Question

我在大型numpy数组中有许多不同的表单，我想使用numpy和scipy来计算它们之间的边缘到边缘的欧氏距离。

注意：我进行了搜索，这与堆栈中之前的其他问题有所不同，因为我希望获得阵列中标记补丁之间的最小距离，而不是像点或其他阵列之间的最小距离问题已经提出。

我目前的方法使用KDTree，但对于大型阵列来说效率非常低。基本上我正在查找每个标记组件的坐标并计算所有其他组件之间的距离。最后，计算平均最小距离作为示例。

我正在寻找一种更智能的方法，使用python，最好没有任何额外的模块。

import numpy
from scipy import spatial
from scipy import ndimage

# Testing array
a = numpy.zeros((8,8), dtype=numpy.int)
a[2,2] = a[3,1] = a[3,2] = 1
a[2,6] = a[2,7] = a[1,6] = 1
a[5,5] = a[5,6] = a[6,5] = a[6,6] = a[7,5] = a[7,6] = 1    

# label it
labeled_array,numpatches = ndimage.label(a)

# For number of patches
closest_points = []
for patch in [x+1 for x in range(numpatches)]:
# Get coordinates of first patch
    x,y = numpy.where(labeled_array==patch)
    coords = numpy.vstack((x,y)).T # transform into array
    # Built a KDtree of the coords of the first patch
    mt = spatial.cKDTree(coords)

    for patch2 in [i+1 for i in range(numpatches)]:
        if patch == patch2: # If patch is the same as the first, skip
            continue
        # Get coordinates of second patch
        x2,y2 = numpy.where(labeled_array==patch2)
        coords2 = numpy.vstack((x2,y2)).T

        # Now loop through points
        min_res = []
        for pi in range(len(coords2)):
            dist, indexes = mt.query(coords2[pi]) # query the distance and index
            min_res.append([dist,pi])
        m = numpy.vstack(min_res)
        # Find minimum as closed point and get index of coordinates
        closest_points.append( coords2[m[numpy.argmin(m,axis=0)[0]][1]] )


# The average euclidean distance can then be calculated like this:
spatial.distance.pdist(closest_points,metric = "euclidean").mean()

修改刚刚测试了@morningsun提出的解决方案，这是一个巨大的速度提升。但是返回的值略有不同：

# Consider for instance the following array
a = numpy.zeros((8,8), dtype=numpy.int)
a[2,2] = a[2,6] = a[5,5] = 1  

labeled_array, numpatches = ndimage.label(cl_array,s)

# Previous approach using KDtrees and pdist
b = kd(labeled_array,numpatches)
spatial.distance.pdist(b,metric = "euclidean").mean()
#> 3.0413115592767102

# New approach using the lower matrix and selecting only lower distances
b = numpy.tril( feature_dist(labeled_array) )
b[b == 0 ] = numpy.nan
numpy.nanmean(b)
#> 3.8016394490958878

编辑2

啊，想通了。 spatial.distance.pdist不返回适当的距离矩阵，因此值是错误的。

Answer 1

这是一种完全矢量化的方法来查找标记对象的距离矩阵：

import numpy as np
from scipy.spatial.distance import cdist

def feature_dist(input):
    """
    Takes a labeled array as returned by scipy.ndimage.label and 
    returns an intra-feature distance matrix.
    """
    I, J = np.nonzero(input)
    labels = input[I,J]
    coords = np.column_stack((I,J))

    sorter = np.argsort(labels)
    labels = labels[sorter]
    coords = coords[sorter]

    sq_dists = cdist(coords, coords, 'sqeuclidean')

    start_idx = np.flatnonzero(np.r_[1, np.diff(labels)])
    nonzero_vs_feat = np.minimum.reduceat(sq_dists, start_idx, axis=1)
    feat_vs_feat = np.minimum.reduceat(nonzero_vs_feat, start_idx, axis=0)

    return np.sqrt(feat_vs_feat)

这种方法需要O（N ²）存储器，其中N是非零像素的数量。如果这个要求太高，你可以沿着一个轴“去矢量化”它（添加一个for循环）。

numpy数组中标记组件之间的最小边缘到边缘欧氏距离

1 个答案: