Borrar filtros
Borrar filtros

Adding two arrays of different sizes together evenly without messing cumulative sum.

12 visualizaciones (últimos 30 días)
I have 2 arrays.
A = [0 1 2 3 4 5 6 . . . . n] % Size 2402 x 1, assume in this case n number is 58
B = [18] % Size 1x1
C = [Cumulative output of A+B] % expected result starting from 0 and ending at 76 @ size of 2402 *1
Now
A is a cummulative sum of energy consumed starting from 0 and ending at 58 kWh during 2402 seconds.
B is total auxiliary energy consumed during 2402 seconds. Please note B is not a cumulative energy. Its just a number in 1 x 1 array.
What I am trying to do is to add B to A evenly such that the result C can be a cumulative result starting from 0 and ending at 76.
I tried a lot of logics but nothing worked for me.
Kindly guide me in correct direction to acheive this result.

Respuesta aceptada

Akira Agata
Akira Agata el 24 de Oct. de 2019
Based on the question, C should be a cumulative result starting from 0. So, it should be:
C = A + linspace(0,B,length(A));
  2 comentarios
Shahab Khan
Shahab Khan el 24 de Oct. de 2019
Hi, Thanks its works. However, the C array is in the form of 2402 x 2402. I actually need it in the form of 2402 x 1 or 1 x 2402.
Shahab Khan
Shahab Khan el 24 de Oct. de 2019
Editada: Shahab Khan el 24 de Oct. de 2019
Ok I managed to do it myself in this maner.
First I tried to understand linespace function. When I understood how it works. I write this code to acheive the above goal.
A = [0 1 2 3 4 . . . . n]; % Size 2402 x 1
B = [18];
D = linspace(0,B,length(A));
E = D';
C = A + E; % Size 2402 x 1
Thanks for the help.

Iniciar sesión para comentar.

Más respuestas (1)

Walter Roberson
Walter Roberson el 24 de Oct. de 2019
Editada: Walter Roberson el 24 de Oct. de 2019
C = cumsum(A + B / length(A));

Categorías

Más información sobre Matrix Indexing en Help Center y File Exchange.

Productos


Versión

R2019a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by