Can I use a vector as index variable for a loop?

Hello, I have a question about the for loop. Why isn't it possible to use an element of a vector as the index variable?

Here is an example:

i_max=[3;2;4];
n=size(i_max,1);
i=zeros(n);
j=1;

for i(2)=0:1:i_max(1)
    A(:,j)=i
    j=j+1
end

Everytime I try this I get the error message "Unbalanced or unexpected parenthesis or bracket."

Is there any other way I can use the vector "i(x)" as index variable? I don't want to define a new variable like:

i_new=i(2)

This wouldn't really help me since my code must be very dynamic.

Thanks.

Answer 1

[michael]

You have to think MATLAB.

Your code is messy and hard to read.

i_max=[3;2;4];    <== column vector of size 1x3
n=size(i_max,1);  <== you can use length instead of size
i=zeros(n);    <=== i is 3x3 matrix of zeros
j=1;

for i(2)=0:1:i_max(1)   if you would like to update the i matrix, do it in the loop. This statement is illegal 
    A(:,j)=i    <== you can't assign matrix to vector
    j=j+1
end

I'd suggest to set your requirements of the code. if you would like to use array of indexes, you can do it like that:

i=(1:10)*2
idx=[1,3,6]
i(idx)

ans =

       2     6    12

example for the for loop:

for n = 1:40
   r(n) = rank(magic(n));
end

[Rene Stiller]

Well, thank you. I am a quite new to programming. That's why my code is messy.

I'll try your solution.

And btw i should be a 3x1 matrix. I messed something up there.

[michael]

if you would like to have a zero matrix of the same size as i_max

i=zeros(size(i_max))

Answer 2

[Joe Yeh]

Well, you can't use loop variable that way. You can use nested for loop to assign new loop variable for the inner for loop. For example :

i_max=[3;2;4];
n=size(i_max,1);
i=zeros(n);
j=1;

for x = 1:length(i)
  ii = i(x);
  for ii = 0:1:i_max(1)
    A(:,j)= ii;
    j=j+1;
  end
end

Note, however, that nested for loop is usually slow. You should try to vectorize your code. I can provide you with further example if you define more clearly what your programming problem is.

[Rene Stiller]

Thank you for you solution. I'll try that, too.

Here is the problem I want to solve. I've got the vector v with the length n. For example with n=3:

v=[3;4;2]

And I want to find all possible combinations of this vector like:

000
001
002 
010
...
242
342

However the amount of solutions equals (v(1)+1)*(v(2)+1)*...*(v(n)+1). I've written this code for n=3 so far.

v=[3;2;4]     
n=length(v)
i=zeros(n,1);

for i(n)=0:1:v(3)                
    for i(n-1)=0:1:v(2)
        for i(n-2)=0:1:v(1)
            A(:,j)=i;
            j=j+1;
        end    
    end
end

I want my code to generate solutions for all n>0. So I wrote the Code like for i(n)=.... to make it dependable from n. I know my code is messy since I am new to programming at all.

[michael]

are you familiar with binary numbers? it is same idea:

0=000

1=001

2=010

3=011

4=100

5=101

6=110

7=111

I think you can do it with single for loop (or at most 2 for loops):

1 pass: go over all numbers. if there are more numbers, just copy the data x times where x is count of 0 to v(2)

2 pass: do the same, but each time write each number x times, where x is the v(2). if there are more numbers, just copy the data x times where x is count of 0 to v(3)

3 pass: do the same, but each time write each number x times, where x is the v(3). if there are more numbers, just copy the data x times where x is count of 0 to v(4)

...

I hope that this algo would be fine, but may have some wrong statements. please use it as guideline.

Try to think about your code before you write it.

[Rene Stiller]

Thank you for the suggestion.

I'll try that tomorrow.

[Rene Stiller]

I havn't thought that it was this easy.

v=[1;1;2];                  %The vector I want to know all combinations off
n=length(v);                %Number of Elements of "v"
k=prod(v+ones(n,1));        %Amount of all possible combinations

x=k;                        %Number of times the Matrix "a" is replicated
y=1;                        %Number of times the Kronecker tensor product is used
A=zeros(n, k);              %Matrix of all solutions  
for i=1:1:n
    x=x/(v(i)+1);             %Divides the times the Matrix a is replicated
    B=ones(1, y);             %Vector for the Kronecker
    a=kron(0:1:v(i),B);       %Kronecker tensor product - makes [0,1,2] and [1,1] to 0,0,1,1,2,2 f.ex. 
    A(i,:)=repmat(a,1,x);     %Write a x-times to A
    y=y*(v(i)+1);             %Multiplies the times the Kronecker product is used  
end

It gives me my solution really fast:

   0 1 0 1 0 1 0 1 0 1 0 1
   0 0 1 1 0 0 1 1 0 0 1 1
   0 0 0 0 1 1 1 1 2 2 2 2

Thanks again for the hint with the binary numbers!