我有一個(gè) numpy 數(shù)組，它代表一個(gè)圖像。該圖像有 3 種顏色：橙色（背景）、藍(lán)色（對(duì)象 1）和綠色（對(duì)象 2）。我使用 3 個(gè)值（0、1 和 2）來指示 numpy 數(shù)組中的 3 種顏色。兩個(gè)對(duì)象不重疊。我的問題是：如何知道哪個(gè)物體更接近圖像的中心（紅點(diǎn)）？（這里，較近是指該物體到一個(gè)物體圖像中心的最近距離小于該物體到另一個(gè)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(nèi)存（大約 30 GB）。如果你想在你的電腦上重現(xiàn)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(shí)現(xiàn)我的目標(biāo)？
2 回答
繁花不似錦
TA貢獻(xiàn)1851條經(jīng)驗(yàn) 獲得超4個(gè)贊
我解決了這個(gè)問題。
因?yàn)閮蓚€(gè)3D數(shù)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(xù)：
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(guān)寶盒
TA貢獻(xiàn)1772條經(jīng)驗(yàn) 獲得超5個(gè)贊
這可能不是最終的解決方案或最佳的運(yùn)行時(shí)間，必須用實(shí)際數(shù)據(jù)進(jìn)行測(cè)試。為了表達(dá)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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