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How to use an index of a vector as a value in another matrix?

조회 수: 1 (최근 30일)
Syed Fahad Hassan
Syed Fahad Hassan 2017년 8월 3일
마감: MATLAB Answer Bot 2021년 8월 20일
Hi, I have a vector x=[1 1 2 2 3] and a matrix [1 3 ; 1 4 ; 2 3 ; 2 4]
The logic that I want to create says that the index of the latter of the similar values is extracted. Then, wherever the matrix has that value, that row becomes zero. For example, the indices of first two similar values in vector x are 1 and 2. I want those rows in the matrix that contains the value 2, should become zero. Same is the case for 3 and 4.
In the end, the matrix should only contain a single row i.e. [1 3]
  댓글 수: 3
Matthew Eicholtz
Matthew Eicholtz 2017년 8월 3일
I think when you say "becomes zero", you mean "is removed", correct?
Syed Fahad Hassan
Syed Fahad Hassan 2017년 8월 4일
Yes Matthew, I want the row to be removed

답변 (3개)

dpb
dpb 2017년 8월 3일
>> x=[1 1 2 2 3] ;
>> m= [1 3 ; 1 4 ; 2 3 ; 2 4];
>> m(any(ismember(m,find(diff(x))),2),:)=[]
m =
1 3
>>

Alex Kerzner
Alex Kerzner 2017년 8월 3일
I'm sure someone can come up with something more clever and efficient, but here's something hack-y that might give you some ideas for improvements. Assuming the matrix is called M :
repeats = [];
while ~isempty(x)
if numel(find(x == x(1))) >= 2 %If we have multiple of them
repeats(end+1) = max(find(x==x(1))); %add the largest index to the list
x(find(x==x(n))) = []; %remove all the duplicates
else
x(1) = []; %If no duplicates, just get rid of it
end
end
for r = repeats
[i,~] = find(M == r);
M(i,:) = [];
end

Matthew Eicholtz
Matthew Eicholtz 2017년 8월 3일
How about this?
Assume you have the inputs:
x = [1,1,2,2,3];
y = [1,3; 1,4; 2,3; 2,4];
Then, you can find indices to remove by:
[~,ind,~] = unique(fliplr(x));
ind = length(x)-ind+1;
Then, remove the rows containing any of those indices:
y(any(ismember(y,ind),2),:) = [];
  댓글 수: 1
Matthew Eicholtz
Matthew Eicholtz 2017년 8월 3일
Note, my approach will remove indices of values that exist only once in x (e.g., the 3). It is unclear from the question whether this should be allowed or not.

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