inserting rows in a matrix

1 visualización (últimos 30 días)
Ali Ekhtiari
Ali Ekhtiari el 3 de Sept. de 2019
Comentada: Andrei Bobrov el 4 de Sept. de 2019
The matrix has 365 rwos and one column(365*1).
I want to add 24 zero rows below the every row. If I want to explain more, I would say, I have daily average of a year (365 days), then I want to change this 365 days to 8760 rwos to put each number with 24 rows distance in between of each number in new matrix.
How can I do this?
Thanks

Respuestas (4)

Andrei Bobrov
Andrei Bobrov el 3 de Sept. de 2019
kron(yourmatrix(:),[1;zeros(24,1)]);

Steven Lord
Steven Lord el 3 de Sept. de 2019
Are you trying to turn daily data into hourly data? If so, consider making datetime vectors for each day and each hour and passing those (along with your daily data) into interp1, like the "Interpolation of Dates and Times" example on the interp1 documentation page shows. Alternately if you're storing your data in a timetable call retime on it.

Walter Roberson
Walter Roberson el 3 de Sept. de 2019
reshape([YourMatrix.'; zeros(24, 365)], [], 1)

madhan ravi
madhan ravi el 3 de Sept. de 2019
Editada: madhan ravi el 4 de Sept. de 2019
Wanted=zeros(365*25,1);
Wanted(1:25:end) = yourmatrix
  2 comentarios
Walter Roberson
Walter Roberson el 3 de Sept. de 2019
I think you have an off-by-one error. According to the description, they want 24 rows of zeros below each row, which would make a total of 25 for the group.
madhan ravi
madhan ravi el 4 de Sept. de 2019
Ah thank you sir Walter, I was confused when the OP mentioned 8760 and then reading it again the first line states 24 zeros rows after each ..

Iniciar sesión para comentar.

Categorías

Más información sobre Matrices and Arrays en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by