1.计算成对距离矩阵
1.计算列和或行和

1. argmin查找medoid索引
   即numpy.argmin(distMatrix.sum(axis=0))或类似。

所以我接受了这里的答案，但我想如果其他人试图做类似的事情，我会提供我的实现：
(1)这是距离函数：

def fractional(p_coord_array, q_coord_array):
  # f is an arbitrary value, but must be greater than zero and 
  # less than one. In this case, I used 3/10. I took advantage
  # of the difference of cubes in this case, so that I wouldn't
  # encounter an overflow error.

  a = np.sum(np.array(p_coord_array, dtype=np.float64))
  b = np.sum(np.array(q_coord_array, dtype=np.float64))
  a2 = np.sum(np.power(p_coord_array, 2))
  ab = np.sum(p_coord_array) * np.sum(q_coord_array)
  b2 = np.sum(np.power(p_coord_array, 2))
  diffab = a - b
  suma2abb2 = a2 + ab + b2

  temp_dist = abs(diffab * suma2abb2)
  temp_dist = np.power(temp_dist, 1./10)

  dist = np.power(temp_dist, 10./3)
  return dist

(2)medoid函数（如果数据集的长度小于6000 [如果大于6000，我会遇到溢出错误.我仍然在努力工作，这一点是完全诚实的...]）：

def medoid(dataset):

  point = []
  w = len(dataset)

  if(len(dataset) < 6000):
    h = len(dataset)
    dist_matrix = [[0 for x in range(w)] for y in range(h)]

    list_combinations = [(counter_1, counter_2, data_1, data_2) for counter_1, data_1 in enumerate(dataset) for counter_2, data_2 in enumerate(dataset) if counter_1 < counter_2]

    for counter_3, tuple in enumerate(list_combinations):
      temp_dist = fractional(tuple[2], tuple[3])
      dist_matrix[tuple[0]][tuple[1]] = abs(temp_dist)
      dist_matrix[tuple[1]][tuple[0]] = abs(temp_dist)

任何问题，请随时发表评论！

如果你不介意使用暴力，这可能会有所帮助：

def calc_medoid(X, Y, f=2):
    n = len(X)
    m = len(Y)
    dist_mat = np.zeros((m, n))
    # compute distance matrix
    for j in range(n):
        center = X[j, :]
        for i in range(m):
            if i != j:
                dist_mat[i, j] = np.linalg.norm(Y[i, :] - center, ord=f)

    medoid_id = np.argmin(dist_mat.sum(axis=0))  # sum over y

    return medoid_id, X[medoid_id, :]

下面是一个用欧氏距离计算单个聚类的中心点的例子。

import numpy as np, pandas as pd, matplotlib.pyplot as plt
a, b, c, d = np.array([0,1]), np.array([1, 3]), np.array([4,2]), np.array([3, 1.5])
vCenroid = np.mean([a, b, c, d], axis=0)

def GetMedoid(vX):
  vMean = np.mean(vX, axis=0)                               # compute centroid
  return vX[np.argmin([sum((x - vMean)**2) for x in vX])]   # pick a point closest to centroid

vMedoid = GetMedoid([a, b, c, d])

print(f'centroid = {vCenroid}')
print(f'medoid = {vMedoid}')

df = pd.DataFrame([a, b, c, d], columns=['x', 'y'])
ax = df.plot.scatter('x', 'y', grid=True, title='Centroid in 2D plane', s=100);
plt.plot(vCenroid[0], vCenroid[1], 'ro', ms=10);   # plot centroid as red circle
plt.plot(vMedoid[0], vMedoid[1], 'rx', ms=20);     # plot medoid as red star

还可以使用以下包为一个或多个集群计算medoid

!pip -q install scikit-learn-extra > log
from sklearn_extra.cluster import KMedoids
GetMedoid = lambda vX: KMedoids(n_clusters=1).fit(vX).cluster_centers_
GetMedoid([a, b, c, d])[0]

一个简单的（但简单的）方法是仅仅平均集群中的向量（质心），然后找到集群中接近质心的向量吗？

import numpy as np
import faiss

vectors = np.array([...], dtype=np.float32)

# Using FAISS to build the vector index

vector_dimension = vectors.shape[1] # assume vectors
index = faiss.IndexFlatL2(vector_dimension)
faiss.normalize_L2(vectors)
index.add(vectors)

# find the mediod, closest to centroid?

centroid = np.array(vectors).mean(axis=0)
faiss.normalize_L2(centroid)
distances, indices = index.search(centroid, 2)

medoid = vectors[indices[0][0]]

我会说你只需要计算中位数。
np.median(np.asarray(points), axis=0)
你的中位数是最大的中心点。
注意：如果你使用的距离不同于欧几里德，这是不成立的。