satellite image pre-processing

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Booker
Booker 2012 年 1 月 22 日
Hello,
I have a satellite image with 365 bands. I intend to get the mean of every 8 bands(that is, (b1+b2+b3+b4+b5+b6+b7+b8)/8, then (b9+b10+b11....b16)/8, and so on? does somebody have a code for this?
Thanks, Booker.

採用された回答

Bjorn Gustavsson
Bjorn Gustavsson 2012 年 1 月 23 日
Brute looping of this kind of problem shouldn't be too bad.
sz = size(SImg);szMean = sz;
szMean(3) = ceil(szMean(3)/8);
BandMean = zeros(szMean);
for i1 = 1:szMean(3)-1,
BandMean(:,:,i1) = mean(SImg(:,:,8*(i1-1)+[1:8]),3);
end
BandMean(:,:,end) = mean(SImg(:,:,(8*(i1)+1):end),3);
HTH,

その他の回答 (1 件)

Walter Roberson
Walter Roberson 2012 年 1 月 22 日
Some image file formats are able to store multiple bands per image, and imread() can read all the bands in at the same time, either as an X by Y by 3 by Bands or X by Y by Bands array (depending on how the data is stored in the image file.)
You can
s = size(ImageMatrix);
By8 = reshape(ImageMatrix, [s(1:end-1), 8, s(end)/8]);
nd = ndims(By8);
mean(By8,nd)
However, you will have a problem because the number of bands you have is not evenly divisible by 8. How do you want to handle the final 5 bands?
  1 件のコメント
Booker
Booker 2012 年 1 月 23 日
I increased the bands of my image to 368 so should be divisible by 8, but the answer is not correct. I am getting a image with 8 bands (i.e. 55,97,8) at the end instead of 46 bands? as in (55,97,46).

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