MATLAB: For loop structure

Question

This is a basic program but since I'm new to MATLAB, I'm not able to figure out the solution.

I have a column vector "Time" in which I want to print value "1" in first 147 cells, followed by "2" in 148 to 2*147 cells and so on. For that, I have written the following script:

 Trial>> c=1;
 Trial>> k=0;
 Trial>> for i = c:146+c
            Time(i,1)=1+k;
            c=i;
            k=k+1;
            end

I know I need to iterate the loop over "Time(i,1)=1+k;" before it executes the next statement. I tried using break but that's not supposed to work. Can anyone suggest me the solution to get the desired results?(It was quite simple in C with just the use of curly braces.)

Answer 1

I am sure you don't want to run c=i; in every iteration.

My code should work for you:

x = 10; % Replace 10 by the max number you need in your array.
k = 1;
for i = 1 : x * 147
    Time(i, 1) = k;
    if rem(i, 147) == 0
        k = k + 1;
    end
end

Answer 2

This is the prime example of a piece of code that should be vectorized can help you understand vectorization. Your code can be written like this:

n = 147;
reps = 10; %% Replace this by the maximum number you want your matrix to have

Time = reshape(bsxfun(@plus, zeros(n,1), 0:reps), 1, []);

Explanation:

Let A be a column vector (1 column, n rows), and B be a row vector (1 row, m columns.

What bsxfun(@plus, A, B) will do here is to add all elements in A with all elements in B, like this:

A(1)+B(1)  A(1)+B(2)  A(1)+B(3) ... A(1)+B(m)
A(2)+B(1)  A(2)+B(2)  ............. A(2)+B(m)
............................................
A(n)+B(1)  A(n)+B(2) .............. A(n)+B(m)

Now, for the two vectors we have: zeros(n,1), and 0:reps, this will give us;

0+0        0+1        0+2           0+reps
0+0        0+1        0+2           0+reps
% n rows of this

So, what we need to do now is place each column underneath each other, so that you will have the column with zeros first, then the row with ones, ... and finally the one with reps (147 in your case).

This can be achieved by reshaping the matrix:

reshape(bsxfun(@plus, zeros(n,1), 0:reps), [], 1);
^       ^                                  ^   ^
|       |                                  |   Number of rows in the new matrix. When [] is used, the appropriate value will be chosen by Matlab
|       |                                  Number of rows in the new matrix
|       matrix to reshape
reshape command   

---

Another approach is using kron:

kron(ones(reps+1, 1) * 0:(n-1)

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For the record, a review of your code:

You should always preallocate memory for matrices that are created inside loops. In this case you know it will become a matrix of dimensions ((reps+1)*n-by-1). This means you should do Time = zeros((reps+1)*n, 1);. This will speed up your code a lot.

You shouldn't use i and j as variable names in Matlab, as they denote the imaginary unit (sqrt(-1)). You can for instance do: for ii = 1:(n*147) instead.

You don't want c=i inside the loop, when the loop is supposed to go from c to c + 146. That doesn't make much sense.

Answer 3

You can use repmat,

x = 10; % Sequence length (or what ever it can be called)
M = repmat(1:x,147,1); % Replicate array 1:x for 147 columns
M = M(:); % Reshape the matrix so that is becomes a column vector.

I can assume that this is a task to practice for loops, but this will work.

Answer 4

An alternative solution may be to do

n = 147;
reps = 10;
a = ceil( (1:(n*reps)) / n);

You first construct an array with the length you want. Then you divide, and round of upwards. 1 to 147 will then become 1.