python – numpy数组上的边框边 *** 作

概述假设我有一个非零值的非numpy数组,并且“background”= 0.作为一个例子,我将采用随机值的范围： array = np.random.randint(1, 5, size = (100,100,100))z,y,x = np.ogrid[-50:50, -50:50, -50:50]mask = x**2 + y**2 + z**2<= 20**2array[np.invert 假设我有一个非零值的非numpy数组,并且“background”= 0.作为一个例子,我将采用随机值的范围：

array = np.random.randint(1,5,size = (100,100,100))z,y,x = np.ogrID[-50:50,-50:50,-50:50]mask = x**2 + y**2 + z**2<= 20**2array[np.invert(mask)] = 0

首先,我想找到“边界体素”(所有非零值在其3x3x3邻域内都为零).其次,我想用非零邻居的平均值替换所有边界体素.到目前为止,我尝试以下列方式使用scipy的通用过滤器：

适用于每个元素的功能：

def borderCheck(values):    #check if the footprint center is on a nonzero value    if values[13] != 0:        #replace border voxels with the mean of nonzero neighbours        if 0 in values:            return np.sum(values)/np.count_nonzero(values)        else:            return values[13]    else:        return 0

通用过滤器：

from scipy import ndimageresult = ndimage.generic_filter(array,borderCheck,footprint = np.ones((3,3,3)))

这是处理这个问题的正确方法吗？我觉得我试图在这里重新发明轮子,并且必须有一个更短,更好的方法来实现结果.我还可以使用其他合适的(numpy,scipy)功能吗？

编辑

我搞砸了一件事：我想用非零和非边界邻居的平均值替换所有边界体素.为此,我试图从ali_m的代码中清理邻居(2D情况)：

#for each neighbour voxel,check whether it also appears in the border/edgesnon_border_neighbours = []for each in neighbours:    non_border_neighbours.append([i for i in each if nonzero_IDx[i] not in edge_IDx])

现在我无法弄清楚为什么non_border_neighbours会变回空？

此外,纠正我,如果我错了,但不是tree.query_ball_point与半径1只加上6个下一个邻居(欧几里德距离1)？我应该将sqrt(3)(3D case)设置为半径以获得26邻域吗？

解决方法 我认为最好先从2D案例开始,因为它可以更容易地被可视化：

import numpy as npfrom matplotlib import pyplot as pltA = np.random.randint(1,size=(100,100)).astype(np.double)y,-50:50]mask = x**2 + y**2 <= 30**2A[~mask] = 0

要找到边缘像素,您可以对蒙版执行二进制侵蚀,然后使用蒙版对结果进行异或

# rank 2 structure with full connectivitystruct = ndimage.generate_binary_structure(2,2)erode = ndimage.binary_erosion(mask,struct)edges = mask ^ erode

找到每个边缘像素的最近非零邻居的一种方法是使用scipy.spatial.cKDTree：

from scipy.spatial import cKDTree# the indices of the non-zero locations and their corresponding valuesnonzero_IDx = np.vstack(np.where(mask)).Tnonzero_vals = A[mask]# build a k-D treetree = cKDTree(nonzero_IDx)# use it to find the indices of all non-zero values that are at most 1 pixel# away from each edge pixeledge_IDx = np.vstack(np.where(edges)).Tneighbours = tree.query_ball_point(edge_IDx,r=1,p=np.inf)# take the average value for each set of neighboursnew_vals = np.hstack(np.mean(nonzero_vals[n]) for n in neighbours)# use these to replace the values of the edge pixelsA_new = A.astype(np.double,copy=True)A_new[edges] = new_vals

一些可视化：

fig,ax = plt.subplots(1,figsize=(10,4),sharex=True,sharey=True)norm = plt.normalize(0,A.max())ax[0].imshow(A,norm=norm)ax[0].set_Title('Original',Fontsize='x-large')ax[1].imshow(edges)ax[1].set_Title('Edges',Fontsize='x-large')ax[2].imshow(A_new,norm=norm)ax[2].set_Title('Averaged',Fontsize='x-large')for aa in ax:    aa.set_axis_off()ax[0].set_xlim(20,50)ax[0].set_ylim(50,80)fig.tight_layout()plt.show()

这种方法也将推广到3D案例：

B = np.random.randint(1,100)).astype(np.double)z,-50:50]mask = x**2 + y**2 + z**2 <= 20**2B[~mask] = 0struct = ndimage.generate_binary_structure(3,3)erode = ndimage.binary_erosion(mask,struct)edges = mask ^ erodenonzero_IDx = np.vstack(np.where(mask)).Tnonzero_vals = B[mask]tree = cKDTree(nonzero_IDx)edge_IDx = np.vstack(np.where(edges)).Tneighbours = tree.query_ball_point(edge_IDx,p=np.inf)new_vals = np.hstack(np.mean(nonzero_vals[n]) for n in neighbours)B_new = B.astype(np.double,copy=True)B_new[edges] = new_vals

测试你的版本：

def borderCheck(values):    #check if the footprint center is on a nonzero value    if values[13] != 0:        #replace border voxels with the mean of nonzero neighbours        if 0 in values:            return np.sum(values)/np.count_nonzero(values)        else:            return values[13]    else:        return 0result = ndimage.generic_filter(B,footprint=np.ones((3,3)))print(np.allclose(B_new,result))# True

我确信这不是最有效的方法,但它仍然比使用generic_filter快得多.

更新

通过减少被视为边缘像素/体素的候选邻居的点的数量,可以进一步改善性能：

# ...# the edge pixels/voxels plus their immediate non-zero neighbourserode2 = ndimage.binary_erosion(erode,struct)candIDate_neighbours = mask ^ erode2nonzero_IDx = np.vstack(np.where(candIDate_neighbours)).Tnonzero_vals = B[candIDate_neighbours]# ...

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