Shrinking image by averaging

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Oah Joan
Oah Joan 2018 年 10 月 17 日
回答済み: Sean de Wolski 2021 年 1 月 15 日
I want to create a function: shrinkbyaveraging(imagematrix) that returns an image that is half as large in both height and width. Each pixel in the returned image should be the average of 4 pixels from the input image. Shrinkbyaveraging only needs to work for images having an even number of rows/columns. However, I do not want to use imresize.
function shrink = shrinkbyaveraging(imagematrix) [row,colm] = size(imagematrix); x = row/2; y = colm/2;
but im not sure what the question is asking since i cannot use imresize. :/
  4 件のコメント
Rik
Rik 2020 年 11 月 20 日
編集済み: John Kelly 2021 年 1 月 15 日
Rik
Rik 2020 年 11 月 20 日
編集済み: John Kelly 2021 年 1 月 15 日
Also some deleted comments :
OP responding to Kalyan:
function shrink = shrinkbyaveraging(imagematrix) [row,colm] = size(imagematrix); x = row/2; y = colm/2;
but im not sure what the question is asking since i cannot use imresize. :/
OP responding to Image Analyst:
I am taking a first year introductory computing science course and I dont think it is that hard. But, I am stuck on it.

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回答 (2 件)

Michael Hawks
Michael Hawks 2018 年 10 月 17 日
Try this:
Nrow = 12; Ncol = 15;
input = rand(Nrow,Ncol);
shifts = [0 0; 0 1; 1 0; 0 -1];
A = zeros(Nrow,Ncol,4);
for ii = 1:4;
out(:,:,ii) = circshift( input, shifts(ii,:) );
end
out = sum( out, 3);
out = out(2:2:end,2:2:end);
In the for loop, we're building up an mxnx4 array of shifted copies of the input array. Then finding the mean over the neighborhood is as easy as averaging along dimension 3. Then the last line downsamples the array to the size you want.

Sean de Wolski
Sean de Wolski 2021 年 1 月 15 日
This works
img = repmat(1:6,4,1)
shrink = @(x)reshape(sum(reshape(sum(reshape(x,2,1,[])),size(x,1)/2,2,[]),2)./4,size(x,1)/2,[])
shrink(img)

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