Please how can I store values of variable size vectors inside a loop for

image processing

[jonas]

cell array

[Stephen23]

Probably in a cell array, or a structure, or a table, or...

But you question does not have enough information to know exactly what you are doing and what you would need.

[Mohamed ouahidi]

I would like doing a histogram partition based on local minima. so after derminated local minima points I want to take the portion between 2 points that is to say a vector of variable size each time we increment in the loop

[jonas]

@Mohamed: If you want a detailed and precise answer then you need to figure out how to formulate the question accordingly. It still sounds like cell arrays will be useful for you, but it's difficult to say more than that based on the limited information.

Take a look at this guide on how to ask questions:

https://se.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

[Stephen23]

Mohamed ouahidi's "Answer" moved here and formatted correctly:

The probleme exist in workspace the values of vectors subH(2) ....subH(n-1) are not stored

a=imread('pout.tif');
imshow(a); title('Original Image');
x = imhist(a);% vecteur d histogrm
maxHeight = max(x); % used to set limits for the histogram  
% Split the original histogram into  lenght of (idxX)
%idxX vector contains the minimas points of histogram image a 
% first sub_histogram
subH_1=x(  idxX(1):  idxX(2) );
% Display the original and  1 sub-histograms
figure, stem(0:255, x,'.'); title('Original Histogram')
axis([0 260 0 maxHeight])
figure, stem( idxX(1): idxX(2,1)  ,  subH_1, '.'); title('Sub-histogram 1');
axis([0 260 0 maxHeight])
% Display others sub-histograms
% n is known
for  k= 2 : (n-1) 
subH(k)= x(  idxX(k)  + 1   :    idxX((k+1)) ); 
%figure;
%stem( idxX(k,1): idxX((k+1),1)  ,  subHk, '.'); 
%axis([0 260 0 maxHeight])
end

[Stephen23]

@Mohamed ouahidi: what values does n have? If it is less than three then that loop will not run.

[Mohamed ouahidi]

I use image 'cameraman.tif' in this case n = 44

Answer 1

[Amir Xz]

" x( idxX(k) + 1 : idxX((k+1)) )" is a vector, so it should store in a vector too.

Use this one:

subH(k,:) = x( idxX(k) + 1 : idxX((k+1)) )';

[Mohamed ouahidi]

''Amir Xz'' thank you for your answer,

I tried but it does not work

when I declare outside the loop

 subH-2 = x( (idxX(2) + 1) : idxX(3));
°
°
 subH-43 = x( (idxX(43) + 1) : idxX(44));

it works .

[Amir Xz]

Okay, could you tell me what is "subH-2" 's size?

[m,n] = size(subH-2);

What is the m and n?

What about "subH-43"? Are all same?

[Walter Roberson]

 subH-2 = x( (idxX(2) + 1) : idxX(3));

You cannot use a minus sign as part of a variable name. You could use

subH_2 = x( (idxX(2) + 1) : idxX(3));

or

subH{2} = x( (idxX(2) + 1) : idxX(3));

[Mohamed ouahidi]

yes I mean subH_2 = x( (idxX(2) + 1) : idxX(3));

[Mohamed ouahidi]

[M2 N2]=[2 1]....[M3 N3]=[4 1]....
[M4 N4]=[3 1] ...........
[M5 N5]=[2 1].............
[M42 N42]=[4 1].............
[M43 N43]=[2 1]

.... The size of vectors is not the same

[Amir Xz]

Use cell array:

subH{k-1} = x( idxX(k) + 1 : idxX((k+1)) );

[Mohamed ouahidi]

the problem is solved Thank you so much