Find the exact multiple of a number by zero padding values.

Say

 A = 13848 x 1

and I want to reshape into a 173(matters a lot) by 80 or 81(this doesn't matter) but that isn't possible as 13848 isn't divible by 173.

So I want a way where it reads A and calculates the next perfect value which is 14013 by zero padding A.

I could just remove 8 from A but I don't loose data but I don't mind zero padding to A towards the end.

[CalebJones]

I could just manually zero pad A, but I want it to be as automated as possible.

Answer 1

[Stephen23]

Method One: create the output matrix first, then use indexing to add the values:

R = 173;
B = zeros(R,ceil(numel(A)/R));
B(1:numel(A)) = A(:);

Method Two: pad some extra zeros and then remove the superfluous elements:

R = 173;
X = 1:R*fix((1+numel(A))/R)
B = [A;zeros(R,1)];
B = reshape(B(X),R,[]);

[CalebJones]

Did you try with the matfile I attached for previous answer?

I'm getting few errors.

[CalebJones]

A = size(right_pc,1);
R = 173;
X = array2table(1:R*fix((1+A)/R))
B = [right_pc(:,1);array2table(zeros(R,1),'VariableNames',{'HbO'})];
B = reshape(B(X),R,[]);

My system just froze for like 30 seconds.

Anyways, B is of dimention 14021 which isn't divisible by 173.

14013 is the closest divisibe thing by 173 to 13848.

[Stephen23]

Why did you convert this index into a table?:

X = array2table(1:R*fix((1+A)/R))

X is supposed to be a numeric index vector (just like in my answer). Converting it into a table is unlikely to work, and just makes the code pointlessly complex.

Apparently your data are in a table, something that you omitted to mention in your question. I doubt that you have really considered what it means to reshape a table: a table is a container class which contains multiple different variables, it is not a simple array class that can be reshaped, as the variables are not one contiguous array stored in memory. In any case, to reshape you should probably get the appropriate array out of the table first:

https://www.mathworks.com/help/matlab/matlab_prog/access-data-in-a-table.html

[CalebJones]

A = size(right_pc,1);
R = 173;
X = 1:R*fix(1+A/R);
B = [right_pc.HbO ; zeros(R,1)];
B = reshape(B(X),R,[]);

Ah yes, SWEEET!

Fixed !

Thanks bud.

Answer 2

[KSSV]

A = rand(13848,1) ; 
n = length(A) ;   % length of A 
r = 80 ;          % to reshape by this dimension 

N = mod(-mod(n,r),r);   % get the number of zeros to be appended 
A = [A ; zeros(N,1)] ;  % append zeros 
iwant = reshape(A,r,[]) ; % reshape 

[Stephen23]

CalebJones's "Answer" moved here:

getting an error.

n =  size(right_pc,1) ;   % length of A 
r = 80;          % to reshape by this dimension 
N = mod(-mod(n,r),r);   % get the number of zeros to be appended 
A = [right_pc(:,1) ; zeros(N,1)] ;  % append zeros 
iwant = reshape(A,r,[]) ; % reshape 
when i tried to run...this
N = mod(-mod(n,r),r);
Index in position 1 exceeds array bounds (must not exceed 173).

I have atteched a mat file as well.

[Stephen23]

You have created a variable named mod. Clear that variable from the workspace.