Accesing data in a three dimensional matrix using a for loop

Hi, I want to create a for loop to go through each of the pages (20) of a matrix 400x600x20 (called polygons). The matrices 400x600 are binary matrices with ones(1) and zeros(0), with different polygons in each of the 20 pages. I have an original matrix (called main01) of 400x600 from where I want to use the data to copy the ones(1) and zeros(0) and locate it in the polygons in each of the 20 pages of the other variable. So I try this creating a variable of only the first page of the matrix polygons:

polygon1(polygon1==1)=main01(polygon1==1);

And it works very well. However, when I set this into a for loop it doesn't give me the zeros and ones of the main01 matrix, it just return the same matrix as polygons.

count=20;
  for u=1:1:count;
     if polygons(:,:,u)==1
        polygons=main01(polygons(:,:,u)==1;
     end
  end

I will appreciate some help with this for loop.

Thank you. Martha

Answer 1

[Guillaume]

First off, the big error in your code is:

if polygons(:,:,u)==1

This will only be true when polygons(:,:,u) is all ones. Furthermore, I'm not sure what you were trying to achieve with this if.

A simple way to do what you want would be to use a temporary variable to store the page you're working on:

count = 20;
for u=1:count  %don't need semicolon at the end. don't need increment when it is 1.
   polygon = polygons(:,:,u); %temporary 2d variable
   polygon(polygon == 1) = main01(polygon == 1); %as you wrote
   polygons(:,:,u) = polygon; %update page with temporary
end

But actually, the operation you're doing is a simple logical AND, thus you can simplify the code with:

for u=1:count  
   polygons(:,:,u) = polygons(:,:, u) & main01; 
end

If the operation you're trying to perform between polygons and main01 are actually more complex, I would change a few things. First, I would declare polygons and main01 as logical. That way you don't have to do the == 1 anymore.

Secondly, instead of storing the polygons as a 3d matrix, I would store them as a cell array of 2d matrix. If I rewrite my initial example that way:

count = 20;
%first two lines just convert existing variables. Better to declare them
%that way in the first place
polygons = arrayfun(@(u) logical(polygons(:,:,u), 1:count, 'uni', false); %polygons is a cell array of 2d logical matrices
main01 = logical(main01);
for u=1:count
  polygons{u}(polygons{u}) = main01(polygons{u});
  %or:
  %polygons{u} = polygons{u} & main01;
end

[Guillaume]

Also, just realised you can also avoid the loop altogether,

for u=1:count  
 polygons(:,:,u) = polygons(:,:, u) & main01; 
end

is equivalent to:

polygons = bsxfun(@and, polygons, main01);

[Martha]

Thank you very much Guillaume, I tried all the three ways and it works very well.. I will give it a go creating a cell array as well.