我正在尝试在MATLAB中实现以下Minimum Error Thresholding(由J. Kittler和J. Illingworth提供)方法。
您可以查看PDF:
我的代码是:
function [ Level ] = MET( IMG )
%Maximum Error Thresholding By Kittler
% Finding the Min of a cost function J in any possible thresholding. The
% function output is the Optimal Thresholding.
for t = 0:255 % Assuming 8 bit image
I1 = IMG;
I1 = I1(I1 <= t);
q1 = sum(hist(I1, 256));
I2 = IMG;
I2 = I2(I2 > t);
q2 = sum(hist(I2, 256));
% J is proportional to the Overlapping Area of the 2 assumed Gaussians
J(t + 1) = 1 + 2 * (q1 * log(std(I1, 1)) + q2 * log(std(I2, 1)))...
-2 * (q1 * log(q1) + q2 * log(q2));
end
[~, Level] = min(J);
%Level = (IMG <= Level);
end
目标是提取字母的二进制图像(希伯来字母)。 我将代码应用于图像的子块(40 x 40)。 然而,我得到的结果不如K-Means Clustering method。
我错过了什么吗? 任何人都有更好的主意吗?
感谢。
P.S。 是否有人会在主题标签中添加“自适应阈值处理”(我不能因为我是新手)。
答案 0 :(得分:4)
阈值处理是一项相当棘手的工作。多年来我一直在对图像进行阈值处理,但我还没有找到一种总是表现良好的单一技术,而且我已经开始不相信CS期刊中普遍出色的表现。
最大错误阈值方法仅适用于很好的双峰直方图(但它适用于那些)。您的图像看起来像信号和背景可能无法清楚地分开,以使此阈值处理方法起作用。
如果你想确保代码工作正常,你可以创建一个这样的测试程序,并检查你是否获得良好的初始分段,以及代码分解的“双峰”级别。
答案 1 :(得分:2)
我认为你的代码并不完全正确。您可以使用图像的绝对直方图而不是纸张中使用的相对直方图。此外,由于每个可能的阈值计算两个直方图,因此您的代码效率相当低。我自己实现了算法。也许,有人可以利用它:
function [ optimalThreshold, J ] = kittlerMinimimErrorThresholding( img )
%KITTLERMINIMIMERRORTHRESHOLDING Compute an optimal image threshold.
% Computes the Minimum Error Threshold as described in
%
% 'J. Kittler and J. Illingworth, "Minimum Error Thresholding," Pattern
% Recognition 19, 41-47 (1986)'.
%
% The image 'img' is expected to have integer values from 0 to 255.
% 'optimalThreshold' holds the found threshold. 'J' holds the values of
% the criterion function.
%Initialize the criterion function
J = Inf * ones(255, 1);
%Compute the relative histogram
histogram = double(histc(img(:), 0:255)) / size(img(:), 1);
%Walk through every possible threshold. However, T is interpreted
%differently than in the paper. It is interpreted as the lower boundary of
%the second class of pixels rather than the upper boundary of the first
%class. That is, an intensity of value T is treated as being in the same
%class as higher intensities rather than lower intensities.
for T = 1:255
%Split the hostogram at the threshold T.
histogram1 = histogram(1:T);
histogram2 = histogram((T+1):end);
%Compute the number of pixels in the two classes.
P1 = sum(histogram1);
P2 = sum(histogram2);
%Only continue if both classes contain at least one pixel.
if (P1 > 0) && (P2 > 0)
%Compute the standard deviations of the classes.
mean1 = sum(histogram1 .* (1:T)') / P1;
mean2 = sum(histogram2 .* (1:(256-T))') / P2;
sigma1 = sqrt(sum(histogram1 .* (((1:T)' - mean1) .^2) ) / P1);
sigma2 = sqrt(sum(histogram2 .* (((1:(256-T))' - mean2) .^2) ) / P2);
%Only compute the criterion function if both classes contain at
%least two intensity values.
if (sigma1 > 0) && (sigma2 > 0)
%Compute the criterion function.
J(T) = 1 + 2 * (P1 * log(sigma1) + P2 * log(sigma2)) ...
- 2 * (P1 * log(P1) + P2 * log(P2));
end
end
end
%Find the minimum of J.
[~, optimalThreshold] = min(J);
optimalThreshold = optimalThreshold - 0.5;