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我有一组数据，它是 1000 个同源蛋白质序列的距离矩阵。

我已经设法为此计算了亲和力矩阵（简单计算：1 - 距离，在我的例子中）。

基本上，如果在 Excel 中查看数据，没有标题行，第一列是序列名称，然后接下来的 1000 列是距离值。

我已经修改了 sklearn 的 Affinity Propagation 站点上提供的代码。这就是它现在的样子：

print __doc__

import numpy as np
from sklearn.cluster import AffinityPropagation
from sklearn import metrics
from sklearn.datasets.samples_generator import make_blobs
import csv

##############################################################################
f = open('ha-sequences-sample-distmat2.csv', 'rU')
csvreader = csv.reader(f)

sequence_names = []
distance_matrix = []
full_data = []

for row in csvreader:
#   print row

    sequence_names.append(row[0])
    distance_matrix.append(row[1:])
    full_data.append(row)

f.close()

distmat = np.array([row for row in distance_matrix]).astype(np.float)

# print distmat

affinity_matrix = np.array([1 - row for row in distmat]).astype(np.float)

full_matrix = zip(sequence_names, affinity_matrix)

# print affinity_matrix, sequence_names




##############################################################################
# Compute Affinity Propagation
af = AffinityPropagation(affinity='precomputed').fit(affinity_matrix)
cluster_centers_indices = af.cluster_centers_indices_
labels = af.labels_

n_clusters_ = len(cluster_centers_indices)

print 'Estimated number of clusters: %d' % n_clusters_
print "Homogeneity: %0.3f" % metrics.homogeneity_score(sequence_names, labels)
print "Completeness: %0.3f" % metrics.completeness_score(sequence_names, labels)
print "V-measure: %0.3f" % metrics.v_measure_score(sequence_names, labels)
print "Adjusted Rand Index: %0.3f" % \
    metrics.adjusted_rand_score(sequence_names, labels)
print("Adjusted Mutual Information: %0.3f" %
      metrics.adjusted_mutual_info_score(sequence_names, labels))
print("Silhouette Coefficient: %0.3f" %
      metrics.silhouette_score(affinity_matrix, labels, metric='sqeuclidean'))

##############################################################################
# Plot result
import pylab as pl
from itertools import cycle

pl.close('all')
pl.figure(1)
pl.clf()

colors = cycle('bgrcmykbgrcmykbgrcmykbgrcmyk')
for k, col in zip(range(n_clusters_), colors):
    class_members = labels == k
    cluster_center = affinity_matrix[cluster_centers_indices[k]]
    pl.plot(affinity_matrix[class_members, 0], affinity_matrix[class_members, 1], col + '.')
    pl.plot(cluster_center[0], cluster_center[1], 'o', markerfacecolor=col,
            markeredgecolor='k', markersize=14)
    for x in affinity_matrix[class_members]:
        pl.plot([cluster_center[0], x[0]], [cluster_center[1], x[1]], col)

pl.title('Estimated number of clusters: %d' % n_clusters_)
pl.show()

我遇到的问题是：我不知道如何输出与每个集群对应的序列名称。如果我可以将聚集在一起的序列输出到 shell 并在绘图上显示簇编号，那将是最好的，但即使我不在绘图上显示东西，那也很酷。

有人知道怎么做这个吗？

4

1 回答 1

5

您有序列名称列表 (sequence_names) 和一个集群标签数组 (af.labels_)。因此，您可以遍历集群标签数组并从集群标签列表中保留一个映射序列名称。例如

#for a simple example, assume the names and cluster labels are predefined
sequence_names = ["a", "b", "c", "d"]
labels = [0,1,1,0]

from collections import defaultdict
clusternames = defaultdict(list)

for i, label in enumerate(labels):
    clusternames[label].append(sequence_names[i])

#clusternames now holds a map from cluster label to list of sequence names
#Print out the label with the list 
for k, v in clusternames.items():
    print k, v
于 2013-04-12T13:58:15.277 回答