필터 지우기
필터 지우기

From a matrix, Pick one element from each row and not from same column for all possibilities. Only for non zero elements.

조회 수: 2 (최근 30일)
Hi, From a matrix, I would like to pick one element from each row and not from same column for all possibilities.
Only for non zero elements.
Example, from A = [1 2 0; 1 2 0; 0 0 3], I want to obtain (1 2 3 and 2 1 3).
Thank you

채택된 답변

Puru Kathuria
Puru Kathuria 2020년 7월 22일
Hi,
Your requirements are answered in a similar question mentioned in this link. Also, selecting set of elements from a matrix such that each element comes from distinct row and column can also be modeled as a Matching problem. (graph theory)

추가 답변 (1개)

Bruno Luong
Bruno Luong 2020년 7월 23일
편집: Bruno Luong 2020년 7월 23일
A dynamic programming for fun
function p = nzperm(A,c)
if nargin < 2
c = 1;
end
r = find(A(:,c));
if isempty(r)
p = [];
elseif c==size(A,2)
p = A(r,c); % p = r(:);
else
p = [];
for i=r.'
Ai = A;
Ai(i,:) = 0;
q = nzperm(Ai,c+1);
a = A(i,c); % a = i
q = [a+zeros(size(q,1),1) q];
p = [p; q]; %#ok
end
end
Result
>> A=rand(5); A = A.*(rand(size(A))>0.6)
A =
0 0.6468 0.7881 0.5598 0.8008
0.5607 0 0 0 0.8961
0 0 0 0.9394 0.5975
0 0.4756 0.1335 0 0
0 0.3625 0.0216 0 0.9437
>> nzperm(A)
ans =
0.5607 0.6468 0.1335 0.9394 0.9437
0.5607 0.4756 0.7881 0.9394 0.9437
0.5607 0.4756 0.0216 0.5598 0.5975
0.5607 0.4756 0.0216 0.9394 0.8008
0.5607 0.3625 0.1335 0.5598 0.5975
0.5607 0.3625 0.1335 0.9394 0.8008
Note: this code will be slow on large array, since MATLAB make copies of matrix at every step. This algorithm could be implemented without copy using C mex.
  댓글 수: 5
MeMariah
MeMariah 2020년 9월 20일
@Bruno, you are right. large array take long. Unfortunately, I keep getting error when I try to generate the C mex file. Am using this app for the first time. Are able to help? error points to the line with
for i= r.'
Thanks.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Characters and Strings에 대해 자세히 알아보기

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by