Finding rows in a matrix
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I have a matrix A = [1 2; 2 1; 1 2; 2 2; 1 1; 2 2]
I want to count how many times the row [1 2] appears in above matrix A. Here for my counting purpose [1 2] would appear 3 times as [1 2] or [2 1].
Thanks guys
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采纳的回答
Matt Fig
2012-10-6
编辑:Matt Fig
2012-10-7
Another:
% The given matrix
A = [1 2; 2 1; 1 2; 2 2; 1 1; 2 2];
% Now find the counts.
[I,J,K] = unique(sort(A,2),'rows'); % I has the unique rows.
C = histc(K,1:max(K)); % This has the corresponding counts.
% Now that we have found the counts, display them:
fprintf('Row [%i %i] appears %i times. \n',[I C]')
If you want to only get the counts for the one type, this will do it quickly:
cnt = sum(all(bsxfun(@eq,sort(A,2),[1,2]),2));
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更多回答(2 个)
Azzi Abdelmalek
2012-10-6
编辑:Azzi Abdelmalek
2012-10-6
A = randi(2,10,2)
idx=find(any(repmat([1 2],size(A,1),1)-sort(A,2),2)==0)
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Image Analyst
2012-10-6
Here's a brute force approach:
A = [1 2; 2 1; 1 2; 2 2; 1 1; 2 2]
sortedA = sort(A, 2)
uniqueRows = unique(sortedA, 'rows')
counts = zeros(1, size(uniqueRows, 1)); % Preallocate.
for row = 1 : size(uniqueRows, 1)
for rowA = 1 : size(A, 1)
matches = sortedA(rowA, :) == uniqueRows(row, :);
if all(matches)
counts(row) = counts(row) + 1;
end
end
end
counts
In the command window:
A =
1 2
2 1
1 2
2 2
1 1
2 2
sortedA =
1 2
1 2
1 2
2 2
1 1
2 2
uniqueRows =
1 1
1 2
2 2
counts =
1 3 2
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