How can already used elements be eliminated one by one in 2D matrix while moving upwards? | Efficienlty
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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.
0 件のコメント
採用された回答
James Tursa
2018 年 11 月 3 日
編集済み: James Tursa
2018 年 11 月 3 日
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
その他の回答 (1 件)
Image Analyst
2018 年 11 月 3 日
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)
4 件のコメント
Walter Roberson
2018 年 11 月 3 日
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;]
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