将自动阈值矩方法从图像J转移到Python

Question

我想将我的图像处理从image J(斐济(转移到Python
在图像J中，我将图像分割为HSB，然后在B通道上使用Moments Auto阈值。在Python上，我希望获得阈值t的值，从该值进行分割。由于我找不到任何关于这个主题的帮助或代码，我来了。
从"时刻保持阈值：一种新方法，Tsai"(这里(的论文中，我做了这个：

import numpy as np
import cv2
import skimage
from skimage import io
img = io.imread("C:\Users\Image.tif")
hsv_img = cv2.cvtColor(filt_img, cv2.COLOR_RGB2HSV)    
H, S, B = cv2.split(img) # spliting the image into HSB
B_histo = skimage.exposure.histogram (B) # making the histogram
pix_sum = B.shape[0]*B.shape[1] # calculating the sum of the pixels
#from the paper, calculating the 4 first odrers m0, m1, m2, m3 to get to p0. The name of the further variables stems from the paper. 
pj = B_histo[0] / pix_sum 
pj_z1 = np.power(B_histo[1], 1) * pj
pj_z2 = np.power(B_histo[1], 2) * pj
pj_z3 = np.power(B_histo[1], 3) * pj
m0 = np.sum(pj)
m1 = np.sum(pj_z1)
m2 = np.sum(pj_z2)
m3 = np.sum(pj_z3)
cd = (m0*m2) - (m1*m1)
c0 = ((-m2*m2) - (-m3*m1))/cd
c1 = ((m0*-m3) - (m1*-m2))/cd
z0 = 0.5 *(-c1 - (np.power(np.power(c1, 2) - 4*c0, 1/2)))
z1 = 0.5 *(-c1 + (np.power(np.power(c1, 2) - 4*c0, 1/2)))
pd = z1 - z0
p0 = (z1 - m1) / pd # p0 should be the percentage of the pixels to which the value 0 is attributed
# using cumulative histogram and comparing it to a target value by calculating the difference. When the difference is the lowest, the index indicates the value of the threshold t
cum_pix = np.cumsum(B_histo[0]) 
target_value = p0 * pix_sum
#td = cum_pix
diff = [abs(i - target_value) for i in cum_pix]
cum_pix[1]
diff[0]
t = [abs(i - target_value) for i in cum_pix].index(np.min(diff))
print(t)

抱歉代码太乱了。无论如何，我在Python上计算的值与在Image J上计算的不同。你知道问题可能来自哪里吗？或者Python中有一个函数可以检索时刻自动阈值的值？我将非常感谢提示或提示挖掘，谢谢

Answer 1

力矩和阈值比例的计算很好，但存在缩减仓(即使指定了256个仓(的撇除直方图问题，以及t的计算方式。以下是问题的解决方案：

#Instead of skimage.exposure.histogram (B, nbins=256)   
grey_value = np.arange(256)
B_freq = cv2.calcHist([S], [0],None,[256],[0,255])
B_freq = B_freq.reshape((256,))
B_freq = np.int64(B_freq)
B_histo = (B_freq, grey_value)
pix_sum = B.shape[0]*B.shape[1] # calculating the sum of the pixels
#from the paper, calculating the 3 first odrers
pj = B_histo[0] / pix_sum 
pj_z1 = np.power(B_histo[1], 1) * pj
pj_z2 = np.power(B_histo[1], 2) * pj
pj_z3 = np.power(B_histo[1], 3) * pj
m0 = np.sum(pj)
m1 = np.sum(pj_z1)
m2 = np.sum(pj_z2)
m3 = np.sum(pj_z3)
cd = (m0*m2) - (m1*m1)
c0 = ((-m2*m2) - (-m3*m1))/cd
c1 = ((m0*-m3) - (m1*-m2))/cd

z0 = 0.5 *(-c1 - (np.power(np.power(c1, 2) - 4*c0, 1/2)))
z1 = 0.5 *(-c1 + (np.power(np.power(c1, 2) - 4*c0, 1/2)))
pd = z1 - z0
p0 = (z1 - m1) / pd # p0 should be the percentage of the pixels to which the threshold t should be done
# using cumulative histogram and comparing it to a target value by calculating the difference. When the difference is the lowest, the index indicates the value of the threshold t
cum_pix = np.cumsum(B_freq) 
target_value = p0 * pix_sum
diff = [(i - target_value) for i in cum_pix]
def find_nearest(array, value):
array = np.asarray(array)
idx = (np.abs(array - value)).argmin()
return array[idx]
t = diff.index(find_nearest(diff, 0))
print(t)

所以这个代码给出了图像J的矩自动阈值的阈值。