數位影像處理–空間變換

上次講了數位影像處理的一題，今天再貼一題

Geometric transform (test image: fig3.tif)

Develope geometric transform program that will rotate, translate, and scale an imageby specified amounts, using the nearest neighbor and bilinear interpolationmethods, respectively.

 

背景

在對映像進行空間變換的過程中，典型的情況是在對映像進行放大,旋轉處理的時候，映像會出現失真的現象。這是由於在變換之後的映像中，存在著一些變換之前的映像中沒有的像素位置。處理這一問題的方法被稱為映像灰階級插值。常用的插值方式有三種：最近鄰域插值、雙線性插值、雙立方插值。理論上來講，最近鄰域插值的效果最差，雙立方插值的效果最好，雙線性插值的效果介於兩者之間。不過對於要求不是非常嚴格的映像插值而言，使用雙線性插值通常就足夠了。

最近領域演算法Matlab代碼

sourcePic=imread('fig3.tif');%以下為了彩色映像%[m,n,o]=size(sourcePic);%grayPic=rgb2gray(sourcePic);grayPic=sourcePic;[m,n]=size(grayPic);%比例係數為0.2-5.0K = str2double(inputdlg('請輸入比例係數(0.2 - 5.0)', '輸入比例係數', 1, {'0.5'}));%驗證範圍if (K < 0.2) && (K > 5.0)    errordlg('比例係數不在0.2 - 5.0範圍內', '錯誤');    error('請輸入比例係數(0.2 - 5.0)');endfigure;imshow(grayPic);width = K * m;                     height = K * n;resultPic = uint8(zeros(width,height));widthScale = m/width;heightScale = n/height;for x = 5:width - 5                              for y = 5:height - 5       xx = x * widthScale;                           yy = y * heightScale;       if (xx/double(uint16(xx)) == 1.0) && (yy/double(uint16(yy)) == 1.0)       % if xx and yy is integer,then J(x,y) <- I(x,y)           resultPic(x,y) = grayPic(int16(xx),int16(yy));       else                                     % xx or yy is not integer           a = double(round(xx));               % (a,b) is the base-dot           b = double(round(yy));           resultPic(x,y) = grayPic(a,b);                     % calculate J(x,y)       end    endendfigure;rotate(resultPic,-20);imshow(resultPic);

雙線性插值Matlab演算法

sourcePic=imread('fig3.tif');%以下為了彩色映像%[m,n,o]=size(sourcePic);%grayPic=rgb2gray(sourcePic);grayPic=sourcePic;[m,n]=size(grayPic);%比例係數為0.2-5.0K = str2double(inputdlg('請輸入比例係數(0.2 - 5.0)', '輸入比例係數', 1, {'0.5'}));%驗證範圍if (K < 0.2) or (K > 5.0)    errordlg('比例係數不在0.2 - 5.0範圍內', '錯誤');    error('請輸入比例係數(0.2 - 5.0)');endfigure;imshow(grayPic);%輸出圖片長寬width = K * m;                          height = K * n;resultPic = uint8(zeros(width,height));widthScale = n/width;heightScale = m/height;for x = 5:width-5                                for y = 5:height-5       xx = x * widthScale;                           yy = y * heightScale;       if (xx/double(uint16(xx)) == 1.0) && (yy/double(uint16(yy)) == 1.0)       % if xx and yy is integer,then J(x,y) <- I(x,y)           resultPic(x,y) = grayPic(int16(xx),int16(yy));       else                                           % xx or yy is not integer           a = double(uint16(xx));                    % (a,b) is the base-dot           b = double(uint16(yy));           x11 = double(grayPic(a,b));                % x11 <- I(a,b)           x12 = double(grayPic(a,b+1));              % x12 <- I(a,b+1)           x21 = double(grayPic(a+1,b));              % x21 <- I(a+1,b)           x22 = double(grayPic(a+1,b+1));            % x22 <- I(a+1,b+1)                     resultPic(x,y) = uint8( (b+1-yy) * ((xx-a)*x21 + (a+1-xx)*x11) + (yy-b) * ((xx-a)*x22 +(a+1-xx) * x12) );       end    endendfigure;resultPic = imrotate(resultPic,-20);imshow(resultPic);

效果如下

最近領域演算法放大2倍並順時針旋轉20度

雙線性插值演算法放大2倍並順時針旋轉20度

體會

該實驗表明雙線性插值得到的映像效果是比較好的。能夠避免採用最近領域插值方式時可能存在的映像模糊、塊狀失真等問題。但雙線性插值也存在問題，在放大倍數比較高的時候，映像失真將會比較嚴重，此時應該考慮使用更高階的插值演算法。