Array Exponentials without For Loop?

Hello, I hope you are doing well! I am not entirely sure how to succinctly describe the question I have, so I will instead demonstrate it.

Let's say I have two arrays:

A = [1 2 3];
B = [2 3];

I would like to do the operation A^B such that I would receive a 2D array C containing the solution of A.^B(1) and A.^B(2).

C(1,:) = A.^B(1);
C(2,:) = A.^B(2);
C = [1 4 9; 1 8 27];

I understand that I could write a For loop like this:

% Full Loop Code
A = [1 2 3];
B = [2 3];
C = zeros(length(B), length(A));

for i = 1: length(B)
    C(i, :) = A.^B(i);
end

But I am interested in knowing how to do this using only vectors and not using a For loop. When I do:

C = A.^B

I receive this error:

Error using .^

Matrix dimensions must agree.

Any suggestions? I am using MATLAB 2016b.

Thank you!

Answer 1

[Jan]

A = [1 2 3];
B = [2 3];
C = A .^ (B.')

[Alec Huynh]

Thank you so much for the quick answer! May I ask what the significance of the '.' in your code is? When I put B.' and B' in the console I get the same answer.

B.'
B'

ans = 
     2.0000e+000
     3.0000e+000

[Steven Lord]

If an array is real, the transpose (.') and conjugate transpose (') operators give the same results.

A = magic(3)

A = 3×3

8 1 6 3 5 7 4 9 2

B1 = A.'

B1 = 3×3

8 3 4 1 5 9 6 7 2

B2 = A'

B2 = 3×3

8 3 4 1 5 9 6 7 2

isequal(B1, B2) % Yes they are equal

ans = logical
   1

If the array is complex they do not give the same result.

format shortg
C = A + 1i*(10-A)

C = 
            8 +          2i            1 +          9i            6 +          4i
            3 +          7i            5 +          5i            7 +          3i
            4 +          6i            9 +          1i            2 +          8i

D1 = C.'

D1 = 
            8 +          2i            3 +          7i            4 +          6i
            1 +          9i            5 +          5i            9 +          1i
            6 +          4i            7 +          3i            2 +          8i

D2 = C'

D2 = 
            8 -          2i            3 -          7i            4 -          6i
            1 -          9i            5 -          5i            9 -          1i
            6 -          4i            7 -          3i            2 -          8i

isequal(D1, D2) % these are not equal

ans = logical
   0

Note the signs of the imaginary parts of D1 and D2.

[Jan]

@Alec Huynh: Thanks, @Steven Lord, for the explanation. An addition, why I've used .' in this case: For real arrays .' and ' produce the same result. It is a good programming practice to avoid assumptions of the inputs, so if I want to transpose the input, .' is the correct operator. Then I do not have to guess, if you (or any other use) is working with real or complex values.

This is the same strategy as e.g. specifying the dimension to operate on:

x = rand(randi(1:3, 2));  % e.g. [1,3], [2,1] or [3,2]
y = sum(x)     % ?!? critical: input might be a column or row vector
y = sum(x, 1)  % Avoid assumptions

The less assumptions the programmer includes in the code, the less likely is producing a bug or misleading output.

[Alec Huynh]

@Steven Lord, @Jan Thank you for the succinct explanations!