只要在可以这样做的地方使用nltk即可，例如:

from nltk.cluster.kmeans import KMeansClusterer
NUM_CLUSTERS = <choose a value>
data = <sparse matrix that you would normally give to scikit>.toarray()

kclusterer = KMeansClusterer(NUM_CLUSTERS, distance=nltk.cluster.util.cosine_distance, repeats=25)
assigned_clusters = kclusterer.cluster(data, assign_clusters=True)

Sklearn Kmeans使用欧几里得距离。它没有度量参数。也就是说，如果你在聚类时间序列，你可以使用tslearn python包，当你可以指定一个度量(dtw, softdtw，欧几里得)。

只要在可以这样做的地方使用nltk即可，例如:

from nltk.cluster.kmeans import KMeansClusterer
NUM_CLUSTERS = <choose a value>
data = <sparse matrix that you would normally give to scikit>.toarray()

kclusterer = KMeansClusterer(NUM_CLUSTERS, distance=nltk.cluster.util.cosine_distance, repeats=25)
assigned_clusters = kclusterer.cluster(data, assign_clusters=True)

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。

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

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