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How to remove repeating rows without unique function?

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Andrew Poissant
Andrew Poissant 2017 年 9 月 25 日
コメント済み: Cedric 2017 年 9 月 26 日
I have a 624x2 matrix of ordered pairs, of which there are some repeating rows that I must remove. Is there a way to remove these repeating rows without using the unique function? Maybe using a for loop? The reason I can't use the unique function is because when I implement this matlab script in Simulink's Matlab Function block, the matrix must be sorted first before using the unique function. I can't sort the matrix because I will lose my ordered pairs and sortrows doesn't solve the issue either because I get the same error from Simulink.

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Cedric
Cedric 2017 年 9 月 25 日
編集済み: Cedric 2017 年 9 月 25 日
Use the 'stable' option:
>> A = randi( 3, 10, 2 )
A =
2 3
1 1
3 1
3 1
3 1
3 3
3 3
2 1
2 3
1 1
>> unique( A, 'rows' )
ans =
1 1
2 1
2 3
3 1
3 3
>> unique( A, 'rows', 'stable' )
ans =
2 3
1 1
3 1
3 3
2 1
  8 件のコメント
Andrew Poissant
Andrew Poissant 2017 年 9 月 26 日
It worked! Thank goodness. Really appreciate all of the help you gave me, you rock!
Cedric
Cedric 2017 年 9 月 26 日
My pleasure. The amount of unsupported features seems to be a real pain, good luck with the rest!

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その他の回答 (1 件)

Jan
Jan 2017 年 9 月 26 日
編集済み: Jan 2017 年 9 月 26 日
I still do not understand, why you cannot simply sort the data. But this might be useful:
!!! UNTESTED !!!
function Data = UniqueRows2(Data)
nRow = size(Data, 1);
keep = true(nRow, 1);
for k1 = 1:nRow
if keep(k1)
for k2 = k1 + 1:nData
if keep(k2)
% For rows of length 2:
keep(k2) = (Data(k2, 1) ~= Data(k1, 1)) | (Data(k2, 2) ~= Data(k1, 2));
% General row length:
% keep(k2) = any(bsxfun(@ne, Data(k2, :), Data(k1, :)), 2);
end
end
end
end
Data = Data(keep, :);
end
This might be faster with vectorization:
function Data = UniqueRows(Data)
nRow = size(Data, 1);
keep = true(nRow, 1);
for k1 = 1:nRow
if keep(k1)
keep(k1 + 1:nData) = and(keep(k1 + 1:nData), ...
any(bsxfun(@ne, Data(k2, :), Data(k1, :)), 2));
end
end
Data = Data(keep, :);
end
  1 件のコメント
Andrew Poissant
Andrew Poissant 2017 年 9 月 26 日
Thanks! This also worked well for me!

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