How to find the number of occurrences of each pair in a matrix?
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Suppose my
A = [1 2 3;
2 3 1;
2 1 3]
% where any digit won't repeat in the same row.
% I need to find out the number of occurrences of each pair.
Expected output:
Pair(1,2) = 1 occurence;
Pair(2,1) = 1;
Pair(2,3) = 2;
Pair(3,2) = 0;
Pair(3,1) = 1;
Pair(1,3) = 1;
I was trying this code
for n = 2:3
[j,i]=ind2sub(fliplr(size(A)), strfind(reshape(A.',1,[]),[1 n]).');
C = [i,j];
d = numel(C(j));
T9 = table(1, n, d)
end
error occurs when the second row ends with 1 and the third row begins with 2.
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采纳的回答
Bruno Luong
2018-10-31
A = [1 2 3;
2 3 1;
2 1 3];
u = unique(A);
P = nchoosek(u,2);
P = [P;fliplr(P)];
[~,J] = ismember([A(:,1:2); A(:,2:3)],P,'rows');
n = accumarray(J(:),1,[size(P,1) 1]);
%
for k=1:size(P,1)
fprintf('Pair %s, %d time\n',mat2str(P(k,:)),n(k));
end
Output
Pair [1 2], 1 time
Pair [1 3], 1 time
Pair [2 3], 2 time
Pair [2 1], 1 time
Pair [3 1], 1 time
Pair [3 2], 0 time
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更多回答(1 个)
David Goodmanson
2018-10-31
编辑:David Goodmanson
2018-10-31
Hi MSK,
nrow = size(A,1);
B = A(:);
s1 = full(sparse(B(1:end-nrow),B(nrow+1:end),1))
puts instances if i,j into entry i,j. This should work with any number n of columns, assuming numbers 1 through n are used in each row.
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