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1

Delete number in two matrices with same indices/position

Asked by lhyin5277 on 3 Feb 2012
Hi... How can i delete numbers in two different matrices with same position? Example: A = [1 2 4 5;2 4 5 6; 3 7 8 2]; B = [3 4 5 6;0 10 4 1;4 12 34 56];
then if i want remove value 2 in matrix A, in the same time i need to delete the value at same position in matrix B.
the answer will get are: A = [1 4 5;4 5 6; 3 7 8]; B = [3 5 6;10 4 1;4 12 34];
Thank you...

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2 Answers

Answer by Walter Roberson
on 3 Feb 2012

At = A .';
Bt = B .';
pos = find(At(:) == 2);
At(pos) = [];
Bt(pos) = [];
A = reshape(At, size(A,2), []) .';
B = reshape(Bt, size(B,2), []) .';
And do not be surprised if the reshape step fails, as in the general case there is no reason to believe that their will be exactly the same number of matches per row.
Note: when you delete an element in MATLAB, MATLAB "pulls up" elements from below in the same column . You want to "pull over" from the same row. That is why the two transposes are needed.

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Answer by Kevin Moerman on 3 Feb 2012

If A and B are the save size then instead of the "find" function (which produces linear or subscript indices) use logic indexing for speed e.g.:
L=A==2; A(L)=[]; B(L)=[];
Reshape is not required (and is slow for large arrays). The approach above and that of Mr. Roberson only work if the empty entries of [] lead to the removal of say always n colums or n rows. In other cases you'll get an error. You could use sparse arrays and set the entries to zero instead or use NaN.
Good luck, Kevin

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True, logical indexing should be a bit faster.
The reshaping is required, however. Deleting assorted array elements given by logical index or linear index, _always_ produces a vector output.
You also did not take in to account that indexing proceeds down columns so entries move "up" to fill the holes, but the poster's example clearly wants elements to move "left" to fill the holes.
"Reshape is not required (and is slow for large arrays)."
This is VERY not true! RESHAPE() is one of the fastest functions around. All it does is change header information in the way a file is stored. The only time it is _at all_ time consuming is with large sparse arrays. Let's do an example:
A = rand(1000,1000,200);
tic
for ii =1:100;
A = reshape(A,numel(A),1);
A = reshape(A,200,1000,1000);
A = reshape(A,1000,200,1000);
A = reshape(A,100,100,[]);
end
toc
%{
Elapsed time is 0.001147 seconds.
Elapsed time is 0.001181 seconds.
%}
So it did 400 reshapes of 1.6Gb array in less than two thousands of a second!!
(I originally had an 8Gb array but it dumped my system ;) )

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