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我正在尝试在 MATLAB 中实现以下最小错误阈值（由 J. Kittler 和 J. Illingworth 编写）方法。

你可以看看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

我在下面的图片上试过了： 信件

原始尺寸图像。

目标是提取字母的二进制图像（Hebrew Letters）。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？有人有更好的主意吗？

谢谢。

PS有人会在主题标签中添加“自适应阈值”吗（我不能，因为我是新手）。

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2 回答 2

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阈值是一个相当棘手的业务。多年来，我一直在对图像进行阈值处理，但我没有找到一种始终表现良好的技术，而且我开始不相信 CS 期刊中普遍表现出色的说法。

最大误差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没有足够清晰地分离，以使这种阈值方法起作用。

如果您想确保代码工作正常，您可以创建一个这样的测试程序，并检查您是否获得了良好的初始分割，以及代码在什么级别的“双峰”分解。

于 2010-01-13T19:58:00.263 回答
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我认为您的代码并不完全正确。您使用图像的绝对直方图而不是论文中使用的相对直方图。此外，您的代码效率相当低，因为它为每个可能的阈值计算两个直方图。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;
于 2011-06-08T14:16:54.537 回答