The Affinity propagation algorithm from the sklearn library allows you to pass the similarity matrix instead of the samples. So, you can use your metric to compute the similarity matrix (not the dissimilarity matrix) and pass it to the function by setting the "affinity" term to "precomputed".https://scikit-learn.org/stable/modules/generated/sklearn.cluster.AffinityPropagation.html#sklearn.cluster.AffinityPropagation.fit In terms of the K-Mean, I think it is also possible but I have not tried it. However, as the other answers stated, finding the mean using a different metric will be the issue. Instead, you can use PAM (K-Medoids) algorthim as it calculates the change in Total Deviation (TD), thus it does not rely on the distance metric. https://python-kmedoids.readthedocs.io/en/latest/#fasterpam

Spectral Python的k-means允许使用L1 (Manhattan)距离。

def distance_metrics(dist_metrics):
    kmeans_instance = kmeans(trs_data, initial_centers, metric=dist_metrics)

    label = np.zeros(210, dtype=int)
    for i in range(0, len(clusters)):
        for index, j in enumerate(clusters[i]):
            label[j] = i

是的，在当前稳定版本的sklearn (scikit-learn 1.1.3)中，您可以轻松地使用自己的距离度量。你所要做的就是创建一个继承自sklearn.cluster.KMeans的类，并覆盖它的_transform方法。

下面的例子是IOU与Yolov2论文的距离。

import sklearn.cluster
import numpy as np

def anchor_iou(box_dims, centroid_box_dims):
    box_w, box_h = box_dims[..., 0], box_dims[..., 1]
    centroid_w, centroid_h = centroid_box_dims[..., 0], centroid_box_dims[..., 1]
    inter_w = np.minimum(box_w[..., np.newaxis], centroid_w[np.newaxis, ...])
    inter_h = np.minimum(box_h[..., np.newaxis], centroid_h[np.newaxis, ...])
    inter_area = inter_w * inter_h
    centroid_area = centroid_w * centroid_h
    box_area = box_w * box_h
    return inter_area / (
        centroid_area[np.newaxis, ...] + box_area[..., np.newaxis] - inter_area
    )

class IOUKMeans(sklearn.cluster.KMeans):
    def __init__(
        self,
        n_clusters=8,
        *,
        init="k-means++",
        n_init=10,
        max_iter=300,
        tol=1e-4,
        verbose=0,
        random_state=None,
        copy_x=True,
        algorithm="lloyd",
    ):
        super().__init__(
            n_clusters=n_clusters,
            init=init,
            n_init=n_init,
            max_iter=max_iter,
            tol=tol,
            verbose=verbose,
            random_state=random_state,
            copy_x=copy_x,
            algorithm=algorithm
        )

    def _transform(self, X):
        return anchor_iou(X, self.cluster_centers_)

rng = np.random.default_rng(12345)
num_boxes = 10
bboxes = rng.integers(low=0, high=100, size=(num_boxes, 2))

kmeans = IOUKMeans(num_clusters).fit(bboxes)

是的，你可以使用差分度量函数;然而，根据定义，k-means聚类算法依赖于每个聚类均值的欧几里得距离。

你可以使用不同的度量，所以即使你仍然在计算平均值你也可以使用像mahalnobis距离这样的东西。

python/ c++中有pyclustering(所以它很快!)，可以让你指定一个自定义度量函数

from pyclustering.cluster.kmeans import kmeans
from pyclustering.utils.metric import type_metric, distance_metric

user_function = lambda point1, point2: point1[0] + point2[0] + 2
metric = distance_metric(type_metric.USER_DEFINED, func=user_function)

# create K-Means algorithm with specific distance metric
start_centers = [[4.7, 5.9], [5.7, 6.5]];
kmeans_instance = kmeans(sample, start_centers, metric=metric)

# run cluster analysis and obtain results
kmeans_instance.process()
clusters = kmeans_instance.get_clusters()

实际上，我还没有测试这段代码，但它拼凑在一起从一个票和示例代码。