我有一個 numpy 數組，它代表一個圖像。該圖像有 3 種顏色：橙色（背景）、藍色（對象 1）和綠色（對象 2）。我使用 3 個值（0、1 和 2）來指示 numpy 數組中的 3 種顏色。兩個對象不重疊。我的問題是：如何知道哪個物體更接近圖像的中心（紅點）？（這里，較近是指該物體到一個物體圖像中心的最近距離小于該物體到另一個物體圖像中心的最近距離）我的代碼是這樣的：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上面的代碼可以工作，但是速度很慢，并且需要很大的內存（大約 30 GB）。如果你想在你的電腦上重現我的代碼，你可以使用更小的形狀而不是 (3200, 1200, 1200)。有沒有更有效的方法來實現我的目標？
2 回答
繁花不似錦
TA貢獻1851條經驗 獲得超4個贊
我解決了這個問題。
因為兩個3D數組太大了。所以首先我用最近鄰法將它們采樣到更小的尺寸。然后繼續：
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
月關寶盒
TA貢獻1772條經驗 獲得超5個贊
這可能不是最終的解決方案或最佳的運行時間，必須用實際數據進行測試。為了表達我的想法，我選擇了較小的矩陣大小并且僅使用 2D 情況
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()
輸出
Distance to first object: 35.38361202590826
Distance to second object: 35.17101079013795
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