如何从numpy数组中删除重叠的块？

如何解决如何从numpy数组中删除重叠的块？

我正在使用cv2.goodFeaturesToTrack函数在图像中查找特征点。最终目标是提取特定大小的正方形块，并以特征点为这些块的中心。

但是，许多特征点彼此靠近，因此这些块是重叠的，这不是我想要的。

这是所有特征点（中心）的示例：

array([[3536.,1419.],[2976.,1024.],[3504.,1400.],[3574.,1505.],[3672.,1453.],[3671.,1442.],[3489.,1429.],[3108.,737.]])

比方说，我想找到前n个blockRadius = 400块，它们不重叠。关于如何实现这一目标的任何想法？

解决方法

您可以更进一步scipy.spatial.KDTree-尽管它不支持查询由不同数量的点组成的块。因此，它可以与另一个库python-igraph结合使用，该库允许快速查找闭合点的连接组件：

from scipy.spatial import KDTree
import igraph as ig

data = np.array([[3536.,1419.],[2976.,1024.],[3504.,1400.],[3574.,1505.],[3672.,1453.],[3671.,1442.],[3489.,1429.],[3108.,737.]])
edges1 = KDTree(data[:,:1]).query_pairs(r=400)
edges2 = KDTree(data[:,1:]).query_pairs(r=400)
g = ig.Graph(n = len(data),edges=edges1 & edges2)
i = g.clusters()

因此，簇对应于某种内部类型igraph的块点的索引序列。快速预览：

>>> print(i)
Clustering with 8 elements and 2 clusters
[0] 0,2,3,4,5,6
[1] 1,7
>>> pal = ig.drawing.colors.ClusterColoringPalette(len(i)) #number of colors used
color = pal.get_many(i.membership) #list of color tags
ig.plot(g,bbox = (200,100),layout=g.layout('circle'),vertex_label=g.vs.indices,vertex_color = color,vertex_size = 12,vertex_label_size = 8)

用法示例：

>>> [data[n] for n in i] #or list(i)
[array([[3536.,1429.]]),array([[2976.,737.]])]

备注： 此方法允许使用成对的封闭点代替n*n矩阵，这在某些情况下在内存中效率更高。

,

您将需要迭代执行此操作，因为这样的经常性辍学无法向量化。我想这样的事情会起作用

from scipy.spatial.distance import pdist,squareform

c = np.array([[3536.,737.]])

dists = squareform(pdist(c,metric = 'chebyshev'))     # distance matrix,chebyshev here since you seem to want blocks
indices = np.arange(c.shape[0])  # indices that haven't been dropped (all to start)
out = [0]                        # always want the first index
while True:
    try:
        indices = indices[dists[indices[0],indices] > 400] #drop indices that are inside threshhold
        out.append(indices[0])   # add the next index that hasn't been dropped to the output
    except:
        break   # once you run out of indices,you'll get an IndexError and you're done
print(out)
[0,1]

让我们尝试一些要点：

np.random.seed(42)
c = np.random.rand(10000,2) * 800
dists = squareform(pdist(c,checbyshev here since you seem to want squares
indices = np.arange(c.shape[0])  # indices that haven't been dropped (all to start)
out = [0]                        # always want the first index
while True:
    try:
        indices = indices[dists[indices[0],you'll get an IndexError and you're done
print(out,pdist(c[out],metric = 'chebyshev'))
[0,6,17] [635.77582886 590.70015659 472.87353138 541.13920029 647.69071411
 476.84658995]

因此，有4个点（因为4个400x400块以4个图块组成一个800x800的空间是有意义的），大部分都是低值（17 400