How to sum up row values in a matrix?

Dear All

I have matrix A

A = [ 1 2 3 5;    
    3 4 5 4;     
   ];

I want to add row values like that using a loop ( without manual input)

A(1,1) + A(1,2) = B1 
A(1,3) + A(1,4) = B2
A(2,1) + A(2,2) = B3
A(2,3) + A(2,4) = B4
B= [ B1 B2;    
     B3 B4    
   ];

How can I do that any tips

Many Thanks in advance

[Image Analyst]

Looks like you've done it (almost) except that you need to flip your B's to the other side:

B1 = A(1,1) + A(1,2);
B2 = A(1,3) + A(1,4);
B3 = A(2,1) + A(2,2);
B4 = A(2,3) + A(2,4);
B= [ B1 B2;    
     B3 B4    
   ];

No loop needed (for such a small array). Why do you want to use a loop?

[Yuli Hartini]

Can you tell how to use a loop? Sorry for the late comment

[Image Analyst]

m = randi(4, 3, 5); % Sample data in a 3 by 5 matrix
[rows, columns] = size(m) % Get dimensions of the matrix.
% Preallocate space for the sums of the rows.
rowSums = zeros(rows, 1);
for row = 1 : rows
  % Get the sum for this row across all columns in this row.
  for col = 1 : columns
    rowSums(row) = rowSums(row) + m(row, col);
  end
  % Print the sums to the command window
  fprintf('For row #%d the sum over the columns = %f.\n', row, rowSums(row));
end

[Yuli Hartini]

What if I have matrix M, M = [1 2 0.2; 2 3 0.1; 3 4 0.4] And I want values like this.. Values= [0.2; (0.2+01); (0.2+0.1+0.4)]

[Yuli Hartini]

Help me please

[Image Analyst]

I'm not sure of your rule, but it looks like you might be doing

Values = cumsum(M(:, end))

Answer 1

[Renda Mohammedjuhar]

I have a matrix like [1 2 3 4] I want an output [1 3 6 10]

Answer 2

[Azzi Abdelmalek]

A = [ 1 2 3 5;3 4 5 4]
res=reshape(sum(reshape(A',1,2,[])),2,2)'

%or

res=A(:,[1 3])+A(:,[2 4])

%or

n=size(A,2)/2
res=[sum(A(1,1:n)) sum(A(1,n+1:end)); sum(A(2,1:n)) sum(A(2,n+1:end))]

Answer 3

[Image Analyst]

Here's one way:

A = [ 1 2 3 5;
    3 4 5 4]

% Get the sliding sum.
a2 = conv2(A, [1 1], 'valid');
% Extract just the first and last column.
output = [a2(:,1) a2(:,3)]

Answer 4

[Sayanta]

Hi Image analyst

I have bigger matrix.

A=

    0.0018    0.0008    0.0000    0.0000    0.2304    0.7345    0.0159    0.0166
    0.0024    0.0016    0.0001    0.0000    0.2161    0.7441    0.0165    0.0192
    0.0029    0.0027    0.0002    0.0000    0.2084    0.7475    0.0169    0.0214
    0.0034    0.0040    0.0003    0.0000    0.2041    0.7479    0.0172    0.0230
    0.0038    0.0055    0.0005    0.0001    0.2016    0.7468    0.0175    0.0243
    0.0041    0.0072    0.0007    0.0001    0.1999    0.7450    0.0177    0.0253
    0.0044    0.0090    0.0009    0.0001    0.1988    0.7429    0.0178    0.0261

 I want to do the operation like your code
% Get the sliding sum
a2 = conv2(A, [1 1], 'valid');

how can I do that

Here I want have to add

A(1,1) + A(1,2) + A(1,3)+ A(1,4) = B1

A(1,5) + A(1,6) + A(1,7)+ A(1,8) = B2

A(2,1) + A(2,2) + A(2,3)+ A(2,4) = B3

A(2,5) + A(2,6) + A(2,7)+ A(2,8) = B4

B = [ B1 B2

         B3 B4]

Thanks

[Azzi Abdelmalek]

B=A(1:2,:)
n=size(B,2)/2
res=reshape(sum(reshape(B',1,n,[])),2,2)'

[Image Analyst]

See my other answer. I was wondering - but usually when it needs to be general for some variable number of rows or columns, people will say that in advance so they get the general answer the first time.

[Azzi Abdelmalek]

or simpler

n=size(A,2)/2
res=[sum(A(1,1:n)) sum(A(1,n+1:end)); sum(A(2,1:n)) sum(A(2,n+1:end))]

[Image Analyst]

Yeah, that's probably better - more direct - as long as he has a 2 row array. In his example here (which he incorrectly posted as an answer), he has a 7 row by 8 column array. See my build on your solution for when it has any number of rows.

Answer 5

[Image Analyst]

A=[...
    0.0018    0.0008    0.0000    0.0000    0.2304    0.7345    0.0159    0.0166
    0.0024    0.0016    0.0001    0.0000    0.2161    0.7441    0.0165    0.0192
    0.0029    0.0027    0.0002    0.0000    0.2084    0.7475    0.0169    0.0214
    0.0034    0.0040    0.0003    0.0000    0.2041    0.7479    0.0172    0.0230
    0.0038    0.0055    0.0005    0.0001    0.2016    0.7468    0.0175    0.0243
    0.0041    0.0072    0.0007    0.0001    0.1999    0.7450    0.0177    0.0253
    0.0044    0.0090    0.0009    0.0001    0.1988    0.7429    0.0178    0.0261]

[rows columns] = size(A)
% Get the sliding sum
a2 = conv2(A, ones(1, columns/2), 'valid')
% Extract just the first and last column.
B = [a2(:,1) a2(:,end)]

Or, building off Azzi's solution and making it work for a 2D array of any number of rows:

B = [sum(A(:,1:columns/2), 2) sum(A(:,(columns/2)+1:end), 2)]

This is probably the most direct way. And it's only 1 line of code instead of 2.