Removing repeated numbers in a matrix

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AA
AA 2020 年 8 月 23 日
コメント済み: Bruno Luong 2020 年 8 月 23 日
Ex: In a matrix
A = [9 9 1 1 2 2 2 2 2 0 3 3; 7 7 4 4 4 5 5 6 6 6 6 6 ; 8 8 7 7 7 8 8 8 9 9 9 5]
, i want to eliminate all repeating elements and retain non-repeating elements
i.e. from the above example
A = [9 1 2 0 3 ; 7 4 5 6 ; 8 7 8 9 5]
To prevent a dimensions error because of different output results, the remaining entries on the left can be filled by zero like here
A = [9 1 2 0 3 ; 7 4 5 6 0 ; 8 7 8 9 5]

採用された回答

Stephen23
Stephen23 2020 年 8 月 23 日
編集済み: Stephen23 2020 年 8 月 23 日
>> A = [9,9,1,1,2,2,2,2,2,0,3,3;7,7,4,4,4,5,5,6,6,6,6,6;8,8,7,7,7,8,8,8,9,9,9,5]
A =
9 9 1 1 2 2 2 2 2 0 3 3
7 7 4 4 4 5 5 6 6 6 6 6
8 8 7 7 7 8 8 8 9 9 9 5
>> X = diff(A(:,[1,1:end]),1,2)~=0;
>> X(:,1) = true;
>> S = size(A);
>> [R,~] = ndgrid(1:S(1),1:S(2));
>> C = cumsum(+X,2);
Method one: accumarray:
>> B = accumarray([R(:),C(:)],A(:),[],@mode,0)
B =
9 1 2 0 3
7 4 5 6 0
8 7 8 9 5
Method two: indexing:
>> B = zeros(S(1),max(C(:)));
>> B(sub2ind(S,R,C)) = A
B =
9 1 2 0 3
7 4 5 6 0
8 7 8 9 5

その他の回答 (1 件)

KSSV
KSSV 2020 年 8 月 23 日
編集済み: KSSV 2020 年 8 月 23 日
A = [9 9 1 1 2 2 2 2 2 0 3 3; 7 7 4 4 4 5 5 6 6 6 6 6 ; 8 8 7 7 7 8 8 8 9 9 9 5] ;
[m,n] = size(A) ;
C = cell(m,1) ;
for i = 1:m
C{i} = unique(A(i,:)) ;
end
L = cellfun(@length,C) ;
B = zeros(m,max(L)) ;
for i = 1:m
B(i,1:L(i)) = C{i} ;
end
  4 件のコメント
Stephen23
Stephen23 2020 年 8 月 23 日
編集済み: Stephen23 2020 年 8 月 23 日
"Easy to fix with 'stable' option"
Nope, not fixed. Lets try it:
>> c = arrayfun(@(r) unique(A(r,:),'stable'), 1:size(A,1), 'unif', 0);
>> n = max(cellfun(@length,c));
>> B = cell2mat(cellfun(@(x) [x,zeros(1,n-length(x))], c, 'unif', 0)')
B =
9 1 2 0 3
7 4 5 6 0
8 7 9 5 0
Note that the last row differs from the expected output given in the original question:
9 1 2 0 3
7 4 5 6 0
8 7 8 9 5
Bruno Luong
Bruno Luong 2020 年 8 月 23 日
You are absolutely right.

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