在python numpy数组中，如何知道哪个对象更接近一张图像中的一个点？

Question

我有一个 numpy 数组，它代表一个图像。该图像有 3 种颜色：橙色（背景）、蓝色（对象 1）和绿色（对象 2）。我使用 3 个值（0、1 和 2）来指示 numpy 数组中的 3 种颜色。两个对象不重叠。我的问题是：如何知道哪个物体更接近图像的中心（红点）？（这里，较近是指该物体到一个物体图像中心的最近距离小于该物体到另一个物体图像中心的最近距离）我的代码是这样的：import numpy as npfrom scipy import spatialimport timesub_image1 = np.ones((30, 30, 30))sub_image2 = np.ones((20, 10, 15))# pad the two sub_images to same shape (1200, 1200, 1200) to simulate my 3D medical dataimg_1 = np.pad(sub_image1, ((1100, 70), (1100, 70), (1100, 70)))img_2 = np.pad(sub_image1, ((1100, 80), (1130, 60), (1170, 15)))def nerest_dis_to_center(img):    position = np.where(img > 0)    coordinates = np.transpose(np.array(position))  # get the coordinates where the voxels is not 0    cposition = np.array(img.shape) / 2  # center point position/coordinate    distance, index = spatial.KDTree(coordinates).query(cposition)    return distancet1 = time.time()d1 = nerest_dis_to_center(img_1)d2 = nerest_dis_to_center(img_2)if d1 > d2:    print("img2 object is nearer")elif d2 > d1:    print("img1 object is nearer")else:    print("They are the same far")t2 = time.time()print("used time: ", t2-t1)# 30 seconds上面的代码可以工作，但是速度很慢，并且需要很大的内存（大约 30 GB）。如果你想在你的电脑上重现我的代码，你可以使用更小的形状而不是 (3200, 1200, 1200)。有没有更有效的方法来实现我的目标？

Answer 1

我解决了这个问题。

因为两个3D数组太大了。所以首先我用最近邻法将它们采样到更小的尺寸。然后继续：

import numpy as np

from scipy import spatial

import time

sub_image1 = np.ones((30, 30, 30))

sub_image2 = np.ones((20, 10, 15))

# pad the two sub_images to same shape (1200, 1200, 1200) to simulate my 3D medical data

img_1 = np.pad(sub_image1, ((1100, 70), (1100, 70), (1100, 70)))

img_2 = np.pad(sub_image1, ((1100, 80), (1130, 60), (1170, 15)))

ori_sz = np.array(img_1.shape)

trgt_sz = ori_sz / 4

zoom_seq = np.array(trgt_sz, dtype='float') / np.array(ori_sz, dtype='float')

img_1 = ndimage.interpolation.zoom(img_1, zoom_seq, order=0, prefilter=0)

img_2 = ndimage.interpolation.zoom(img_2, zoom_seq, order=0, prefilter=0)

print("it cost this secons to downsample the nearer image" + str(time.time() - t0))  # 0.8 seconds

def nerest_dis_to_center(img):

    position = np.where(img > 0)

    coordinates = np.transpose(np.array(position))  # get the coordinates where the voxels is not 0

    cposition = np.array(img.shape) / 2  # center point position/coordinate

    distance, index = spatial.KDTree(coordinates).query(cposition)

    return distance

t1 = time.time()

d1 = nerest_dis_to_center(img_1)

d2 = nerest_dis_to_center(img_2)

if d1 > d2:

    print("img2 object is nearer")

elif d2 > d1:

    print("img1 object is nearer")

else:

    print("They are the same far")

t2 = time.time()

print("used time: ", t2-t1)

# 1.1 seconds

Answer 2

这可能不是最终的解决方案或最佳的运行时间，必须用实际数据进行测试。为了表达我的想法，我选择了较小的矩阵大小并且仅使用 2D 情况

import numpy as np

import matplotlib.pyplot as plt

sub_image1 = np.ones((30, 30))  # 1st object

sub_image2 = np.ones((20, 10)) * 2  # 2nd object

# pad the two sub_images to same shape (120, 120)

img_1 = np.pad(sub_image1, ((110, 60), (60, 110)))

img_2 = np.pad(sub_image2, ((100, 80), (130, 60)))

final_image = img_1 + img_2  # creating final image with both objects in a background of zeros

image_center = (np.array([final_image.shape[0], final_image.shape[1]]) / 2).astype(np.int)

# mark the center

final_image[image_center[0], image_center[1]] = 10

# find the coordinates of where the objects are

first_obj_coords = np.argwhere(final_image == 1)  # could be the most time consuming operation

second_obj_coords = np.argwhere(final_image == 2) # could be the most time consuming 

# find their centers

first_obj_ctr = np.mean(first_obj_coords, axis=0)

second_obj_ctr = np.mean(second_obj_coords, axis=0)

# turn the centers to int for using them to index

first_obj_ctr = np.floor(first_obj_ctr).astype(int)

second_obj_ctr = np.floor(second_obj_ctr).astype(int)

# mark the centers of the objects

final_image[first_obj_ctr[0], first_obj_ctr[1]] = 10

final_image[second_obj_ctr[0], second_obj_ctr[1]] = 10

# calculate the distances from center to the object center

print('Distance to first object: ', np.linalg.norm(image_center - first_obj_ctr))

print('Distance to second object: ', np.linalg.norm(image_center - second_obj_ctr))

plt.imshow(final_image)

plt.show()

输出

Distance to first object:  35.38361202590826

Distance to second object:  35.17101079013795