Function hierarchical

pattern/vector/__init__.py:1610–1645

Returns a Cluster containing k items (vectors or clusters with nested items). With k=1, the top-level cluster contains a single cluster.

(vectors, k=1, iterations=1000, distance=COSINE, **kwargs)

        yield i; i=f(i)

def hierarchical(vectors, k=1, iterations=1000, distance=COSINE, **kwargs):
    """ Returns a Cluster containing k items (vectors or clusters with nested items).
        With k=1, the top-level cluster contains a single cluster.
    """
    id = sequence()
    features  = kwargs.get("features", _features(vectors))
    clusters  = Cluster((v for v in shuffled(vectors)))
    centroids = [(id.next(), v) for v in clusters]
    map = {}
    for _ in range(iterations):
        if len(clusters) <= max(k, 1): 
            break
        nearest, d0 = None, None
        for i, (id1, v1) in enumerate(centroids):
            for j, (id2, v2) in enumerate(centroids[i+1:]):
                # Cache the distance calculations between vectors.
                # This is identical to DistanceMap.distance(),
                # but it is faster in the inner loop to use it directly.
                try:
                    d = map[(id1, id2)]
                except KeyError:
                    d = map[(id1, id2)] = _distance(v1, v2, method=distance)
                if d0 is None or d < d0:
                    nearest, d0 = (i, j+i+1), d
        # Pairs of nearest clusters are merged as we move up the hierarchy:
        i, j = nearest
        merged = Cluster((clusters[i], clusters[j]))
        clusters.pop(j)
        clusters.pop(i)
        clusters.append(merged)
        # Cache the center of the new cluster.
        v = centroid(merged.flatten(), features)
        centroids.pop(j)
        centroids.pop(i)
        centroids.append((id.next(), v))
    return clusters

#v1 = Vector(wings=0, beak=0, claws=1, paws=1, fur=1) # cat
#v2 = Vector(wings=0, beak=0, claws=0, paws=1, fur=1) # dog