the dimensions of matrix and the size is different?

I reshaped the dimensions of my matrix in my code. When i run this:

x=[M,N] y=[M2,P]

i get the following answer x =

   335    80

y =

   335     2

However, when i check the size with this command sz1= size(x) sz2=size(y)

i get the following answer: sz1 =

     1     2

sz2 =

     1     2

my code is this

function d = disteu(x, y) % DISTEU Pairwise Euclidean distances between columns of two matrices % % Input: % x, y: Two matrices whose each column is an a vector data. % % Output: % d: Element d(i,j) will be the Euclidean distance between two % column vectors X(:,i) and Y(:,j) % % Note: % The Euclidean distance D between two vectors X and Y is: % D = sum((x-y).^2).^0.5

 %  D = sum((x-y).^2).^0.5
 [M, N] = size(x);
[M2, P] = size(y);
if (M ~= M2)
C=padarray(x,[21,0],0,'post');
sz=size(C);
[M,N]= size(C);

end x=[M,N] y=[M2,P]

sz1= size(x) sz2= size(y)

d = zeros(N, P);

if (N < P) copies = zeros(1,P); for n = 1:N d(n,:) = sum((x(:, n+copies) - y) .^2, 1); end else copies = zeros(1,N); for p = 1:P d(:,p) = sum((x - y(:, p+copies)) .^2, 1)'; end end

d = d.^0.5;

[M]

If

x =

   335    80

then it is a vector of dimension 1x2 : one row, two columns. So the size that you get does correspond. What is the problem ?

[Jan]

Please read http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup#answer_18099 for formatting code in the forum.

SQRT is faster than the power operation .^0.5

Answer 1

[Jan]

This is exactly the expected behavior.

x = [M,N]
y = [M2,P]

This concatenates the scalars M and N, as well as M2 and P horizontally. Then both variables have the dimensions [1, 2], as the size() command tells you. The contents of x and y can be [335, 80] and [335, 2], but this is no contradiction.

[Bisma Waheed]

Actually , the padarray command is changing the size of my matrix. The size of the first two lines of my code is

 [M, N] = size(x);
[M2, P] = size(y);
sz=size(x)
sz2=size(y)

this means that it has 335 elements in their rows.

sz =

   335    80

sz2 =

   335     2

[M]

So in your first post, you were printing the result of:

size(sz)

which is a 1 by 2 vector.

Here, x is a matrix of dimension 335 by 80.

[Stephen23]

@Bisma Waheed: So far there is nothing that you have shown us that is unexpected: you create two 2x1 vectors by concatenating some scalar values like this: x=[M,N] y=[M2,P], where M, N, M2, and P are all scalar values, and then you measure the size of those two 2x1 vectors, which shows that they indeed have size 2x1. It is not clear what you expect to happen.

If you define a 2x1 vector what size do you expect it to have?

In your comment you measure the size of some different variables x and y: clearly different variables can have different sizes. You know that they are different variables because you wrote the code that redefines them:

function d = disteu(x, y)       % Input x & y
...
end x=[M,N] y=[M2,P]            % redefine x & y
...

[Bisma Waheed]

I am not able to convey my problem. My code is really long hence i am not able to upload it. could you tell me this my matrix dimensions are A= 335 by 16 B=314 by 1 How can i add zeroes to make B= 335 by 16

[Stephen23]

See new question here:

https://www.mathworks.com/matlabcentral/answers/389667-how-to-add-zeroes-to-a-matrix-to-make-the-dimensions-same

[Jan]

@Bisma:

B          = rand(314, 1);  % For testing only, use your data
B(355, 16) = 0

fills all needed elements with zeros automatically.