numpy数组中标记组件之间的最小边到边欧几里得距离

Question

我在大型numpy 数组中有许多不同的形式，我想使用numpy 和scipy 计算它们之间的边到边欧几里得距离。

注意：我进行了搜索，这与堆栈上之前的其他问题不同，因为我想获得数组中标记的补丁之间的最小距离，而不是像其他点或单独的数组之间的距离问题已提出。

我当前的方法使用 KDTree，但对于大型数组来说效率非常低。本质上，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。最后以平均最小距离为例进行计算。

我正在寻找一种使用 python 的更智能的方法，最好不要使用任何额外的模块。

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()

---

编辑 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 没有返回正确的距离矩阵，因此值是错误的。

Answer 1

这是一种完全矢量化的方法来找到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(N²) 内存，其中 N 是非零像素的数量。如果这要求太高，您可以沿一个轴“去矢量化”它（添加一个 for 循环）。