## Creating a matrix from a lagrer matrix

Asked by Benedict Egbon

### Benedict Egbon (view profile)

on 7 Mar 2019
Latest activity Commented on by Benedict Egbon

### Benedict Egbon (view profile)

on 7 Mar 2019
Accepted Answer by Akira Agata

### Akira Agata (view profile)

Hello everyone.
Please I need help in automatically dividing the matrix A below into three 5by 1 matrix.
A=[1
2
3
4
5
6
8
9
10
11
12
13
14
15]
Such that the first matrix created will be say
B=[1
2
3
4
5]
Thanks.

### Akira Agata (view profile)

Answer by Akira Agata

### Akira Agata (view profile)

on 7 Mar 2019

How about the following?
In this case, C{1}, C{2}, ..., C{151} are 200-by-1 matrices each, extracted from original 30194-by-1 matrix.
% Sample 30194-by-1 data
A = rand(30194,1);
% Padding with "nan"
n = 200;
N = numel(A);
A = [A;nan(ceil(N/n)*n-N,1)];
% Divide A into 200-by-1 segments
C = mat2cell(A,repelem(n,ceil(N/n)));

### madhan ravi (view profile)

on 7 Mar 2019
reshape() is much faster
Benedict Egbon

### Benedict Egbon (view profile)

on 7 Mar 2019
Hi Akira.
Thanks for the respone.
This actually worked well for me as desired.
Regards.

### Kevin Phung (view profile)

Answer by Kevin Phung

### Kevin Phung (view profile)

on 7 Mar 2019

A = reshape(1:15,5,3); %reshaped into a 5x3 matrix.
B =A(:,1); %one of the column vectors that you desire
reading up on indexing can help:

Benedict Egbon

### Benedict Egbon (view profile)

on 7 Mar 2019
Thank you very much for the response as this worked very well for this case.
At the moment, I have a 30194 by 1 matrix, and I want to make many 200 by 1 matrices from it.
Any suggestion on how to automate it, instead of doing B =A(:,1), C =A(:,2) etc.
Thanks.