How can already used elements be eliminated one by one in 2D matrix while moving upwards? | Efficienlty

조회 수: 1 (최근 30일)
Input
input = [1 2 0;
2 3 4;
4 5 6];
Output
Output= [1 0 0;
2 3 0;
4 5 6];
Why efficiently : For inputting a big matrix of say 360000 x 36.

채택된 답변

James Tursa
James Tursa 2018년 11월 3일
편집: James Tursa 2018년 11월 3일
r = rot90(inputMatrix,-1); % turn rows into columns with last row as first column
[~,ix] = unique(r); % get the unique linear indexing with preference to leftmost columns
result = zeros(size(r)); % initialize result as all 0's
result(ix) = r(ix); % fill in the unique values with preference to leftmost columns
result = rot90(result); % rotate back to original orientation

추가 답변 (1개)

Image Analyst
Image Analyst 2018년 11월 3일
Use unique():
inputMatrix = [1 2 0;
2 3 4;
4 5 6]
[~, ia, ic] = unique(inputMatrix);
output = zeros(size(inputMatrix)); % Initialize
output(ia) = inputMatrix(ia)
  댓글 수: 4
Walter Roberson
Walter Roberson 2018년 11월 3일
That output contradicts the original pattern. For that "should be" output to be true, then the output for
input = [1 2 0;
2 3 4;
4 5 6];
should be
[1 2 0;
3 4 0;
5 6 0;]
JAI PRAKASH
JAI PRAKASH 2018년 11월 3일
편집: JAI PRAKASH 2018년 11월 3일
@James Tursa
"Moving elements to the left" is not a mandatory requirement.
inputMatrix = [1 2 0;
4 3 1;
4 5 6]
outputCouldBe =
0 2 0
0 3 1
4 5 6
Above output matrix also holds good :)
At the end, more priority given to bottom rows is what required.
I mean, the last row's elements are unchanged. 2nd last row's common elements turn to zero. 3rd last row's common elements(with 2nd last and very last rows) turn to zero.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Programming에 대해 자세히 알아보기

태그

제품


릴리스

R2018b

Community Treasure Hunt

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

Start Hunting!

Translated by