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How can I find and remove the nonzero duplicates in each column of a matrix

3 vues (au cours des 30 derniers jours)
X is a n-by-n matrix of integers ranging from 0 to n. I want to find nonzero duplicate entries in each column and remove them.
  2 commentaires
Image Analyst
Image Analyst le 4 Juil 2016
Modifié(e) : Image Analyst le 4 Juil 2016
Unless there is exactly the same number of elements to remove in each column, you can't. For example, you can't "remove" 3 elements from column 1 and 8 elements from column 2. Can you give an example of input and output and how you used unique() to try to solve it?
Matt Talebi
Matt Talebi le 6 Juil 2016
Modifié(e) : per isakson le 6 Juil 2016
The number of duplicates in each column is either 1 or none. Also the duplicate, if exists, is always the same integer as the column number. Example:
X = [ 1 2 3 4 5
2 9 5 3 8
7 5 4 0 1
6 7 3 2 0
3 1 6 7 9 ];
As can be seen in the 3rd column, 3 is a duplicate. Also in my real data set, the first row is always the column number (similar to this example). If it's not too much to ask I want Y to be:
Y = [ 1 2 5 4 5
2 9 4 3 8
7 5 6 0 1
6 7 0 2 0
3 1 0 7 9 ];
where duplicates removed by shifting one element up and adding two zeros at the end to balance the matrix (the order of numbers should be preserved). Otherwise, I think it should be still fine for me to be able to just identify columns with duplicate and then remove them manually. Thank you for your time!

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Réponse acceptée

per isakson
per isakson le 6 Juil 2016
Modifié(e) : per isakson le 6 Juil 2016
Given
  • "matrix of integers"
  • "the first row is always the column number"
  • "the duplicate, if exists, is always the same integer as the column number"
Try this
X = [ 1 2 3 4 5
2 9 5 3 8
7 5 4 0 1
6 7 3 2 0
3 1 6 7 9 ];
Y = nan( size(X) );
for jj = 1 : size( X, 2)
isdub = X( :, jj ) == jj;
if any( isdub(2:end) )
col = X(:,jj);
col( isdub ) = [];
Y(:,jj) = cat( 1, col, zeros(sum(isdub),1) );
else
Y(:,jj) = X(:,jj);
end
end
result
>> Y
Y =
1 2 5 4 5
2 9 4 3 8
7 5 6 0 1
6 7 0 2 0
3 1 0 7 9
>>
This code trades performance for readability.
&nbsp
Requirement of comment: "modify the codes ... keep the one in the first row and only remove the other one"
Y = nan( size(X) );
for jj = 1 : size( X, 2)
col = X(2:end,jj);
isdub = col == jj;
if any( isdub )
col( isdub ) = [];
Y(:,jj) = cat( 1, jj, col, zeros(sum(isdub),1) );
else
Y(:,jj) = X(:,jj);
end
end
result
>> Y
Y =
1 2 3 4 5
2 9 5 3 8
7 5 4 0 1
6 7 6 2 0
3 1 0 7 9
  2 commentaires
Matt Talebi
Matt Talebi le 6 Juil 2016
Thanks Per, it works flawlessly! Just as a minor modification can you possibly modify the codes such that, once a duplicate detected in a column, keep the one in the first row and only remove the other one (to preserve the column numbers).
Matt Talebi
Matt Talebi le 7 Juil 2016
All good now, thanks a lot!

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