Random but unique values in matrix

Please help me! I want to generate random but unique row as given in code. I want generate complete 100*13 matrix. When this code run it stuck after around 29th iterations. Please help me and give suggestions. If my question is not understandable please let me know. I will explain more in detail.

 j=1;
 pop_size=100;
 V=13;
 while j:pop_size
    y=randi([0 1],j,V);
    y(:,2)=~y(:,1);
    y(:,3)=~y(:,4);
    y(:,5)=~y(:,6);
    y(:,12)=~y(:,13);
    y = unique(y, 'rows');
    j=size(y,1)+1;
 end

[Stephen23]

This line is highly unusual, where you use a vector as the logical condition:

 while j:pop_size

what is it supposed to do? The while documentation states, "An expression is true when its result is nonempty and contains only nonzero elements (logical or real numeric). Otherwise, the expression is false." Because you are constructing this vector with pop_size at the end it will only become false when the vector is empty. Is this what you need to achieve?

Or are you actually trying to loop from 1 to 100 ?

What does "it stuck after around 29th iterations" mean? How is the loop "stuck" ? Do you get an error, does the loop complete correctly, or something else ?

[Asad Abbas]

I am trying to loop 1 to 100. Not actually stuck. but it update unique values and running. if value in new row is not unique then it replace again and again. I am trying to get non-repetition rows thats why i mentioned 'unique'. It was running from last three hours but no any end result or error.

[John D'Errico]

that is a FOR loop, not a while.

[Stephen23]

What are these for?:

    y(:,2)=~y(:,1);
    y(:,3)=~y(:,4);
    y(:,5)=~y(:,6);
    y(:,12)=~y(:,13);

You compare some values, but never allocate their output to anything or use them for anything at all.

[John D'Errico]

I've looked at your code, and I have no idea what your goal is here. So I have no idea if you are doing something that is easy to do otherwise, but just doing it inefficiently.

Can you please explain your goal here?

[Asad Abbas]

these are constraints. For example column 1 should not be equal column 2. means if column 1 is 0 then column 2 should be 1.

[Asad Abbas]

Dear Sir John D'Errico, I tried with for loop but it not giving me complete 100*13 matrix. It giving me some times 91*13, 89*13, 92*13, 84*13 some thing like this. I want generate 100*13 unique value matrix

[Asad Abbas]

Sir, I want generate 100*13 random matrix with 0 or 1 value and each row should be unique means no repetition should occur. I have given so me constraints also. These constraints are. For example column 1 should not be equal column 2. means if column 1 is 0 then column 2 should be 1. y(:,2)=~y(:,1); y(:,3)=~y(:,4); y(:,5)=~y(:,6); y(:,12)=~y(:,13);

[Steven Lord]

So generate a random 100-by-8 matrix. Check if all its rows are unique using the unique function with the 'rows' flag. Once you have a suitable matrix, expand it to 12 columns. [Once columns 1, 3, 5, and 12 are fixed they determine columns 2, 4, 6, and 13.]

[Stephen23]

My answer exactly implements the concept that Steve Lord suggests.

Answer 1

[dpb]

>> z=randi([0 1],100,13);  % Generate trial array
>> [u,ia,ib]=unique(z,'rows','stable');  % check for uniqueness between rows
>> length(ia)    % dang!, one short...
ans =
  99
>> find(ismember(z,z(setdiff([1:100],ia),:),'rows'))  % locate which are the offending rows
ans =
  34
  92
>> 
>> [z(34,:);z(92,:)]  % indeed, they are identical...
ans =
   0     0     1     0     0     1     1     0     1     0     1     0     0
   0     0     1     0     0     1     1     0     1     0     1     0     0
>>

OK, NOW you can begin a loop looking to generate a new row that is different than z(34,:) and also not duplicate another already...

[Thorsten]

Or you generate a trial array with many more rows than actually needed and hope that you'll end up with more than 100 unique rows. See my answer below.

BTW: You forgot the ensure the four constraints

 y(:,2)=~y(:,1);
 y(:,3)=~y(:,4);
 y(:,5)=~y(:,6);
 y(:,12)=~y(:,13);

[Asad Abbas]

Thank you so much Sir, Its working. I am so happy.

Answer 2

[Stephen23]

A complete working example:

X = randi([0,1],100,9);
chk = true;
while chk
    [~,idx] = unique(X,'rows');
    idy = setdiff(1:100,idx);
    X(idy,:) = randi([0,1],numel(idy),9);
    chk = numel(idy)>0;
end
Y = X(:,[1,1,2,2,3,3,4:end,end]);
idz = [1,3,5,12];
Y(:,idz) = ~Y(:,1+idz);

and tested:

 >> size(unique(Y,'rows'))
 ans =
    100    13
 >> Y(:,1:2) % check that two columns are always different
ans =
   0   1
   1   0
   1   0
   1   0
   0   1
   1   0
   1   0
   1   0
   0   1
   1   0
   1   0
 etc

 >> Y(:,12:13)
 ans =
   1   0
   1   0
   0   1
   1   0
   1   0
   0   1
   1   0
   1   0
   1   0
   0   1
   0   1
   0   1
   1   0
   0   1
 etc

[Asad Abbas]

Sir Its working exactly. thank you much.

[Stephen23]

@Asad Abbas: you can also vote for my answer, if it solves your task. That is an easy way for you to say "thank you" to us volunteers.

[antlhem]

Hi Stephen Cobeldick, how would you suggest me to use this method in order to create a bigger array (1500x1500). With all the rows unique among each other?

Answer 3

[Ori Golani]

Hi all, I know that this is an old thread, but I found an efficient non-iterative solution for this.

Complete working example:

 I= 2; % maximal integer in the matrix (values will be [0...I-1])
 M= 3; % Matrix rows
 N= 10;  % matrix columns

 word_indices= randperm(I^N,M);
 matrix_in_char= dec2base(word_indices,I);
 random_matrix=  reshape(base2dec(matrix_in_char(:),I),M,N);

Tested:

random_matrix =

     1     1     1     0     1     1     1     0     1     0
     0     0     1     1     0     1     0     0     0     1
     0     0     1     0     1     0     1     1     1     0

[Stephen23]

+1 neat idea.

[antlhem]

Hi Ori Golani, do you know how to create a bigger matrix with this method? I tested with M=N=1500, but is not passing this test. There is an error from MATLAB saying:

Error using randperm

N must be less than 2^53.

Any suggestions, to be able to create bigger random-unique matrix?

Answer 4

[Thorsten]

The trick is to get more rows then actually needed, like 2*pop_size. Then you'll get your y almost always in the first trial:

 pop_size=100;
 V=13;

 trials = 1;
 while 1
   y = randi([0 1], 2*pop_size, V);

   y(:,2)= 1- y(:,1);
   y(:,3)= 1- y(:,4);
   y(:,5)= 1- y(:,6);
   y(:,12)= 1- y(:,13);
   [yu, xi] = unique(y, 'rows');
   if size(yu, 1) >= pop_size
     y = y(xi(1:pop_size),:);
     break;
   else
     trials = trials + 1;
   end
 end

 % check
 size(unique(y, 'rows'), 1)

[Asad Abbas]

Thank you so much for your response. But when I run this command at the end of your given code y=unique(y,'rows'); It is not giving 100*13 unique rows, It giving 89*13, 92*13 something like this.

[Thorsten]

You're right. I corrected the code.