Rearrange matrix terms is a challenge

조회 수: 1(최근 30일)
Roger Breton
Roger Breton 2022년 5월 3일
편집: Roger Breton 2022년 5월 3일
I am not finding exactly the right parameters to rearrange the terms of a matrix? It starts off as an RGB image I later convert to Lab, and want to store the Lab values to a text file in a particular rows x columns order, such that I end up with 49 rows by 33 columns. This is the code I have so far :
img = imread('Image 8-bit aRGBD65.tif');
imshow(img);
img_double = im2double(img);
img2Lab = rgb2lab(img_double, 'ColorSpace','adobe-rgb-1998', 'WhitePoint','d50');
LabIMG = reshape(permute(img2Lab,[3 1 2]),3,[]); % 3 x 1617
The LabIMG give mes a 3 rows x 1617 columns like this :
What I would like to have is for the data to be rearranged this way :
A1 = 42.29 66.60 0.69
B1 = some data (not shown)
C1 = some other data (not shown either)
... up to column 49 (AW1)
Then, the second row would follow with :
A2 = 47.57 58.15 1.54
B2 = ...
B3 = ...
.... Up to BW1
Third row :
C1 = 53.84 46.90 -0.31
C2 = ...
C3 = ...
... Up to CW1
Until I have 33 rows by 49 data points. I tried Transpose to no avail :
Transposed_CIELab = transpose(LabIMG);
I enclosed the resultant text file.... in case
I'm going in circles... Any help is appreciated.
  댓글 수: 5
Roger Breton
Roger Breton 2022년 5월 3일
I was coming to a solution on my own, using a loop. It's not elegant :
LabIMG = reshape(permute(img2Lab,[3 1 2]),3,[]); % 3 x 1617
% Note: Les données sont en ordre de Colonnes dans LabIMG
Position = 1;
RowStep = 1;
for i = 1:1617
Index(i) = Position;
Position = Position + 49;
reste = rem(i,33);
if (reste == 0)
RowStep = RowStep + 1;
Position = RowStep;
end
end
IndexColumn = transpose(Index);
CIE_L = transpose(LabIMG(1,:,:)); % --> k'th slice along x direction
CIE_a = transpose(LabIMG(2,:,:)); % --> k'th slice along y direction
CIE_b = transpose(LabIMG(3,:,:)); % --> k'th slice along z direction
Final = [IndexColumn, CIE_L, CIE_a, CIE_b];
Let me digest your comment...
At this stage, I'm going to read on Matlab "sorting". We'll see where it gets me but glad I thought about the remainder function, so that I can test the position in the row and determine whether I need to break it or not on the base of remainder. Sigh! ...

댓글을 달려면 로그인하십시오.

답변(0개)

제품


릴리스

R2021a

Community Treasure Hunt

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

Start Hunting!

Translated by