Creating a matrix using the colon function

Hi everyone,

Let us assum that I have two arrays as follows:

A = [1 2 3 4 5]
B = [3 4 5 6 7]

I want to use colon funcion to create Matrix C:

C = [1 2 3
     2 3 4
     3 4 5
     4 5 6
     5 6 7]

I used the folloing function, but it does not work. I can use a loop to create this matrix, but I want to know what is the fastest solution. Any idea?

 [1 2 3 4 5]: 1: [3 4 5 6 7]

Thanks,

Amir

Answer 1

[John D'Errico]

An array in MATLAB NEEDS to be rectangular. That is a given.

So if the increment is ALWAYS 1 between elements, then the second vector must always be some integer constant greater than the first, for ALL elements. In this case, the difference is 2.

A = [1 2 3 4 5];
B = [3 4 5 6 7];
B - A
ans =
     2     2     2     2     2

Given that, it is trivial to generate the result you ask for, (AND it is efficiently done too.)

C = A(:) + (0:2)
C =
     1     2     3
     2     3     4
     3     4     5
     4     5     6
     5     6     7

The above works as long as you have release R2016b or later. Earlier releases would efficiently use

C = bsxfun(@plus,A(:),0:2)
C =
     1     2     3
     2     3     4
     3     4     5
     4     5     6
     5     6     7

As you should see, the first solution is much cleaner and easier to read.

If some elements of B were not uniformly 2 greater than the corresponding element in A, then you would need to decide how to deal with it, since arrays in MATLAB are rectangular. You might decide to zero pad, or insert NaNs or use a cell array to store the results.

Answer 2

[madhan ravi]

No other fastest way , there’s arrayfun() I could think of but I don’t think any faster. I have also submitted feature request regarding this , let’s see what the future holds.

Answer 3

[madhan ravi]

 ix = max(diff([A;B]));
 c = cumsum([A;ones(ix, numel(A))]).'; % if the difference between each A and B elements are the same , this would be the last step
 Wanted = c .* (c <= B(:))

[madhan ravi]

You will have to compare the timings , I didn’t check it. Still I bet the loop is the eifficient one.