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

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JAI PRAKASH
JAI PRAKASH 2018-11-2
评论： JAI PRAKASH 2018-11-3
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.

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采纳的回答

James Tursa
James Tursa 2018-11-3
编辑：James Tursa 2018-11-3
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

更多回答（1 个）

Image Analyst
Image Analyst 2018-11-3
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 个评论
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Walter Roberson
Walter Roberson 2018-11-3
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 2018-11-3
编辑：JAI PRAKASH 2018-11-3
@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.

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