How to combine 2 vectors of unequal length

Hi there

If I have:

A = [1 2 3]
B = [5 6 7 8 9 22 13]

How would I be able to create C=[A(1) B(1) A(2) B(2).....]

such that when you run out of elements in one vector, the new vector also contains the remaining elements from the longer vector?

Answer 1

[Star Strider]

One possibility:

A = [1 2 3]; 
B = [5 6 7 8 9 22 13];
LenA =length(A);
C = zeros(1, LenA+length(B));
C(1:2:LenA*2) = A;
C(2:2:LenA*2) = B(1:LenA);
C(LenA*2+1:end) = B(LenA+1:end)

C =
     1     5     2     6     3     7     8     9    22    13

EDIT— Originally reversed orders of ‘A’ and ‘B’ in ‘C’. Correct now.

[Lorraine Williams]

Excellent - thank you so much!

[Star Strider]

My pleasure!

[Lorraine Williams]

So, as this is all new to me, could I impose and ask if you could explain each step for me - I sort of get it but when it gets to LenA*2 then I'm not sure what's going on :)

[Star Strider]

No imposition at all! This is MATLAB Answers!

The LenA call is just the length of the shorter vector, ‘A’. I use that later in the code, so it’s more efficient to do the operation once and then use that result.

The ‘C’ assignment creates the initial vector with a length that is the sum of the lengths of its two components.

This next is an introduction to the colon (:) operator. It defines integer vectors here, although it has other uses. (See Special Characters for details.)

The ‘C(1:2:LenA*2) = A;’ assignment uses the colon operator to begin with 1, step by 2, and end with ‘LenA*2’ because we want ‘C’ to have the elements of ‘A’ as its first, third, and fifth elements. This creates a vector of indices: [1 3 5]. Because this is by definition going to be twice the length of ‘A’, we stop it at ‘LenA*2’.

The next assignment, ‘C(2:2:LenA*2) = B(1:LenA);’ does essentially the same thing, however it starts at C(2) with a step of 2 to fill ‘C([2 4 6])’ with the first three elements of ‘B’. (Note that ‘C([2 4 6])’ is also correct MATLAB code.)

At this point the code have created ‘C(1:6)’, or ‘C(1:LenA*2)’ from all of ‘A’ and the first ‘LenA’ elements of ‘B’, that is ‘B(1:LenA)’. Since the rest of ‘C’ (that is, ‘C(LenA*2+1:end)’) are the remaining elements of ‘B’ (that is, ‘B(LenA+1:end)’) , we assign:

C(LenA*2+1:end) = B(LenA+1:end)

I didn’t assign steps here because the step is assumed to be +1 unless specifically defined as something else. I left off the end semicolon because I wanted to print the output so I could post it.

Answer 2

[Cedric]

Here is one way:

 if numel( A ) < numel( B )
    C = [B; B] ;
    C(1,1:numel( A )) = A ;
 else
    C = [A; A] ;
    C(2,1:numel( B )) = B ;
 end
 C = C(:)' ;

[Cedric]

Seeing Star Strider's answer, I realize that I probably misunderstood the question and that you don't want to use elements of e.g. B for replacing missing elements of A. In other words, you want the output that Star Strider provides, and not:

 C =  1  5  2  6  3  7  8  8  9  9  22  22  13  13

[Lorraine Williams]

Thank you so much for helping and being so responsive! I'm new to this and it's a bit daunting!!