如何在多维数组中有效地查找相似元素的簇

时间:2015-12-17 02:56:48

标签: python arrays python-2.7 multidimensional-array

我有一个多维数组,如下所示:

array = [[a, b, b, a, a, b]
         [a, a, b, a, b, a]]

我想要做的是识别相似元素的集群,即查看每个元素,并根据是否存在另一个元素来找到一组“a' a'在其上方,下方,左侧或右侧,以便程序找到如下数组:

[ a b b a a ,b]

[ a a b a ,b,a ]

它将返回一个像这样的数组[" 0:0"," 1:0"," 1:1"]用于第一个& #39;一个' S

我的问题是,在python中执行此操作的最有效方法是什么?

仅供参考:我使用的是Python 2.7

1 个答案:

答案 0 :(得分:2)

使用scipy.ndimage.measurements

from scipy import ndimage 

def find_clusters(array):
    clustered = np.empty_like(array)
    unique_vals = np.unique(array)
    cluster_count = 0
    for val in unique_vals:
        labelling, label_count = ndimage.label(array == val)
        for k in range(1, label_count + 1):
            clustered[labelling == k] = cluster_count
            cluster_count += 1
    return clustered, cluster_count

clusters, cluster_count = find_clusters(array)
print("Found {} clusters:".format(cluster_count))
print(clusters)

ones = np.ones_like(array, dtype=int)
cluster_sizes = ndimage.sum(ones, labels=clusters, index=range(cluster_count)).astype(int)
com = ndimage.center_of_mass(ones, labels=clusters, index=range(cluster_count))
for i, (size, center) in enumerate(zip(cluster_sizes, com)):
    print("Cluster #{}: {} elements at {}".format(i, size, center))

的产率:

Found 6 clusters:
[[0 3 3 1 1 4]
 [0 0 3 1 5 2]]
Cluster #0: 3 elements at (0.66666666666666663, 0.33333333333333331)
Cluster #1: 3 elements at (0.33333333333333331, 3.3333333333333335)
Cluster #2: 1 elements at (1.0, 5.0)
Cluster #3: 3 elements at (0.33333333333333331, 1.6666666666666667)
Cluster #4: 1 elements at (0.0, 5.0)
Cluster #5: 1 elements at (1.0, 4.0)

要获取每个群集中元素的位置,您可以执行clusters == cluster_id,例如

In [126]:
clusters == 3
Out[126]:
array([[False,  True,  True, False, False, False],
       [False, False,  True, False, False, False]], dtype=bool)

或者,要获取每个群集的边界框,您可以使用同一个SciPy包中的find_objects

In [128]:
# +1 because zeros would be ignored otherwise
scipy.ndimage.measurements.find_objects(clusters+1)
Out[128]:
[(slice(0, 2, None), slice(0, 2, None)),
 (slice(0, 2, None), slice(3, 5, None)),
 (slice(1, 2, None), slice(5, 6, None)),
 (slice(0, 2, None), slice(1, 3, None)),
 (slice(0, 1, None), slice(5, 6, None)),
 (slice(1, 2, None), slice(4, 5, None))]