## inserting rows in a matrix

### Ali Ekhtiari (view profile)

on 3 Sep 2019
Latest activity Commented on by Andrei Bobrov

### Andrei Bobrov (view profile)

on 4 Sep 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

### Andrei Bobrov (view profile)

on 3 Sep 2019

kron(yourmatrix(:),[1;zeros(24,1)]);

### Steven Lord (view profile)

on 3 Sep 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.

Andrei Bobrov

on 4 Sep 2019
+1

### Walter Roberson (view profile)

on 3 Sep 2019

reshape([YourMatrix.'; zeros(24, 365)], [], 1)

on 3 Sep 2019

on 4 Sep 2019

Wanted=zeros(365*25,1);
Wanted(1:25:end) = yourmatrix

Walter Roberson

### Walter Roberson (view profile)

on 3 Sep 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.