如何将这个matlab代码转换为AForge.net+c#代码?
cdf1 = cumsum(hist1) / numel(aa); 我发现Histogram.cumulative方法存在于Accord.net中。但我不知道怎么用。
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)发布于 2017-12-27 02:55:19
实际上,我对Accord.NET不太了解,但是阅读文档,我认为ImageStatistics类就是您要找的(参考在这里)。问题是,它不能为图像建立一个单一的直方图,你必须自己去做。Matlab中的imhist只合并这三个通道,然后计数整个像素出现的情况,所以您应该这样做:
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#中,可以从R、G和B信道值构建RGB值,如下所示:
public static int ChannelsToRGB(Int32 red, Int32 green, Int32 blue)
{
return ((red << 0) | (green << 8) | (blue << 16));
}https://stackoverflow.com/questions/47985519
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