Generate new matrix from old matrix with specific conditions

Hello,

I have matrix R(100*100) and I want to generate new matrix A(100*100) as below conditions

If I have a frequency to add a new condition(let us say frequency = 20). This mean I will have five conditions inside my domain `(100/20)=5

From 1 to 20, from 21 to 40, from 41 to 60, from 61 to 80, and from 81 to 100.

Let us assume my original matrix is

 [1 1 1……..1]

 [2 2 2 …….2]

 [3 3 3 ………3]

And so on

 [100 100 100 …..100]

The new matrix must be

 From 1 to 20 same as old matrix

 [1 1 1……..1]

 [2 2 2 …….2]

 [3 3 3………3]

 From 21 to 40

 A(21,j)=R(21,j)+R(1,j)   (row 21 + row 1)

 A(22,j)=R(22,j)+R(2,j)   (row 22 + row 2)

 A(23,j)=R(23,j)+R(3,j)   (row 23 + row 3)

 And so on

 From 41 to 60

 A(41,j)=R(41,j)+R(21,j) +R(1,j)  (row 41+row 21 + row 1)

 A(42,j)=R(42,j)+R(22,j) +R(2,j)  (row 42+row 22 + row 2)

 And so on
 A(81,j)=R(81,j)+R(61,j)+R(41,j)+R(21,j)+R(1,j)  (row 81+row 61 + row 41 + row 21+ row 1)

At each time that I reach to the frequency, I will have a new condition. My question is there any sufficient way to do that? I wrote my code and its work fine but each time I change frequency I will have a new condition. For the case above I have 5 conditions but if I use frequency 5 I will have 20 conditions and I need to change all equations. I mean I need code to be able to handle any frequency

I wrote below code

 clc;
clear;
prompt = 'Enter Frequency=';       %Frequency=20
N= input(prompt);
Frequency=N;
one_step=1/Frequency;              %Frequency
time=Frequency*one_step;
number_of_steps=time/one_step;     %Number of steps until next condition
total_steps=100;
R1 = rand(100,100);
Number_of_Lines=(total_steps/number_of_steps)+1;  %100/20
A(100,100)=0;
    for i=1:100
    for j=1:100
        if i>=1 && i<=number_of_steps
            A(i,j)=R1(i,j);
        elseif i>number_of_steps && i<= 2*number_of_steps
                A(i,j)=R1(i,j)+R1(i-number_of_steps,j);
        elseif i>2*number_of_steps && i<= 3*number_of_steps
                A(i,j)=R1(i,j)+R1(i-number_of_steps,j)+R1(i-2*number_of_steps,j);
        elseif i>3*number_of_steps && i<= 4*number_of_steps
                A(i,j)=R1(i,j)+R1(i-number_of_steps,j)+R1(i-2*number_of_steps,j)...
                    +R1(i-3*number_of_steps,j);
        elseif i>4*number_of_steps && i< 5*number_of_steps
                A(i,j)=R1(i,j)+R1(i-number_of_steps,j)+R1(i-2*number_of_steps,j)...
                    +R1(i-3*number_of_steps,j)+R1(i-4*number_of_steps,j);

        end
end

 end

Answer 1

[Roger Stafford]

   [m,n] = size(R);
   f = 20; % <-- Check that f is a divisor of m
   A = reshape(cumsum(reshape(R,f,m/f,n),2),m,n);

[Ali Kareem]

Hello,

Thank you so much for your reply. Please, I just have one problem. if the division of (m/f) is not an integer. let us assume f=15. How I can handle this situation?

[Roger Stafford]

One way would be to do this:

   c = ceil(m/f);
   R2 = [R;zeros(f*c-m,n)]; % Expand R to multiple of f rows
   A = reshape(cumsum(reshape(R2,f,c,n),2),f*c,n);
   A = A(1:m,:); % Remove those extra rows