看起来像是块建模的工作。谷歌的“块建模”和前几个点击很有帮助。

假设我们有一个协方差矩阵，其中 N=100，实际上有 5 个集群：

块建模试图做的是找到行的顺序，以便集群变得明显为“块”：

下面是一个代码示例，它执行基本的贪心搜索来完成此操作。对于您的 250-300 个变量来说，这可能太慢了，但这是一个开始。看看你是否可以跟随评论：

import numpy as np
from matplotlib import pyplot as plt

# This generates 100 variables that could possibly be assigned to 5 clusters
n_variables = 100
n_clusters = 5
n_samples = 1000

# To keep this example simple, each cluster will have a fixed size
cluster_size = n_variables // n_clusters

# Assign each variable to a cluster
belongs_to_cluster = np.repeat(range(n_clusters), cluster_size)
np.random.shuffle(belongs_to_cluster)

# This latent data is used to make variables that belong
# to the same cluster correlated.
latent = np.random.randn(n_clusters, n_samples)

variables = []
for i in range(n_variables):
    variables.append(
        np.random.randn(n_samples) + latent[belongs_to_cluster[i], :]
    )

variables = np.array(variables)

C = np.cov(variables)

def score(C):
    '''
    Function to assign a score to an ordered covariance matrix.
    High correlations within a cluster improve the score.
    High correlations between clusters decease the score.
    '''
    score = 0
    for cluster in range(n_clusters):
        inside_cluster = np.arange(cluster_size) + cluster * cluster_size
        outside_cluster = np.setdiff1d(range(n_variables), inside_cluster)

        # Belonging to the same cluster
        score += np.sum(C[inside_cluster, :][:, inside_cluster])

        # Belonging to different clusters
        score -= np.sum(C[inside_cluster, :][:, outside_cluster])
        score -= np.sum(C[outside_cluster, :][:, inside_cluster])

    return score


initial_C = C
initial_score = score(C)
initial_ordering = np.arange(n_variables)

plt.figure()
plt.imshow(C, interpolation='nearest')
plt.title('Initial C')
print 'Initial ordering:', initial_ordering
print 'Initial covariance matrix score:', initial_score

# Pretty dumb greedy optimization algorithm that continuously
# swaps rows to improve the score
def swap_rows(C, var1, var2):
    '''
    Function to swap two rows in a covariance matrix,
    updating the appropriate columns as well.
    '''
    D = C.copy()
    D[var2, :] = C[var1, :]
    D[var1, :] = C[var2, :]

    E = D.copy()
    E[:, var2] = D[:, var1]
    E[:, var1] = D[:, var2]

    return E

current_C = C
current_ordering = initial_ordering
current_score = initial_score

max_iter = 1000
for i in range(max_iter):
    # Find the best row swap to make
    best_C = current_C
    best_ordering = current_ordering
    best_score = current_score
    for row1 in range(n_variables):
        for row2 in range(n_variables):
            if row1 == row2:
                continue
            option_ordering = best_ordering.copy()
            option_ordering[row1] = best_ordering[row2]
            option_ordering[row2] = best_ordering[row1]
            option_C = swap_rows(best_C, row1, row2)
            option_score = score(option_C)

            if option_score > best_score:
                best_C = option_C
                best_ordering = option_ordering
                best_score = option_score

    if best_score > current_score:
        # Perform the best row swap
        current_C = best_C
        current_ordering = best_ordering
        current_score = best_score
    else:
        # No row swap found that improves the solution, we're done
        break

# Output the result
plt.figure()
plt.imshow(current_C, interpolation='nearest')
plt.title('Best C')
print 'Best ordering:', current_ordering
print 'Best score:', current_score
print
print 'Cluster     [variables assigned to this cluster]'
print '------------------------------------------------'
for cluster in range(n_clusters):
    print 'Cluster %02d  %s' % (cluster + 1, current_ordering[cluster*cluster_size:(cluster+1)*cluster_size])

你看过层次聚类吗？它可以处理相似之处，而不仅仅是距离。您可以将树状图切割成 k 个簇的高度，但通常最好目视检查树状图并决定要切割的高度。

如另一个答案所示，分层聚类也经常用于为相似矩阵可视化生成巧妙的重新排序：它将更多相似的条目彼此相邻放置。这也可以作为用户的验证工具！

您是否研究过相关聚类？该聚类算法使用成对的正/负相关信息来自动提出具有明确定义的功能和严格的生成概率解释的最佳聚类数量。

我会过滤一些有意义的（统计显着性）阈值，然后使用 dulmage-mendelsohn 分解来获得连接的组件。也许在您可以尝试消除传递相关之类的问题之前（A 与 B 高度相关，B 与 C，C 与 D，因此有一个包含所有这些的组件，但实际上 D 与 A 的相关性很低）。你可以使用一些基于中介的算法。正如有人建议的那样，这不是一个双聚类问题，因为相关矩阵是对称的，因此没有双元素。