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

1 vue (au cours des 30 derniers jours)
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.

Réponse acceptée

James Tursa
James Tursa le 3 Nov 2018
Modifié(e) : James Tursa le 3 Nov 2018
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

Plus de réponses (1)

Image Analyst
Image Analyst le 3 Nov 2018
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 commentaires
Walter Roberson
Walter Roberson le 3 Nov 2018
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 le 3 Nov 2018
Modifié(e) : JAI PRAKASH le 3 Nov 2018
@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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