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python-3.x - 力矩保持阈值 : Trilevel Case

转载 作者:行者123 更新时间:2023-12-04 13:27:21 26 4
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我目前正在实现 Tsai 提出的分割方法对于 Python 3.6 中 8 位图像中的三级情况。这是代码,考虑到论文附件中解释的数学关系:


### EXAMPLE GIVEN BY TSAI ###

# This is the "dummy" example described in the paper
img = np.array([[10, 8, 10, 9, 20, 21,32,30,40,41,41,40],
[12, 10, 11, 10, 19, 20, 30, 28, 38, 40, 40, 39],
[10, 9, 10, 8, 20, 21, 30, 29, 42, 40, 40, 39],
[11, 10, 9, 11, 19, 21, 31, 30, 40, 42,38, 40]])

bins_hist = list(range(0,257))
histogram = np.zeros((256,4)) # We generate the counts for each grey value, 3rd column probabilities and 4th column cumulative sum
histogram[:,0] = np.arange(256) # First column contains the intensity levels

hist_i = np.histogram(img, bins=bins_hist)
counts = hist_i[0]
histogram[:,1] = np.add(histogram[:,1], counts)

row, col = img.shape
total_voxels = row * col

histogram[:,2] = histogram[:,1]/total_voxels # Relative frequency at each GV
histogram[:,3] = np.cumsum(histogram[:,2]) # Cumulative of the relative frequency


### THRESHOLDS CALCULATION BEGIN ###

# 0 Moment = 1
m0 = 1.0

# 1st Moment
m1 = np.cumsum((histogram[:,0])*histogram[:,2])[-1]

# 2nd Moment
m2 = np.cumsum((histogram[:,0]**2)*histogram[:,2])[-1] # Take the last value

# 3rd Moment
m3 = np.cumsum((histogram[:,0]**3)*histogram[:,2])[-1]

# 4th Moment
m4 = np.cumsum((histogram[:,0]**4)*histogram[:,2])[-1]

# 5th Moment
m5 = np.cumsum((histogram[:,0]**5)*histogram[:,2])[-1]

# Now we must find the value in the binary image that preserves these moments


# We solve the equalities --> For solutions refer to Paper Annex A.2
cd = (m0*m2*m4) + (m1*m3*m2) + (m1*m3*m2) - (m2*m2*m2) - (m1*m1*m4) - (m3*m3*m0)
c0 = ((-m3*m2*m4) + (-m4*m3*m2) + (m1*m3*-m5) - (-m5*m2*m2) - (-m4*m1*m4) - (m3*m3*-m3)) / cd
c1 = ((m0*-m4*m4) + (m1*-m5*m2) + (-m3*m3*m2) - (m2*-m4*m2) - (m1*-m3*m4) - (-m5*m3*m0)) / cd
c2 = ((m0*m2*-m5) + (m1*m3*-m3) + (m1*-m4*m2) - (m2*m2*-m3) - (m1*m1*-m5) - (m3*-m4*m0)) /cd

a1 = c0/2 - c1*c2/6 + (c2**3)/27
a2 = (c0/2 - c1*c2/6 + (c2**3)/27)**2
a3 = (c1/3 - (c2**2)/9)**3

a = (a1 - cmath.sqrt(a2 + a3))**1/3

b = -(c1/3 - (c2**2)/9)/a
w1 = -0.5 + 1j * (math.sqrt(3)/2)
w2 = -0.5 - 1j * (math.sqrt(3)/2)

z0 = -c2/3 - a - b
z1 = -c2/3 - w1*a - w2*b
z2 = -c2/3 - w2*a - w1*b

pd = (z1*z2**2) + (z2*z0**2) + (z0*z1**2) - (z0**2*z1) - (z0*z2**2) - (z1**2*z2)
p0 = ((m0*z1*z2**2) + (m1*z1**2) + (z2*m2) - (m2*z1) - (m1*z2**2) - (z1**2*z2*m0)) /pd
p1 = ((m1*z2**2) + (z0*m2) + (m0*z2*z0**2) - (z0**2*m1) - (z0*m0*z2**2) - (m2*z2)) / pd # Fraction of the below-threshold pixels in the binary histogram

th1 = 0.0 # First threshold in the trimodal histogram
th2 = 0.0 # Second threshold
dist1 = 10000000
dist2 = 10000000

for i in range(254):
for j in range(i+1, 255):
# Select threshold --> closest to p0 from the normlaized histogram
p0_orig = histogram[i,3] # Take the cumulative relative frequency at the value p0
p1_orig = histogram[j, 3] - histogram[i,3]
dist_i = abs(p0 - p0_orig)
dist_j = abs(p1 - p1_orig)

if dist_i < dist1 and dist_j < dist2: # This one was the one mentioned by Tsai ("Minimize the distance")

print(i,j,dist_i, dist_j)
dist1 = dist_i
dist2 = dist_j
th1 = i
th2 = j


然而,当我考虑到图 1 中描述的“虚拟”图像而应用它来重现他的结果时,我没有获得相同的阈值(在我的情况下,我得到 12 和 29,而不是 18 和 30)。
有没有人有这种方法的经验,可以帮助我找出我的代码有什么问题?

最佳答案

可能有一些错别字,例如:a = (a1 - cmath.sqrt(a2 + a3)) ** 1 / 3这实际上应该是:a = (a1 - cmath.sqrt(a2 + a3)) ** (1 / 3)并建议使用numpy线性代数的函数来避免一些错误,比如npmpy.dot(x,y), numpy.vdot(x,y)等等。这也使代码更具可读性。

关于python-3.x - 力矩保持阈值 : Trilevel Case,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/67425932/

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