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How to use an index of a vector as a value in another matrix?
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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]
3 件のコメント
Matthew Eicholtz
2017 年 8 月 3 日
I think when you say "becomes zero", you mean "is removed", correct?
回答 (3 件)
dpb
2017 年 8 月 3 日
>> 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
>>
1 件のコメント
Alex Kerzner
2017 年 8 月 3 日
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
1 件のコメント
Matthew Eicholtz
2017 年 8 月 3 日
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),:) = [];
1 件のコメント
Matthew Eicholtz
2017 年 8 月 3 日
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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