combination

Dear I have an matrix

x =

     1     2     3     4     5
     6     7     8     9    10
    11    12    13    14    15
    16    17    18    19    20

I want all combination of changing each variables in first row with other rows for example give me 6 2 8 4 5 and .....

Answer 1

[Image Analyst]

It seems to me that the brute force way of 5 nested for loops should do it. Each for statement would iterate over all numbers (all rows) in the column. For example (untested)

numberOfRows = size(x, 1);
for row1 = 1 : numberOfRows 
  for row2 = 1 : numberOfRows 
    for row3=1 : numberOfRows 
      for row4 = 1 : numberOfRows 
        for row5 = 1 : numberOfRows 
          fprintf('%d %d %d %d %d', x(row1, 1), x(row2, 2), x(row3, 3), x(row4, 4), x(row5, 5));
        end
      end
    end
  end
end

[Niki]

this one does not have end

[Walter Roberson]

It provably has an end after numberofRows^5 operations. On the other hand, that would have been more clear if IA had included \n at the end of the output format string.

Answer 2

[Andrei Bobrov]

c = mat2cell(x,4,[1 1 1 1 1]);
out = cell(1,5);
[out{:}] = ndgrid(c{:});
outd = cell2mat(cellfun(@(x)x(:),out,'un',0));

[Niki]

this one it was an example if I wanted to do it on 5 variables

if it is more what should I do?

for example the example matrix was 5*4

but now I have 10* 4,

I want it works on any matrix that I have Andrei

[Sven]

Cool, I've always been typing "..., 'UniformOuput',false" for my cellfun() calls. You just saved me a bunch of letters.

[Niki]

andrei, I would like to do that for all rows ,

and I would like it be flexible , for example I want to say only I need 100 combination , or more or less,

and I do not know the dimension of my data

Answer 3

[Sven]

I think from your question you want all permutations of the first column and the rest of the columns as a block. I.e., I think in your question where you typed

[6 2 8 4 5]

you actually meant

[6 2 3 4 5].

If so, here is what I would do:

x = [1     2     3     4     5
     6     7     8     9    10
    11    12    13    14    15
    16    17    18    19    20];

inds = 1:4;
[firstInds, secondInds] = meshgrid(inds,inds);
allXcombinations = [x(firstInds(:),1) x(secondInds,2:end)];
disp(allXcombinations)

     1     2     3     4     5
     1     7     8     9    10
     1    12    13    14    15
     1    17    18    19    20
     6     2     3     4     5
     6     7     8     9    10
     6    12    13    14    15
     6    17    18    19    20
    11     2     3     4     5
    11     7     8     9    10
    11    12    13    14    15
    11    17    18    19    20
    16     2     3     4     5
    16     7     8     9    10
    16    12    13    14    15
    16    17    18    19    20

[Niki]

yes it is true I want permutations, but lets say this time I have a matrix 10*100 (variables)(in the previous example I had 4*5)

now I want to do the same what should I do?

I want something automatically

and also I want to do for not only the first row , then when all combination of the first row is finished I want to go with second and third and so on.

is it possible to say how many combination I want ?

[Sven]

Ah, sorry for the misunderstanding.

As the size of your matrix increases, it will very quickly become impossible to store all combinations in memory. Instead, you will need to loop over each combination and do your processing inside that loop.

There is a similar concept in this question, admittedly with a different end goal:

http://www.mathworks.com/matlabcentral/answers/19651-generate-sequence

[Niki]

Sven, I saw the work that you did is what I want , just there is some problems, first the dimention of the data is not known,

second I want to have the permutation for all rows

third I would like to have a key to say how many permutation I would like

could you please do me a favor and help me with that?

[Sven]

You mean you would like to limit your output to N number of rows (even if more *possible* answers exist)?

[Sven]

And if you're limiting the output to N number of rows, which do you choose? Any N random solutions?

[Niki]

yes, exactly

[Niki]

No, I would like to put a limitation, but also if I do not put the limitation I want to calculate all possible answer

[Sven]

Yes, but if you *do* put a limitation, how will it work?

For example take your small data set provided. Imagine you put a limitation of "5" results. Which 5 would you choose? Would any random 5 be acceptable?

[Niki]

yes, it does not matter which one, just the first 5 for example

Answer 4

[Sven]

Mohammad, try this for a solution. It is a loop that iterates along the columns of x, picking the next available x(row,col). When it reaches the end of the columns, it prints out to the screen the "solution" it just found.

This avoids memory issues because it doesn't store every combination of results, it just prints them. It should work fine on any size input set you provide. If you wanted to do some processing on each solution, you can embed your processing within the loop. Here it is:

First set up "x" and "solutionLimit". You can set solutionLimit to Inf if you want all solutions printed.

x = [1     2     3     4     5
     6     7     8     9    10
    11    12    13    14    15
    16    17    18    19    20];
solutionLimit = 50;

Now for the function itself:

[nRows,nCols] = size(x);
rowSet = zeros(1,nCols);        % Which row of x are we up to at each col?
thisSol = zeros(1,nCols);       % The running list of our current solution
colNo = 1; solutionCount = 0;   % Initialise
while solutionCount < solutionLimit
    rowSet(colNo) = rowSet(colNo)+1;        % Get the next free number
    if rowSet(colNo)>nRows                  % Is this col out of numbers?
        if all(rowSet>=nRows), break; end   % Are we ALL finished?
        rowSet(colNo) = 0;  colNo = colNo - 1;  continue;
    end
    thisSol(colNo) = x(rowSet(colNo), colNo);  % Append to our solution
    % If we're at the last column, print the solution. Otherwise move on.
    if colNo==nCols
        solutionCount = solutionCount+1;
        fprintf('#%d: %s\n',solutionCount, sprintf('%d ',thisSol))
    else
        colNo = colNo + 1;
    end
end