How can I create a specific matrix in a for loop?

Hi,

I have a 4x50 matrix which is defined as x. I'm trying to store each row of x inside each row of b with a formula that is stated in the code below. So, each row of b is predetermined with the corresponding row of x. After creating the matrix b, I will generate arrays with the code ndgrid. I keep getting the error "Subscripted assignment dimension mismatch." when it comes to the for loop. How I can get around this problem? Thank you very much in advance.

[M N] = size(x); % M number of rows, N number of column

b = zeros(M,N);

b(1,:) = x(1,:);

for j=1:1:M-1

    b(j,:) = [min(x(j,:)):(max(x(j,:))-min(x(j,:)))/20:max(x(j,:))];

end

binscenter(1:M,:) = ndgrid(b(1:M,:));

Edit: Also in the last line with the code ndgrid, how can I generate arrays? I don't think my code is going to work, since I didn't specify what binscenter is. The problem is, number of rows and columns of x can change under different circumstances. So I'm trying to write a more compact code which calculates number of rows of x and than binscenter is determined accordingly.

[the cyclist]

I suggest you resolve your first problem, and then post your second question separately.

Answer 1

[the cyclist]

The right-hand side of your assignment statement inside the for loop is creating a vector that is length 21, and the right-hand side is length 50. That is the mismatch.

Instead of "20", you need "N-1" there.

You might still need to be careful in how you constructed that statement, because creating vectors like that can lead to unexpected results when using floating point numbers. I suggest you use the linspace() function to do that operation:

linspace(min(x(j,:)),max(x(j,:)),N)

should do what you want.

[melampyge]

linspace did the trick a couple of minutes ago, but strangely now it doesn't work. I changed it to:

b(1,:) = linspace(min(x(1,:)),max(x(1,:)),N);

for j=1:1:M-1
    b(j,:) = linspace(min(x(j,:)),max(x(j,:)),N);
end

binscenter = ndgrid(b(1:M,:));

[the cyclist]

Are there any rows where all the values are equal, such that the min and max are the same? In that case, your right-hand side will be empty, and therefore you will get the dimension mismatch again.

[the cyclist]

Also, don't you want

for j = 2:1:M

?

[melampyge]

I checked again but there aren't any rows with all the values equal.

[the cyclist]

I suggest you use debug mode to look at the state of your code when you get the error. Here's a document that describes how to do that: http://www.mathworks.com/help/matlab/debugging-code.html

Also, if you could post a very small example of x that exhibits the problem, then maybe I could take a look at that specific example. Just creating the smallest possible example that exhibits the problem may help you find the problem, too.

[melampyge]

Sure, I will try doing debugging. x is a 3x1001 matrix at the moment, I will now try with a smaller matrix than to see what happens.

Answer 2

[David Sanchez]

what's the reason behind the "20" in the 5th line of your code?

[M N] = size(x); % M number of rows, N number of column
b = zeros(M,N);
b(1,:) = x(1,:);
for j=1:M-1
    b(j,:) = min(x(j,:)):(max(x(j,:))-min(x(j,:)))/20:max(x(j,:));
end

if your x matrix is 20x20, this part of the code will work as follows:

x = rand(20); % example matrix
[M N] = size(x); % M number of rows, N number of column
b = zeros(M,N);
b(1,:) = x(1,:);
for j=1:M-1
    b(j,:) = min(x(j,:)):(max(x(j,:))-min(x(j,:)))/19:max(x(j,:));
end

About your last line, is this what you want?

binscenter = ndgrid(b(1:M,:));

[melampyge]

It doesn't have to be 20 per se, I can change it to 50 for example. I'm calculating drift in a stochastic differential equation, so I need to generate as many points as I can. For the last line, that is what I want, thank you!