# Add zeros to matrices same as length of another matrix

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I have to determine the beta weights of an fmri data subset. To do this I think I should perform a multiple regression of the bold signal of the voxels with the Design matrix (convolved with the hrf) - please correct me if I'm wrong in this.

The problem is the bold values are in a 360x3 matrix and the Design is 369x5. This is probably a very long way about it but to be able to regress, I separated each bold voxel into a separate vector and tried to pad the end with zeros which gives me an error everytime. How do I change this code to work for matrices? Or can I add zeros to the bold 360x3 matrix to make it 369x3 in one go?

bold1=[bold1, zeros(1,length(X(:,1))-length(bold1))];

##### 2 Comments

Jos (10584)
on 15 Mar 2018

### Accepted Answer

KL
on 15 Mar 2018

Edited: KL
on 15 Mar 2018

can I add zeros to the bold 360x3 matrix to make it 369x3 in one go?

You say you want to add 9 rows but you are trying to add columns with your code. Probably something like,

bold1=[bold1; zeros(size(X(:,1),1)-size(bold1,1),3)];

...Dimensions of matrices being concatenated are not consistent...

This simply means what it says. For example,

>> A = [1 2;3 4]

A =

1 2

3 4

>> B = [5 6 7; 8 9 0]

B =

5 6 7

8 9 0

>> C = [A; B]

Error using vertcat

Dimensions of matrices being concatenated are not consistent.

You cannot stack them up and down (vertical cancatenation, hence vertcat) because they have different number of columns (hence not consistent!).

But if you want to stack them side by side (horizontal)

>> C = [A B]

C =

1 2 5 6 7

3 4 8 9 0

Now you can, because they have same number of rows. Simple!

##### 5 Comments

Steven Lord
on 15 Mar 2018

You can simplify this:

size(X(:,1),1)

X(:, 1) has the same number of rows as X and 1 column. So size(X(:, 1), 1) returns the number of rows in X. Why go to the trouble of extracting that first column? Just ask for the size of X in the first dimension.

size(X, 1)

### More Answers (2)

Jos (10584)
on 15 Mar 2018

To pad an matrix A with zeros to match a larger or same-sized array B, you can use this:

A = magic(3)

B = ones(3, 5)

newA = zeros(size(B))

newA(1:size(A,1), 1:size(A,2)) = A

If you are sure the new dimensions are larger, this will also work:

A2 = magic(4)

B = ones(5,7) % all dimensions larger than A

A2(size(B,1), size(B,2)) = 0

##### 0 Comments

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