Back track the index of a matrix

2014 1 20
4 个回答
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更新时间：2014 1 21
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0 个投票

Hi everybody, I'm writing a program to output results in a matrix(n x n).Let's call it matrix "A". I need to sort all the elements in matrix A in ascending order and select the first n elements. Let's put this elements in to a vector "B".
How should I write the code to back track the original index (row and column)of the elements of B in A ?
example:
A=[ 1 3 5; 9 7 6; 2 8 4];
B= [7 4 3];
elements of B in A: row= 2 3 1
column = 2 3 2
Thank you in advanced for the attention.
Dulan

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 采纳的回答

Bruno Pop-Stefanov
Bruno Pop-Stefanov 2014-1-20
编辑：Bruno Pop-Stefanov 2014-1-20

2 个投票

The sort function can also output the linear indices of the sorted elements. Use these indices to access the corresponding elements in A. You can transform linear indices into subscript indices using the ind2sub function.
% Input matrix
A = [ 1 3 5; 9 7 6; 2 8 4];
% Sort. Transform A into a vector with (:)
[B,IX] = sort(A(:), 'ascend');
% Convert linear indices to subscript
[I,J] = ind2sub(size(A),IX);
% Display n first elements
for i=1:3
fprintf('Element %d: %d at row %d and column %d\n', i, B(i), I(i), J(i));
end

4 个评论
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Adam Silva
Adam Silva 2014-1-20
Hi Bruno. Thank you for the quick response. This code almost work. But the elements of my matrix A are complex numbers.
sample element set
-0.0154 -45.4596i -0.0164 -45.4599i -0.0159 -45.4598i -0.0158 -45.4597i
-0.0150 -45.4594i -0.0160 -45.4598i -0.0155 -45.4596i -0.0154 -45.4596i
-0.0136 -45.4589i -0.0147 -45.4592i -0.0142 -45.4590i -0.0141 -45.4590i
-0.0160 -45.4599i -0.0171 -45.4602i -0.0165 -45.4600i -0.0164 -45.4600i
-0.0172 -45.4604i -0.0183 -45.4607i -0.0177 -45.4605i -0.0176 -45.4605i
-0.0099 -45.4573i -0.0110 -45.4576i -0.0105 -45.4575i -0.0104 -45.4575i
-0.0058 -45.4556i -0.0069 -45.4559i -0.0063 -45.4558i -0.0062 -45.4558i
-0.0074 -45.4561i -0.0085 -45.4564i -0.0080 -45.4562i -0.0079 -45.4562i
0.0260 -45.4466i 0.0249 -45.4470i 0.0254 -45.4468i 0.0255 -45.4468i
0.0249 -45.4470i 0.0477 -45.4405i 0.0243 -45.4471i 0.0244 -45.4471i
0.0254 -45.4468i 0.0243 -45.4471i 0.0374 -45.4435i 0.0250 -45.4469i
0.0255 -45.4468i 0.0244 -45.4471i 0.0250 -45.4469i 0.0353 -45.4440i
when I used your code the answers were in the very end n elements. How can we change the code to select those elements.
Azzi Abdelmalek
Azzi Abdelmalek 2014-1-20
You have to explain how do you want to sort your complex number, by real part? by modulus? ...
Adam Silva
Adam Silva 2014-1-20
Sorry for the incomplete question.
Complex number should be sort "by modules"
Bruno Pop-Stefanov
Bruno Pop-Stefanov 2014-1-21
编辑：Bruno Pop-Stefanov 2014-1-21
By default, sort orders complex numbers by modulus first, and, if two numbers have same modulus, it orders them by angle.
If you want to sort your list by angle only, then use sort on angle(A(:)) instead:
% Input matrix
A = [-0.0154-45.4596i, -0.0164-45.4599i, -0.0159-45.4598i; ...
-0.0150-45.4594i, -0.0160-45.4598i, -0.0155-45.4596i; ...
-0.0136-45.4589i, -0.0147-45.4592i, -0.0142-45.4590i];
% Sort. Transform A into a vector with (:)
[B,IX] = sort(angle(A(:)), 'ascend');
% Convert linear indices to subscript
[I,J] = ind2sub(size(A),IX);
% Display n first elements
for i=1:3
fprintf('Element %d: %d at row %d and column %d\n', i, B(i), I(i), J(i));
end

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更多回答（3 个）

Azzi Abdelmalek
Azzi Abdelmalek 2014-1-20

1 个投票

A=[ 1 3 5; 9 7 6; 2 8 4];
B= [7 4 3];
idx=[];
idy=[];
for k=1:numel(B)
ij=ismember(A,B(k));
[ii,jj]=find(ij);
idx=[idx ii];
idy=[idy jj];
end
idx
idy
Andrei Bobrov
Andrei Bobrov 2014-1-21
编辑：Andrei Bobrov 2014-1-21

1 个投票

A = [ 1 3 5; 9 7 6; 2 3 4];
B = [7 4 3];
[a,b] = ismember(A,B);
[r,c] = find(a);
out = accumarray(b(a),1:numel(b(a)),[],@(ii){[r(ii),c(ii)]});
Adam Silva
Adam Silva 2014-1-21

0 个投票

Thank you all for quick responses and help. I got the answer for my assignment and learn some new codes in Matlab.

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