我试图过滤使用的图像imfilter.我应该得到相同的结果,imfilter但我一直得到不同的结果.谁能告诉我哪里出错了?
orignal=imread('obj6__17.png');
filter=1/9*[-1 -1 -1 ; -1 17 -1 ; -1 -1 -1];
s=size(orignal);
r=zeros(s(1));
temp = zeros(3);
for i= 2: s(1)-1
for j = 2: s(2)-1
for n= 1: 3
for m= 1:3
temp(n,m)=orignal(i+2-n,j+2-m)*filter(n,m);
end
end
r(i,j)=sum(single(sum(temp)));
end
end
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大小r应该与我认为的原始大小相同.我不明白为什么你转换为单精度使用single.无论如何,我想你想做以下事情:
%# Let's first create a small test image from the built-in peppers image
original = im2double(imread('peppers.png'));
original = original(1:5,1:8,1);
filter = 1/9 * [-1 -1 -1 ; -1 17 -1 ; -1 -1 -1];
s = size(original);
r = zeros(s);
for i = 2:s(1)-1
for j = 2:s(2)-1
temp = original(i-1:i+1,j-1:j+1) .* filter;
r(i,j) = sum(temp(:));
end
end
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结果如下:
r =
0 0 0 0 0 0 0 0
0 0.2336 0.2157 0.2514 0.2436 0.2257 0.2344 0
0 0.2453 0.2444 0.2671 0.2693 0.2418 0.2240 0
0 0.2741 0.2728 0.2397 0.2505 0.2375 0.2436 0
0 0 0 0 0 0 0 0
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并且imfilter,它是:
r2 = imfilter(original, filter)
r2 =
0.3778 0.3325 0.3307 0.3442 0.3516 0.3312 0.3163 0.3856
0.3298 0.2336 0.2157 0.2514 0.2436 0.2257 0.2344 0.3386
0.3434 0.2453 0.2444 0.2671 0.2693 0.2418 0.2240 0.3512
0.3272 0.2741 0.2728 0.2397 0.2505 0.2375 0.2436 0.3643
0.3830 0.3181 0.3329 0.3403 0.3508 0.3272 0.3412 0.4035
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如您所见,除边框上的结果外,结果相同.有一些策略可以计算边界上的图像,将图像镜像到边界外,保持相同,等等.请阅读文档imfilter并选择一种策略.
请注意,filter由于滤波器在两个方向上都是对称的,所以我没有在这里翻转.我建议你避免循环!代码中有深度为4的嵌套循环!
最后,您可以使用二维卷积执行相同的操作imfilter:
r3 = conv2(original, filter, 'same');
r3 =
0.3778 0.3325 0.3307 0.3442 0.3516 0.3312 0.3163 0.3856
0.3298 0.2336 0.2157 0.2514 0.2436 0.2257 0.2344 0.3386
0.3434 0.2453 0.2444 0.2671 0.2693 0.2418 0.2240 0.3512
0.3272 0.2741 0.2728 0.2397 0.2505 0.2375 0.2436 0.3643
0.3830 0.3181 0.3329 0.3403 0.3508 0.3272 0.3412 0.4035
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