将直方图的CDF从matlab转换为c＃？

Question

如何将此matlab代码转换为AForge.net + c＃代码？

cdf1 = cumsum(hist1) / numel(aa); 

我发现Accord.net中存在Histogram.cumulative方法。但我不知道如何使用。

请教如何转换。

% Histogram Matching
%

clear
clc
close all

pkg load image

% 이미지 로딩
aa=imread('2.bmp');
ref=imread('ref2.png');

figure(1); imshow(aa); colormap(gray)
figure(2); imshow(ref); colormap(gray)

M = zeros(256,1,'uint8'); % Store mapping - Cast to uint8 to respect data type
hist1 = imhist(aa); % Compute histograms
hist2 = imhist(ref);

cdf1 = cumsum(hist1) / numel(aa); % Compute CDFs
cdf2 = cumsum(hist2) / numel(ref);

% Compute the mapping
for idx = 1 : 256
[~,ind] = min(abs(cdf1(idx) - cdf2));


M(idx) = ind-1;
end

% Now apply the mapping to get first image to make
% the image look like the distribution of the second image
out = M(double(aa)+1);

figure(3); imshow(out); colormap(gray)

Answer 1

实际上，我对Accord.NET并不了解，但是阅读文档我认为ImageStatistics类是你正在寻找的（reference here）。问题是它无法为图像构建单个直方图，您必须自己完成。 Matlab中的imhist只是将三个通道合并，然后计算整个像素的出现次数，这就是你应该做的：

Bitmap image = new Bitmap(@"C:\Path\To\Image.bmp");

ImageStatistics statistics = new ImageStatistics(image);
Double imagePixels = (Double)statistics.PixelsCount;

Int32[] histR = statistics.Red.Values.ToArray();
Int32[] histG = statistics.Green.Values.ToArray();
Int32[] histB = statistics.Blue.Values.ToArray();
Int32[] histImage = new Int32[256];

for (Int32 i = 0; i < 256; ++i)
    histImage[i] = histR[i] + histG[i] + histB[i];

Double cdf = new Double[256];
cdf[0] = (Double)histImage[0];

for (Int32 i = 1; i < 256; ++i)
    cdf[i] = (Double)(cdf[i] + cdf[i - 1]);

for (Int32 i = 0; i < 256; ++i)
    cdf[i] = cdf[i] / imagePixels;

在C#，RGB值可以从R，G和B通道值建立如下：

public static int ChannelsToRGB(Int32 red, Int32 green, Int32 blue)
{
    return ((red << 0) | (green << 8) | (blue << 16));
}