Question

我已经完成了一项编程，通过使用从000到111的RGB方法，可以在图像中计算多达8种颜色。我需要对其进行一些修改。到目前为止，我宣布如果大于128的数字将等于1＆amp;低于128将为0.它将计算8种颜色。如果我说部分从0-64,65-128,129-192,193-255进行，如何增加颜色数量。

count=zeros(1,8);
for i = 1:w 
    for j = 1:h

        if redChannel(i,j) > 128,
            aredChannel2 = 1;
        else
            aredChannel2=0;
        end
        quantizedImage(i,j,1)=aredChannel2*255;

        if greenChannel(i,j) > 128,
            agreenChannel2 = 1;
        else
            agreenChannel2=0;
        end
        quantizedImage(i,j,2)=agreenChannel2*255;

        if blueChannel(i,j) > 128,
            ablueChannel2 = 1;
        else
            ablueChannel2=0;
        end
        quantizedImage(i,j,3)=ablueChannel2*255;

        bin=4*aredChannel2+2*agreenChannel2+ablueChannel2+1;

        count(bin)=count(bin)+1;

    end
end

figure, imshow(uint8(quantizedImage))

Answer 1

增加间隔数会增加您所计算的基数：而不是2^3=8量化颜色，您将获得4^3=64量化颜色。

rgb = imread( ... ); %// read you image here
qImage = zeros( size(rgb(:,:,1)) ); %// preallocate result
intervals = permute([64, 128, 192, 256], [1 3 2]); %// the quantization intervals
base = numel(intervals);
for ci=1:size(rgb,3) %// for each channel
    whichInterval = bsxfun( @le, rgb(:,:,ci), intervals ); %// select per pixel, which interval is relevant
    [~, q] = max( whichInterval, [], 3 ); %// get index of relevant interval
    qImage = qImage*base + (q-1); %// -1 to convert from matlab's 1-based indexing
end

增加颜色数而不是8

1 个答案: