusters'
            print self.c[-1]
            print 'Distance'
            print self.dist

    def min_dist(self):
        # A faster version of this code...
        # dist = self.dist + 1e10 * np.eye(self.dist.shape[0])
        # i, j = np.where(dist == dist.min())
        # return i[0], j[0]
        # ...is this:
        x = self.dist
        N = x.shape[0]
        # With complete linkage the min distance was sometimes on the diagonal
        # I think it occured on the last merge (one cluster). So I added + 1.
        x.flat[::N+1] = x.max() + 1
        ij = x.argmin()
        i, j = divmod(ij, N)
        return i, j


def test():
    # Example from
    # home.dei.polimi.it/matteucc/Clustering/tutorial_html/hierarchical.html
    label =          ['BA', 'FI', 'MI', 'NA', 'RM', 'TO']
    dist = np.array([[0,    662,  877,  255,  412,  996],
                     [662,  0,    295,  468,  268,  400],
                     [877,  295,  0,    754,  564,  138],
                     [255,  468,  754,  0,    219,  869],
                     [412,  268,  564,  219,  0,    669],
                     [996,  400,  138,  869,  669,  0  ]])
    clust = Cluster(dist, label, linkage='single', debug=True)
    clust.run()

def test2(n):
    x = np.random.rand(n,n)
    x = x + x.T
    c = Cluster(x)
    t1 = time.time()
    c.run()
    t2 = time.time()
    print 'n = %d took %0.2f seconds' % (n, t2-t1)


More information about the Numpy-discussion mailing list