How to multiply each element with next one in the same row, then sum the product

I have a matrix with (2^N,N) dimensions, with N be any integer number, and I want to generate a new matrix with (m,s)=(2^N,2) dimensions, where m is just the sum of the elements for each row, and s is the sum of product of each element with its neighbor in the same row. for example

A=[-1 -1;-1 1;1 -1;1 1]

B=(m,s)=[-2 1;0 -1, 0 -1;2 1]

thank you very much

Answer 1

[Arif Hoq]

according to your example, i think s would be only product of each element of a row.

N=2;
AA=randi(10,2^N,3);  % random matrix generation
m=sum(AA(:,:),2);  % sum of the elements for each row
s=prod(AA(:,:),2);  % product of each element
matrix=[m s]   

matrix = 4×2

18 150 23 420 13 35 11 21

[Arif Hoq]

checking your example:

A=[-1 -1;-1 1;1 -1;1 1]  % first matrix

A = 4×2

-1 -1 -1 1 1 -1 1 1

m=sum(A(:,:),2);
s=prod(A(:,:),2);
matrix=[m s]

matrix = 4×2

-2 1 0 -1 0 -1 2 1

same as B Matrix.

[Adil Najim]

thank you very much, the m value is ok, but s is not, when A is more than two columns, where s must be the sum of product of each element with its neighbor, so for example if the fisrt row of A is

-1 -1 -1 -1

the value must be

m= - 4

s=(-1)*(-1)+(-1)*(-1)+(-1)*(-1)= 3

if the second row of A is

1 -1 -1 -1

then

m= - 2

s=(1)*(-1)+(-1)*(-1)+(-1)*(-1)= 1

and so on, thank you for your help.

[Arif Hoq]

A=[1 2 3 4;4 3 2 1;1 3 4 2];
m=sum(A(:,:),2);
C=cell(3,3);
for i=1:1:size(A,2)
        if i==4
        break
    end
    C{i,1}=A(1,i)*A(1,i+1);
    C{i,2}=A(2,i)*A(2,i+1);
    C{i,3}=A(3,i)*A(3,i+1);
end
matrix=[C{:}];
s=sum(reshape(matrix,3,3))';
output=[m s]

output = 3×2

10 20 10 20 10 23

[Adil Najim]

this is ok, thank you, but how to generalize it for N by M matrix?

[Jan]

Creating a cell array is an expensive indirection usually. You can store the elements directly in a numerical matrix, which is much faster.

You can simplify:

for i=1:1:size(A,2)
   if i==4
      break
   end 

to

for i = 1:size(A,2) - 1

With collecting the values in a numerical vector:

m = sum(A, 2);
s = 0;
for i = 1:size(A, 2) - 1
   s = s + A(:,i) .* A(:, i+1);
end
output = [m, s];

Or omit even the loop - see my answer.

[Arif Hoq]

Thanks @Jan. even I did not think about that

Answer 2

[Jan]

A  = [-1 -1; -1 1; 1 -1; 1 1];
nA = size(A, 2);
B  = [sum(A, 2), sum(A(:, 1:nA-1) .* A(:, 2:nA), 2)]

B = 4×2

-2 1 0 -1 0 -1 2 1

The 2nd column of B is the direct "word to word" translation to Matlab of: "sum of product of each element with its neighbor in the same row".

[Adil Najim]

N is dummy index, may take any value, and the B matrix must be has 2^N rows and just 2 columns

B=[m s]

the value of m is ok, but the value of s is still false

N=3;

C = cell(1,N);

[C{:}] = ndgrid([-1,1]);

C = cellfun(@(a)a(:),C,'Uni',0);

A = [C{:}]

nA = size(A, 2);

B = [sum(A, 2), A(:, 1:nA-1) .* A(:, 2:nA)]

the result must be

A =

-1 -1 -1

1 -1 -1

-1 1 -1

1 1 -1

-1 -1 1

1 -1 1

-1 1 1

1 1 1

B =

-3 2

-1 0

-1 -2

1 0

-1 0

1 -2

1 0

3 2

[Arif Hoq]

 A=[
    -1    -1    -1
     1    -1    -1
    -1     1    -1
     1     1    -1
    -1    -1     1
     1    -1     1
    -1     1     1
     1     1     1];

N=size(A,1);
m=sum(A(:,:),2);
C=cell(N,size(A,2));
for i=1:1:N
   for  j=1:1:size(A,2)
        if j==3
        break
    end
    C{i,j}=A(i,j)*A(i,j+1);
    C{i,j}=A(i,j)*A(i,j+1);
    C{i,j}=A(i,j)*A(i,j+1);
   end
end
matrix=[C{:}];
s=sum(reshape(matrix,N,[]),2);
output=[m s]

output = 8×2

-3 2 -1 0 -1 -2 1 0 -1 0 1 -2 1 0 3 2

[Jan]

"but the value of s is still false" - no, it is not.

Did you see, that there is another "sum()" in my answer?

N=3;
C = cell(1,N);
[C{:}] = ndgrid([-1,1]);
C = cellfun(@(a)a(:),C,'Uni',0);
A = [C{:}]

A = 8×3

-1 -1 -1 1 -1 -1 -1 1 -1 1 1 -1 -1 -1 1 1 -1 1 -1 1 1 1 1 1

nA = size(A, 2);
% B  = [sum(A, 2), A(:, 1:nA-1) .* A(:, 2:nA)]  Nope
B  = [sum(A, 2), sum(A(:, 1:nA-1) .* A(:, 2:nA), 2)]

B = 8×2

-3 2 -1 0 -1 -2 1 0 -1 0 1 -2 1 0 3 2

Exactly as wanted in an efficient one-liner (if you use end instead of nA).

[Adil Najim]

Thank you, very much , it is done

[Jan]

You are welcome.

It is interesting for learning Matlab to compare the loop with the vectorized version:

% Loop:
m = sum(A, 2);
s = 0;
for i = 1:size(A, 2) - 1
   s = s + A(:,i) .* A(:, i+1);
end

% Vectorized:
m = sum(A, 2);
s = sum(A(:, 1:nA-1) .* A(:, 2:nA), 2);