Create all combinations of a vector

Hello,

I need your help to write down the different possible combinations of a vector x used as input for my model.

Basically I have a vector x of i elements that can take the values 0 or 1. The subset of the y last elements of x is "mutually exclusive", which means that only 1 of those elements at a time can have a value of 1.

For every value of x I run my model and then save the results in a file.

Until now I have nested as many for-loops as there are elements in x to depict every case, but I suspect there must be a much more convenient way to do this!

So basically I would like to improve following code (it's ok if you laugh seeing it! :P) and add the mutually exclusive condition:

%% Create the x vector for all possible combinations
for m=1:2 
    x(1)=m-1;
    for n=1:2 
    x(2)=n-1;
    for o=1:2 
    x(3)=o-1;
    for p=1:2
    x(4)=p-1;
    for q=1:2
    x(5)=q-1;
    for r=1:2
    x(6)=r-1;
    for s=1:2
    x(7)=s-1;
    for t=1:2
    x(8)=t-1;
    for u=1:2
    x(9)=u-1;
    for v=1:2
    x(10)=v-1;

%% Run the model
[A,B]=Zmultipleexecution(x);

%% Save the results in a .mat file
file=['C:\Users\XXX\Documents\',num2str(m)-1,num2str(n)-1,num2str(o)-1,num2str(p)-1,num2str(q)-1,num2str(r)-1,num2str(s)-1,num2str(t)-1,num2str(u)-1,num2str(v)-1];
save(file);

    end
    end
    end
    end
    end
    end
    end
    end
    end
end

Thank you very much!

[Adam Danz]

Code from OP with indentation to see its structure.

%% Create the x vector for all possible combinations
for m=1:2
    x(1)=m-1;
    for n=1:2
        x(2)=n-1;
        for o=1:2
            x(3)=o-1;
            for p=1:2
                x(4)=p-1;
                for q=1:2
                    x(5)=q-1;
                    for r=1:2
                        x(6)=r-1;
                        for s=1:2
                            x(7)=s-1;
                            for t=1:2
                                x(8)=t-1;
                                for u=1:2
                                    x(9)=u-1;
                                    for v=1:2
                                        x(10)=v-1;
                                        
                                        %% Run the model
                                        [A,B]=Zmultipleexecution(x);
                                        
                                        %% Save the results in a .mat file
                                        file=['C:\Users\XXX\Documents\',num2str(m)-1,num2str(n)-1,num2str(o)-1,num2str(p)-1,num2str(q)-1,num2str(r)-1,num2str(s)-1,num2str(t)-1,num2str(u)-1,num2str(v)-1];
                                        save(file);
                                        
                                    end
                                end
                            end
                        end
                    end
                end
            end
        end
    end
end

[Adam Danz]

So, x is a 1x10 vector of 1s and 0s, is that correct?

[Lenilein]

x is a 1 x i vector (i is stored in separate model parameter) of 1s and 0s

[Adam Danz]

Then my answer should be what you're looking for.

Answer 1

[Adam Danz]

It looks like input x is a 1x10 vector of 1s and 0s starting from [0 0 0 0 0 0 0 0 0 0] and ending at [1 1 1 1 1 1 1 1 1 1].

This loop produces the 1024 permutations of x in 'permMat' and then loops through each row.

% For 10 variables, 
permMat = dec2bin(0:bin2dec('1111111111'));  % 10 ones.
for i = 1:size(permMat,1)
    inputs = str2double(num2cell(permMat(i,:)));  % Convert to vector of doubles
    [A,B]=Zmultipleexecution(inputs);
    % Whever else you need to do.....
end

The permMat should produce the same sequence as your loop in the same order, too. Here are the first few rows/iterations. Each row is converted to a vector of doubles within the loop.

>> permMat 
ans =
  1024×10 char array
    '0000000000'  % Converted to [0 0 0 0 0 0 0 0 0 0] in loop.
    '0000000001'
    '0000000010'
    '0000000011'
    '0000000100'
    '0000000101'
    '0000000110'
    '0000000111'
    '0000001000'

{updated}

Alternatively, you can convert the char array to a matrix before the loop,

permMat = dec2bin(0:bin2dec('1111111111'));  % 10 ones.
permMatDbl = str2double(num2cell(permMat));
for i = 1:size(permMat,1)
    inputs = permMatDbl(i,:);
    [A,B]=Zmultipleexecution(inputs);
    % Whever else you need to do.....
end

[Lenilein]

Thanks, this is very useful!

Now remains the second part of my question: I would like to remove all elements of permMat for which more than one of the last three digits is equal to 1, any hint?

[Adam Danz]

Within the for-loop, skip iterations that match that condition.

for i = 1:size(permMat,1)
    inputs = str2double(num2cell(permMat(i,:)));  % Convert to vector of doubles
    if sum(inputs(end-2:end)) > 1
        continue
    else
        [A,B]=Zmultipleexecution(x);
    end
end

[Adam Danz]

Alternatively you could convert permMat to a numeric matrix and eliminate rows before the loop.

This is the better method.

permMat = dec2bin(0:bin2dec('1111111111'));  % 10 ones.
permMatDbl = str2double(num2cell(permMat));
removeIdx = sum(permMatDbl(:,end-2:end),2) > 1; 
permMatDbl(removeIdx, :) = [];

for i = 1:size(permMatDbl,1)
    inputs = permMatDbl(i,:)l 
    [A,B]=Zmultipleexecution(inputs);
    % Whever else you need to do.....
end

[Lenilein]

Great, thank you so much!!