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Generate Random Matrix 0's,1's

Mohammad Al ja'idi さんによって質問されました 2019 年 6 月 13 日
最新アクティビティ Mohammad Al ja'idi さんによって コメントされました 2019 年 6 月 14 日
clould you please assist me to solve the following question:
I need to get random of (0,1) matrices with size (n*m) with the following constraints:
  • Each row has only one (1) and the other values are (0's)
as in the following example:
Example :: for matrix (3*2) as we see we have only one (1) in each row and the other columns are 0's
[1 0 0
1 0 0]
---------
[1 0 0
0 1 0]
---------
[1 0 0
0 0 1]
--------
[0 1 0
1 0 0]
----------
[0 1 0
0 1 0]
---------
[0 1 0
0 0 1]
--------
[0 0 1
1 0 0]
--------
[0 0 1
0 1 0]
--------
[0 0 1
0 0 1]

  4 件のコメント

i did this solution before but the problem is the location of 1 is always locate at the same column
John D'Errico
on 13 Jun 2019
I MIGHT answer, except your question is inconsistent. Are you looking for random matrices, or ALL POSSIBILITIES? You say both, in different places.
Note that there are m^n possible such matrices, given a matrix sixze of n rows, m columns. Do you want to create all of them? or only some, at random?
Mohammad Al ja'idi 2019 年 6 月 13 日
Hello mr. John D'Errico
yes i need only some of them at random distribution for 1's which each row has only one (1) and the rest are 0's, this is my code which I suppose 10 random matrices, but the problem that the 1's are always on the same column, how can i make the 1's distributed randomly.
matno = 10; % the number of random possiblities
rows=10;
columns=3;
matrices = zeros(rows,columns,matno);
for i=1:matno
matrices(:,randi(columns),i)=1;
end

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2 件の回答

John D'Errico
回答者: John D'Errico
2019 年 6 月 13 日
編集済み: John D'Errico
2019 年 6 月 13 日
 採用された回答

Easy enough.
matno = 10; % the number of random possiblities
rows=10;
columns=3;
Now we want to create a matrix of size (rows,columns,matno), such that each row has exactly one element that is 1, and the placement of that 1 is random. This is most simply done by creating a matrix of size (matno*rows,columns), where each of those rows has exactly one true element.
The location of those 1's will be:
cloc = randi(columns,[matno*rows,1]);
matrices = zeros(matno*rows,columns);
matrices(sub2ind([matno*rows,columns],(1:(matno*rows))',cloc)) = 1;
matrices = permute(reshape(matrices,matno,rows,columns),[2 3 1]);
matrices
matrices(:,:,1) =
0 0 1
0 0 1
1 0 0
0 1 0
0 1 0
1 0 0
1 0 0
0 0 1
0 0 1
1 0 0
matrices(:,:,2) =
0 1 0
1 0 0
1 0 0
0 0 1
1 0 0
0 0 1
1 0 0
0 1 0
1 0 0
0 0 1
matrices(:,:,3) =
1 0 0
1 0 0
1 0 0
0 1 0
0 1 0
1 0 0
1 0 0
0 1 0
0 0 1
1 0 0
matrices(:,:,4) =
0 1 0
0 0 1
0 1 0
0 1 0
0 0 1
0 0 1
0 1 0
0 1 0
0 1 0
1 0 0
matrices(:,:,5) =
1 0 0
1 0 0
0 1 0
0 0 1
1 0 0
1 0 0
0 0 1
0 1 0
0 1 0
0 1 0
matrices(:,:,6) =
0 0 1
0 0 1
1 0 0
1 0 0
0 0 1
0 1 0
0 1 0
1 0 0
0 0 1
0 1 0
matrices(:,:,7) =
1 0 0
0 0 1
1 0 0
0 1 0
0 0 1
1 0 0
0 0 1
0 0 1
0 1 0
1 0 0
matrices(:,:,8) =
0 0 1
1 0 0
0 1 0
0 0 1
1 0 0
0 1 0
1 0 0
0 1 0
0 1 0
0 0 1
matrices(:,:,9) =
0 0 1
1 0 0
0 0 1
1 0 0
1 0 0
0 1 0
0 1 0
0 0 1
0 0 1
1 0 0
matrices(:,:,10) =
1 0 0
1 0 0
1 0 0
0 0 1
0 1 0
0 1 0
1 0 0
0 1 0
1 0 0
0 1 0
Easy. Almost trivial. To prove that the result has exactly one 1 in each row, this next result should be a 10x10 array of 1's.
squeeze(sum(matrices,2))
ans =
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
As it is.

  2 件のコメント

Great .. thank you so much
madhan ravi
on 14 Jun 2019
+1

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Guillaume
回答者: Guillaume
2019 年 6 月 13 日

Note that a 3x2 is a matrix with 3 rows and 2 columns, not the other way round.
Bearing in mind that there are n^m such matrices of size m x n, which will quickly get out of hand as n and particularly m grows, here is one way:
m = 4; %number of ROWS. demo data
n = 5; %number of COLUMNS. demo data
columns = cell(1, m); %1 x m array to store all possible column combinations
[columns{:}] = ndgrid(1:n); %cartesian product of 1:n across all rows
columns = cat(m+1, columns{:}); %store as one matrix
pages = repmat(1:n^m, 1, m); %matching page for each element of columns
rows = repelem(1:m, n^m); %matching row for each element of columns
result = zeros(m, n, n^m); %destination matrix
result(sub2ind(size(result), rows(:), columns(:), pages(:))) = 1;
result is a m x n x (n^m) matrix where result(:, :, p) is one of your desired matrix.
Note that you could use Matt J's ndSparse function to build a sparse matrix (using the same rows(:), columns(:), pages(:) arguments) instead of a full matrix. It would be a lot more memory efficient.

  5 件のコメント

great thank you so much
madhan ravi
on 14 Jun 2019
+1
what about the duplication in the results

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