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

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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.

回答(2 个)

michael
michael 2016-10-3
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
  2 个评论
Rene Stiller
Rene Stiller 2016-10-3
编辑:Rene Stiller 2016-10-3
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
michael 2016-10-3
if you would like to have a zero matrix of the same size as i_max
i=zeros(size(i_max))

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Joe Yeh
Joe Yeh 2016-10-3
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.
  4 个评论
Rene Stiller
Rene Stiller 2016-10-3
Thank you for the suggestion.
I'll try that tomorrow.
Rene Stiller
Rene Stiller 2016-10-4
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!

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