python kmedoids - 更有效地计算新的中心点

Question

我正在关注一篇优秀的媒体文章：https://towardsdatascience.com/k-medoids-clustering-on-iris-data-set-1931bf781e05 从头开始​​实施 kmedoids。代码中有一个地方计算每个像素到中心点的距离，而且速度非常慢。它在循环中有 numpy.linalg.norm。有没有办法使用 numpy.linalg.norm 或使用 numpy 广播或 scipy.spatial.distance.cdist 和 np.argmin 来优化它来做同样的事情？

###helper function here###
def compute_d_p(X, medoids, p):
    m = len(X)
    medoids_shape = medoids.shape
    # If a 1-D array is provided, 
    # it will be reshaped to a single row 2-D array
    if len(medoids_shape) == 1: 
        medoids = medoids.reshape((1,len(medoids)))
    k = len(medoids)

    S = np.empty((m, k))

    for i in range(m):
        d_i = np.linalg.norm(X[i, :] - medoids, ord=p, axis=1)
        S[i, :] = d_i**p

    return S

这就是减速发生的地方

        for datap in cluster_points:
            new_medoid = datap
            new_dissimilarity= np.sum(compute_d_p(X, datap, p))

            if new_dissimilarity < avg_dissimilarity :
                avg_dissimilarity = new_dissimilarity

                out_medoids[i] = datap

下面的完整代码。所有致谢文章作者。

# Imports
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from sklearn import datasets
from sklearn.decomposition import PCA

# Dataset
iris = datasets.load_iris()
data = pd.DataFrame(iris.data,columns = iris.feature_names)

target = iris.target_names
labels = iris.target

#Scaling
from sklearn.preprocessing import MinMaxScaler
scaler = MinMaxScaler()
data = pd.DataFrame(scaler.fit_transform(data), columns=data.columns)

#PCA Transformation
from sklearn.decomposition import PCA
pca = PCA(n_components=3)
principalComponents = pca.fit_transform(data)
PCAdf = pd.DataFrame(data = principalComponents , columns = ['principal component 1', 'principal component 2','principal component 3'])

datapoints = PCAdf.values
m, f = datapoints.shape
k = 3

def init_medoids(X, k):
    from numpy.random import choice
    from numpy.random import seed

    seed(1)
    samples = choice(len(X), size=k, replace=False)
    return X[samples, :]

medoids_initial = init_medoids(datapoints, 3)

def compute_d_p(X, medoids, p):
    m = len(X)
    medoids_shape = medoids.shape
    # If a 1-D array is provided, 
    # it will be reshaped to a single row 2-D array
    if len(medoids_shape) == 1: 
        medoids = medoids.reshape((1,len(medoids)))
    k = len(medoids)

    S = np.empty((m, k))

    for i in range(m):
        d_i = np.linalg.norm(X[i, :] - medoids, ord=p, axis=1)
        S[i, :] = d_i**p

    return S

S = compute_d_p(datapoints, medoids_initial, 2)


def assign_labels(S):
    return np.argmin(S, axis=1)

labels = assign_labels(S)

def update_medoids(X, medoids, p):

    S = compute_d_p(points, medoids, p)
    labels = assign_labels(S)

    out_medoids = medoids

    for i in set(labels):

        avg_dissimilarity = np.sum(compute_d_p(points, medoids[i], p))

        cluster_points = points[labels == i]

        for datap in cluster_points:
            new_medoid = datap
            new_dissimilarity= np.sum(compute_d_p(points, datap, p))

            if new_dissimilarity < avg_dissimilarity :
                avg_dissimilarity = new_dissimilarity

                out_medoids[i] = datap

    return out_medoids

def has_converged(old_medoids, medoids):
    return set([tuple(x) for x in old_medoids]) == set([tuple(x) for x in medoids])

#Full algorithm
def kmedoids(X, k, p, starting_medoids=None, max_steps=np.inf):
    if starting_medoids is None:
        medoids = init_medoids(X, k)
    else:
        medoids = starting_medoids

    converged = False
    labels = np.zeros(len(X))
    i = 1
    while (not converged) and (i <= max_steps):
        old_medoids = medoids.copy()

        S = compute_d_p(X, medoids, p)

        labels = assign_labels(S)

        medoids = update_medoids(X, medoids, p)

        converged = has_converged(old_medoids, medoids)
        i += 1
    return (medoids,labels)

results = kmedoids(datapoints, 3, 2)
final_medoids = results[0]
data['clusters'] = results[1]

Answer 1

numpy 的广播功能很有可能会有所帮助。让广播在 3+ 维度上工作有点棘手，我通常不得不通过一些试验和错误来获得正确的细节。

这里使用linalg.norm 使事情更加复杂，因为我的代码版本不会为所有输入提供与linalg.norm 相同的结果。但我相信在这种情况下，它会对所有相关输入给出相同的结果。

我在代码中添加了一些 cmets 来解释某些细节背后的想法。

def compute_d_p_broadcasted(X, medoids, p):

    # If a 1-D array is provided, 
    # it will be reshaped to a single row 2-D array

    if len(medoids.shape) == 1: 
        medoids = medoids.reshape((1,len(medoids)))

    # In general, broadcasting n-dim arrays requires that the last
    # dim of the first array be a singleton dimension, and that the
    # first dim of the second array be a singleton dimension. We can
    # quickly accomplish that by slicing with `None` in the appropriate
    # places. (`np.newaxis` is a slightly more self-documenting way
    # of spelling `None`, but I rarely bother.)

    # In this case, the shapes of the other two dimensions also
    # have to align in the same way you'd expect for a dot product.
    # So we pass `medoids.T`.

    diff = np.abs(X[:, :, None] - medoids.T[None, :, :])

    # The last tricky bit is to figure out which axis to sum. Right
    # now, the array is a 3-dimensional array, with the first 
    # dimension corresponding to the rows of `X` and the last 
    # dimension corresponding to the columns of `medoids.T`. 
    # The middle dimension corresponds to the underlying dimensionality
    # of the space; that's what we want to sum for a sum of squares.
    # (Or sum of cubes for L3 norm, etc.)

    return (diff ** p).sum(axis=1)

def compute_d_p(X, medoids, p):
    m = len(X)
    medoids_shape = medoids.shape
    # If a 1-D array is provided, 
    # it will be reshaped to a single row 2-D array
    if len(medoids_shape) == 1: 
        medoids = medoids.reshape((1,len(medoids)))
    k = len(medoids)
    S = np.empty((m, k))
    for i in range(m):
        d_i = np.linalg.norm(X[i, :] - medoids, ord=p, axis=1)
        S[i, :] = d_i**p
    return S

# A couple of simple tests:
X = np.array([[ 1.0,  2,  3],
              [ 4,  5,  6],
              [ 7,  8,  9],
              [10, 11, 12]])

medoids = X[[0, 2], :]

np.allclose(compute_d_p(X, medoids, 2), 
            compute_d_p_broadcasted(X, medoids, 2))
# Returns True
np.allclose(compute_d_p(X, medoids, 3),
            compute_d_p_broadcasted(X, medoids, 3))
# Returns True

当然，这些测试并不能说明这是否真的能显着加速。您必须自己检查相关用例。但我怀疑它至少会有所帮助。