Pdist2 inside for

Hi everybody,

i have two 3D matrix A and B with different lengths. I need to build a for loop to calculate the pdist2 between the first row of A and all the rows of B, the second row of A and all the rows of B, ...., the n row of A and all the rows of B.

An example of the two matrix:

A

X Y Z

0.2 0.4 0.5

1.4 0.6 4.5

0.3 0.5 3.1

B

X Y Z

0.8 0.7 0.5

1.4 0.9 1.5

0.2 3.5 2.1

0.3 0.7 0.5

2.4 0.3 0.5

0.2 3.5 3.1

Thank you very much!

Answer 1

[Jan]

Why? pdist2 works directly with providing a [M1 x N] and a [M2 x N] matrix. So why do you want to write a loop and process one matrix rowwise?

D = pdist2(A, B)

But you write "3D matrix". Maybe you mean "3D array", because matrices are 2D by definition. Then the shown examples are misleading. If A and B are 3D arrays, you for got to mention what you want to do with the 3rd dimension.

[Riccardo Rossi]

Thank's for answering.

Yes i was wrong, i mean 3D arrays. I need a loop because for each row wise result i have to put an "if" - "else if" statement.

So is it possible write a loop and process one array rowwise?

Thank you!

[Jan]

Yes, of course this is possible. Because I cannot know, what "put an "if" - "else if" statement" exactly means, I cannot post some explicit code. But I suggest to call pdist2 for the matrices. not for single rows, and use logical indexing to examine the results.

% Two 3D arrays, the 2nd dimension must be equal:
A = rand(5,  3, 2);
B = rand(10, 3, 7);

for iA = 1:size(A, 3)
    for iB = 1:size(B, 3)
        D     = pdist2(A(:, :, iA), B(:, :, iB));
        match = (D > 0.4);  % Or whatever you need
        ...
    end
end

Note: You neither posted the correct input, nor explained, what you want as output, nor asked a question, nor mentioned, what should be computed actually. This makes it hard to help you. Sorry, if my bold guesses are confusing only.

Answer 2

[Star Strider]

No loop needed, since pdist2 does it all for you:

A = [0.2  0.4  0.5
     1.4  0.6  4.5 
     0.3  0.5  3.1];

B = [0.8  0.7  0.5
     1.4  0.9  1.5 
     0.2  3.5  2.1
     0.3  0.7  0.5
     2.4  0.3  0.5 
     0.2  3.5  3.1];

D = pdist2(A,B);

fprintf(1, 'B:\t\t%d\t\t%d\t\t%d\t\t%d\t\t%d\t\t%d\n',1:size(B,1))
fprintf(1, 'A:%d\t%.3f\t%.3f\t%.3f\t%.3f\t%.3f\t%.3f\n', [(1:size(A,1))' D]')

A1B2 = sqrt((A(1,:)-B(2,:))*(A(1,:)-B(2,:))');          % First Row Of A, Second Row Of B
A3B6 = sqrt((A(3,:)-B(6,:))*(A(3,:)-B(6,:))');          % Third Row Of A, Sixth Row Of B

The ‘proof’ is in the last two lines, that show those distances.