findout the second largest element in each row and its location in a matrix
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navan
am 20 Mai 2015
Beantwortet: Steven Lord
am 28 Okt. 2018
I have a (3*5) matrix i.e A=[1 2 3 4 5, 6 7 9 8 10, 11 12 14 13 15], i would like to find out the second largest element in each row and its location in the matrix. i expect 2 outputs 1)second_maxvalue_A=[4,9,14] 2)location_of_second_maxval=[4,3,3]
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Guillaume
am 20 Mai 2015
One possible way is to find the largest values, replace them with a very small value (-Inf) and search for the new largest values:
A=[1 2 3 4 5; 6 7 9 8 10; 11 12 14 13 15] %your example A was a single row
m = max(A, [], 2); %find the max in each row
A(bsxfun(@eq, A, m)) = -Inf; %replace the max(s) by -Inf
m = max(A, [], 2) %find the new max which is the second largest.
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Bruno Luong
am 28 Okt. 2018
zahra najja: Please read the other ANSWERS bellow, or even better the DOC of MIN
Weitere Antworten (3)
Thomas Koelen
am 20 Mai 2015
Bearbeitet: Thomas Koelen
am 20 Mai 2015
Edit: seems like I misread second largest...
, should be ;.
A=[1 2 3 4 5; 6 7 9 8 10; 11 12 14 13 15];
[m,i] = max(A,[],2)
gives you:
m =
5
10
15
i =
5
5
5
example:
A=rand(3,5);
[m,i] = max(A,[],2)
gives:
0.8147 0.9134 0.2785 0.9649 0.9572
0.9058 0.6324 0.5469 0.1576 0.4854
0.1270 0.0975 0.9575 0.9706 0.8003
m =
0.9649
0.9058
0.9706
i =
4
1
4
3 Kommentare
Guillaume
am 20 Mai 2015
Doesn't this find the second largest element in the matrix, rather than each row?
Secondly, if there are more than one element equal to the max, it still returns the max.
Guillaume
am 20 Mai 2015
A generic way of finding the nth largest value of each row, if it exists:
function nl = nthlargestrow(m, n)
%m: a 2d matrix to search
%n: an integer
%nl: cell array containing the nth largest value of each row. If a row does not have n different values, the cell is empty
nl = cell(size(m, 1), 1);
u = cellfun(@unique, num2cell(m, 2), 'UniformOutput', false);
hasnelements = cellfun(@numel, u) >= n;
nl(hasnelements) = cellfun(@(row) row(end-n+1), u(hasnelements), 'UniformOutput', false)
end
Thorsten
am 20 Mai 2015
A=[1 2 3 4 5; 6 7 9 8 10; 11 12 14 13 15];
% remove highest value in each row
sz = size(A);
[~, ind] = max(A, [], 2);
ind = ind + [0; cumsum(repmat(sz(2), [sz(1)-1 1]))];
A = A';
A(ind) = [];
A = reshape(A, [sz(2)-1 sz(1)])';
% find highest value in each row (i.e., second largest of original matrix)
[max2, ind2] = max(A, [], 2);
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