聚类相关矩阵

我有一个相关矩阵，该矩阵说明每个项目如何与另一个项目相关。因此，对于N个项目，我已经具有N * N个相关矩阵。使用此相关矩阵，如何将N个项目聚类在M个仓中，以便可以说第k个仓中的Nk个项目表现相同。请帮我。所有项目值都是分类的。

谢谢。请让我知道是否需要更多信息。我需要使用Python解决方案，但是任何将我推向要求的帮助都会有很大帮助。

看起来像是块建模的工作。Google对于“块建模”和前几期热门搜索很有帮助。

假设我们有一个协方差矩阵，其中N = 100，实际上有5个簇：

块建模正在尝试做的是找到行的顺序，以使簇变得明显为“块”：

下面是一个执行基本贪婪搜索以完成此操作的代码示例。您的250-300变量可能太慢了，但这只是一个开始。查看您是否可以遵循以下注释：

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])

您是否看过分层集群？它可以具有相似性，而不仅仅是距离。您可以在将树状图拆分成k个簇的高度处切割树状图，但通常最好是目视检查树状图并确定要切割的高度。

分层聚类通常也用于对相似矩阵虚拟化产生巧妙的重新排序，如另一个答案所示：它将更多相似的条目彼此相邻放置。这也可以用作用户的验证工具！

您是否研究了相关性聚类？该聚类算法使用成对的正/负相关信息自动提出具有良好定义的功能和严格的生成概率解释的最佳聚类数。

我将过滤某个有意义的（统计显着性）阈值，然后使用dulmage-mendelsohn分解来获取连接的组件。也许在您尝试消除传递相关性之类的问题之前（A与B，B与C，C与D高度相关，因此有一个包含所有这些成分的组件，但实际上D与A的相关性很低）。您可以使用一些基于中间性的算法。就像有人建议的那样，这不是一个双重问题，因为相关矩阵是对称的，因此不存在二元性。