Moving a greyscale image such that the lowest pixel with an intensity above 0 is at the bottom of the image.

4 ビュー (過去 30 日間)
Cyrus N
Cyrus N 2017 年 6 月 28 日
回答済み: Image Analyst 2017 年 6 月 29 日
I'm trying to translate a greyscale image such that the bottommost pixel with an intensity>0 is at the bottom of the pic, I already have it such that the left most pixel . My code for translating is as follows: lungc_ref = imref2d(size(c)) [row, column] = find(c, 1, 'first'); [row2 column2] = find(c, 1, 'last'); t = [1 0 0; 0 1 0; -(column-1) (256-row2) 1]; tform = affine2d(t); imwarp(c, tform); [c_translated,lungc_ref] = imwarp(c,tform,'OutputView',lungc_ref); imshow(c_translated, [0 1000])

採用された回答

Massimo Zanetti
Massimo Zanetti 2017 年 6 月 29 日
編集済み: Massimo Zanetti 2017 年 6 月 29 日
Use the circshift function, which shifts arrays along one dimension. Here is an example:
%generate an image (matrix)
A = rand(6,9)
%find linear index of the minimum value
[~,ix] = min(A(:));
%convert linear index into subscript index
[r,c] = ind2sub(size(A),ix)
%shift image rows to put the lowest valued pixel at the bottom
Y = circshift(A,size(A,1)-r,1)
%additionally, you can also shift image columns to put the lowest valued pixel at the left
Y = circshift(Y,size(A,2)-c+1,2)
Check it out.

その他の回答 (2 件)

Cyrus N
Cyrus N 2017 年 6 月 29 日
Alternatively I found that lowestRow = find(sum(one, 2)>0, 1, 'last'); also seems to work for finding the bottom row after which you can subtract the value for lowestRow from the image dimension to affix the bottom row of the grey image to the bottom row of the figure.

Image Analyst
Image Analyst 2017 年 6 月 29 日
Simply figure out how much to move the image, then call imtranslate(). It's different than circshift in that it doesn't wrap the image around to the other side.
Description
B = imtranslate(A,translation) translates image A by the translation vector specified in translation. If A has more than two dimensions and translation is a two-element vector, imtranslate applies a 2-D translation to A, one plane at a time.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by