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Hello I have 3 variables i,j and k. I want to run a loop on each of them such that for each value of i, one value of j is considered and iterated over all values of k.Then for same i, take second value of j and iterate over all values of k and so on... Then repeat this process for all the values of i and save the output. For example for i=1,j=1 iterate over all values of k.Then for i=1, j=2 iterate over all values of k so on. Then save the output m for all the values of i and j over k iterations.

for i=1:30
for j=1:5
for k=1:100
    m=i*0.5*j*sin(k);
end
end
end

May be my code is not right as well?

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  • Many thanks for your time but that's not what I am looking for. I have edited my question in more detail. Can you please check if you have time. Many Thanks Commented Nov 12, 2015 at 15:23

3 Answers 3

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Using ndgrid you can accomplish this without any explicit for loops:

[i,j,k] = ndgrid(1:30,1:5,1:100);
m = 0.5*i.*j.*sin(k);

Or with meshgrid if you flip the first two inputs:

[j,i,k] = meshgrid(1:5,1:30,1:100);
m = 0.5*i.*j.*sin(k);
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4 Comments

Many thanks for your time but that's not what I am looking for. I have edited my question in more detail. Can you please check if you have time. Many Thanks
@bilal: This code exactly replicates the output of the code in your question (except that it stores all values for each combination of i, j, and k in a 3-D array). You need to update that too. The updated text in your question is not clear at all.
Perhaps my code is not right. I want to find m for i=1,j=1 over k iterations, then i=1,j=2 and for all values of k and so on. Does this make any sense?
[i,j,k] = ndgrid(1:30,1:5,1:100); m = 0.5*i.*j.*sin(k); then m(1,1,:) is a vector for i=1, j=1, and all values of k (1 to 100). From your description, I don't see what else you could want.
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The easiest thing you can do is to define m as a 3D array; also, since you know in advance the final size of the array you can initialize it before the loops to avoid MatLab allocating the memory at each iteration:

% Initialize the array
m=zeros(30,5,100);
for i=1:30
   for j=1:5
      for k=1:100
         % Set the values in the array "m"
         m(i,j,k)=i*0.5*j*sin(k);
      end
  end
end

EDIT

In case you want, for each i and j the value of m be computed over k iterations, the variable m should be present also on the right of the = sign.

It should, for example, be somenthing like:

m=m+i+j+k

you can try the following modified version of the above code

% Initialize the array
m=zeros(30,5);
for i=1:30
   for j=1:5
      % Initialize the temporary varaible at each iteration
      tmp_var=0;
      for k=1:100
         % Compute the value of "tmp_var" over "k" iteration
         tmp_var=tmp_var+i*0.5*j*sin(k);
      end
      % Assign the value of "tmp_var" computed for a given (i,j) couple
      m(i,j)=tmp_var;
  end
end

In this example you need to define a temporary varaible to be used inside the k loop; at the end of the k loop the value of the temporary variable will be stored in the m matrix.

At the end of the script, you will have a matrix m of size (30 x 5).

Hope this helps.

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5 Comments

Many thanks for your time but that's not what I am looking for. I have edited my question in more detail. Can you please check if you have time. Many Thanks
Unfortunately the additional information you've added are not clear: for a given value of i and j does the value of m at the k iteration somehow depend on the value of m at the (k-1) iteration?
Perhaps my code is not right. I want to find m for i=1,j=1 over k iterations, then i=1,j=2 and for all values of k and so on. Does this make any sense?
I've updated my answer, hope I've undersood you need.
Thanks for your update. But tmp-var value is added on every iteration of k ? I want to get all the values of k . Does it make any sence? Much appreciated
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In cases where the computation in the loop is only depending on the loop index you could just write:

[j,i,k]=meshgrid(1:5,1:30,1:100);
m=.5*i.*j.*sin(k);

meshgrid will create the appropriate 3D arrays of the indices i,j,k and m will be of the same dimension when the elementwise multiplication (.* instead of *) is used.

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1 Comment

Many thanks for your time but that's not what I am looking for. I have edited my question in more detail. Can you please check if you have time. Many Thanks

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