I have a 5x6 matrix. I want to calculate sum of only those values which are consecutive nonzero along each row. How to do that? The desired result given in description box.
3 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Payel
el 24 de Jun. de 2023
Comentada: Image Analyst
el 24 de Jun. de 2023
a = [2 3 1 4 0 0;
0 0 0 0 0 0;
0 0 1 5 7 0;
0 0 0 0 4 0;
4 5 0 0 6 8];
desired_result = [10;0;13;0;23]
4 comentarios
the cyclist
el 24 de Jun. de 2023
I believe a correct re-statement of the rule is, "Sum all non-zeros that are next to at least one other non-zero."
Respuesta aceptada
the cyclist
el 24 de Jun. de 2023
Editada: the cyclist
el 24 de Jun. de 2023
I believe this does what you want:
% Input
a = [2 3 1 4 0 0;
0 0 0 0 0 0;
0 0 1 5 7 0;
0 0 0 0 4 0;
4 5 0 0 6 8];
% Append columns of zeros at both ends, because we are going to check in
% the next line if an element is a "singleton" with zeros on both sides
b = [zeros(height(a),1) a zeros(height(a),1)];
% Identify the non-singleton value (i.e. the ones that do not have zeros on either side)
includeInSum = not((b(:,1:end-2) == 0) & (b(:,3:end)==0));
% Sum the values that should be included
result = sum(a.*includeInSum,2)
0 comentarios
Más respuestas (1)
Image Analyst
el 24 de Jun. de 2023
a = [2 3 1 4 0 0;
0 0 0 0 0 0;
0 0 1 5 7 0;
0 0 0 0 4 0;
4 5 0 0 6 8];
desired_result = [10;0;13;0;23]
desired_result = sum(a, 2)
To learn other fundamental concepts, invest 2 hours of your time here:
2 comentarios
Image Analyst
el 24 de Jun. de 2023
Row 4 has one "consecutive' value, so why not sum it in? Do you want to sum in values only if the run of non-zero values is 2 or more? See @the cyclist's answer below. Is it homework? If so you can't use it.
This looks like a homework problem. If you have any questions ask your instructor or read the link below to get started:
Obviously we can't give you the full solution because you're not allowed to turn in our code as your own.
Ver también
Categorías
Más información sobre Loops and Conditional Statements en Help Center y File Exchange.
Productos
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!