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我有一个 numpy 数组，它代表一个图像。图像有 3 种颜色：橙色（背景）、蓝色（object1）和绿色（object2）。我使用 3 个值（0、1 和 2）来表示 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靠近图像的中心（红点）？（这里，更近是指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 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)))

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)
# 30 seconds

上面的代码可以工作，但速度很慢，而且需要很大的内存（大约 30 GB）。如果您想在您的 PC 中重现我的代码，您可以使用较小的形状而不是 (3200, 1200, 1200)。有没有更有效的方法来实现我的目标？

注意：实际上我的图像是3D CT医学图像，太大无法上传。图像中的对象是随机的，可能是凸的，也可能不是。这就是为什么我的实施速度很慢。这里为了澄清我的问题，我使用2D图像来解释我的方法。

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2 回答 2

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这可能不是最终解决方案或最佳时间，必须使用实际数据进行测试。为了理解我的想法，我选择了更小的矩阵尺寸和只有 2D 的情况

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

在此处输入图像描述

于 2020-10-10T22:13:07.257 回答
0

我解决了这个问题。

因为这两个 3D 数组太大了。所以起初我用最近邻法将它们缩小到更小的尺寸。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
于 2020-10-11T14:15:07.297 回答