基本图像压缩:MATLAB中的“索引超出矩阵尺寸”错误

时间:2013-09-22 02:48:44

标签: matlab image-compression

我正在进行基本的图像压缩。我想在图像中采用8x8窗口,并进一步将该窗口细分为2x2,然后必须找到它们的平均值。在这之后,我必须将该平均值矩阵与阈值平均值(整个图像的平均值或该8x8窗口的平均值)进行比较。如果元素大于或等于阈值,如果不是0,则应分配1。为此,我在下面写了代码。但我自己陷入了第一步。它显示错误“索引超出矩阵尺寸”这一行P=J(i:(i+7),j:(j+7));请帮忙,我不确定如何比较和分配0& 1到最终矩阵。我试过X=bsxfun(@ge,M,thr) ..它不起作用..我的代码似乎很大..我是以写方式做的吗?请建议。我认为它可以变得更简单。我是matlab的新手,请帮我学习。这是我的代码:

I=imread('C:\Users\Prem\Documents\MATLAB\mandrill.jpg');
G=rgb2gray(I);
J=imresize(G,[256 256]);
thr=mean(J(:));
[m,n]=size(J); % Reading the size of the image
for i=1:m 
    for j=1:n

      P=J(i:(i+7),j:(j+7)); % Reading 8x8 window

        % Sub dividing the 8 x 8 window into four 4x4 sub windows 

    tl = P(1:4, 1:4); % top left sub-window
    tr = P(1:4, 5:8); % top right sub-window
    bl = P(5:8, 1:4); % bottom left sub-window
    br = P(5:8, 5:8); % bottom right sub-window

     % Sub dividing the 4 x 4 window into four 2x2 sub windows 

    newtl_1 = tl(1:2, 1:2); % top left sub-window
    newtr_1 = tl(1:2, 3:4); % top right sub-window
    newbl_1 = tl(3:4, 1:2); % bottom left sub-window
    newbr_1 = tl(3:4, 3:4); % bottom right sub-window

     % Sub dividing the 4 x 4 window into four 2x2 sub windows 

    newtl_2 = tr(1:2, 1:2); % top left sub-window
    newtr_2 = tr(1:2, 3:4); % top right sub-window
    newbl_2 = tr(3:4, 1:2); % bottom left sub-window
    newbr_2 = tr(3:4, 3:4); % bottom right sub-window

     % Sub dividing the 4 x 4 window into four 2x2 sub windows 

    newtl_3 = bl(1:2, 1:2); % top left sub-window
    newtr_3 = bl(1:2, 3:4); % top right sub-window
    newbl_3 = bl(3:4, 1:2); % bottom left sub-window
    newbr_3 = bl(3:4, 3:4); % bottom right sub-window

     % Sub dividing the 4 x 4 window into four 2x2 sub windows 

    newtl_4 = br(1:2, 1:2); % top left sub-window
    newtr_4 = br(1:2, 3:4); % top right sub-window
    newbl_4 = br(3:4, 1:2); % bottom left sub-window
    newbr_4 = br(3:4, 3:4); % bottom right sub-window

      % median values of the four sub windows

        m1=mean(newtl_1(:));
        m2=mean(newtr_1(:));
        m3=mean(newbl_1(:));
        m4=mean(newbr_1(:));

         % median values of the four sub windows

        m5=mean(newtl_2(:));
        m6=mean(newtr_2(:));
        m7=mean(newbl_2(:));
        m8=mean(newbr_2(:));

         % median values of the four sub windows

        m9=mean(newtl_3(:));
        m10=mean(newtr_3(:));
        m11=mean(newbl_3(:));
        m12=mean(newbr_3(:));

        % median values of the four sub windows

        m13=mean(newtl_4(:));
        m14=mean(newtr_4(:));
        m15=mean(newbl_4(:));
        m16=mean(newbr_4(:));

     M=[m1 m2 m3 m4; m5 m6 m7 m8; m9 m10 m11 m12; m13 m14 m15 m16];

     %X=bsxfun(@ge,M,thr)
    end
end
    imshow(M)

2 个答案:

答案 0 :(得分:0)

  1. 当循环中的i从1变为图像大小时,i+7将最终对应于超出图像大小的值。调整循环的限制(到imageSize-7),或在运行代码之前填充图像。

  2. 如果要创建一个具有大于或等于阈值的数组的数组,否则为零,则编写X = M>thr。这是Matlab的一个很好的功能。

  3. 请注意,mean取平均值,median取中位数。

答案 1 :(得分:0)

最后,我在matlab中使用blockproc函数完成了它。点击下面的链接查看代码。

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