Replace elements of matrix

13 vues (au cours des 30 derniers jours)
Ionut  Anghel
Ionut Anghel le 3 Mar 2013
Commenté : Mike Lynch le 24 Nov 2020
I have the folowing:
vector=[1 3 8 9];
matrix=[ 100 1 5 9 6; 100 10 13 3 8; 100 9 10 1 4; ];
% I want to search and replace the vector element with "0"in the matrix (i.e new matrix should be : Newmatrix=[ 100 0 5 0 6; 100 10 13 0 0; 100 0 10 0 4; ]; )
The script is:
Newmatrix=zeros(size(matrix));
for i=1:numel(matrix)
for j=1:length(vector)
valvect=vector(j);
if matrix(i)==valvect
Newmatrix(i)=0;
else
Newmatrix(i)=matrix(i);
end
end
end
The results is not the desired one but:
Newmatrix=100 1 5 0 6
100 10 13 3 8
100 0 10 1 4
So what I'm doing wrong?
Thank you
  1 commentaire
Mike Lynch
Mike Lynch le 24 Nov 2020
The accepted answer or changem are cleaner and more compact, but for the code you wrote the addition of a "break" should fix the problem.
if matrix(i)==valvect
Newmatrix(i)=0;
break
else ...

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Réponse acceptée

Youssef  Khmou
Youssef Khmou le 3 Mar 2013
hi, try ;
F=matrix;
for i=1:length(vector)
F(F==vector(i))=0;
end
  5 commentaires
Joel Bly
Joel Bly le 11 Avr 2019
@BJAnderson how do I change the 1s that come from the ismember function back to the values they are referring to, without changing the zeros?
Walter Roberson
Walter Roberson le 12 Avr 2019
If you use ismember() with two outputs, then the second output is the index at which the element in the first parameter appears in the second parameter. In places that the element does not occur, then the returned value will be 0 there. You can select just the valid indices by indexing the second output by the first output.
[is_it_there, idx] = ismember(A, B);
idx(is_it_there)

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Plus de réponses (4)

Metin Ozturk
Metin Ozturk le 1 Août 2018
The more vectorized and easier way to do this could be as follows:
new_matrix = changem(matrix,zeros(length(vector),1),vector);
  2 commentaires
BJ Anderson
BJ Anderson le 12 Mar 2019
YES YES YES. It looks like hardly anyone knows about changem...I see similar questions asked and typically the "solution" involves a loop. A loop in MATLAB is the first sign you're doing something wrong.
changem covers so many situations, and it is elegant and concise, not to mention avoids the risk of re-reassigning an element.
changem is almost always the right anwer. Let the world know!
Thanks Metin for helping to spread the word.
BJ Anderson
BJ Anderson le 12 Mar 2019
A quick update on changem:
Sadly, if one inspects the actual code within changem, it functions as a loop. While it is a handy one-liner, it does not have the time-savings of moving from a looped function to an matrix-operation function.

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Bruno Luong
Bruno Luong le 12 Mar 2019
F = matrix .* ~ismember(matrix,vector)
  1 commentaire
Bruno Luong
Bruno Luong le 13 Mar 2019
A variant is:
matrix(ismember(matrix,vector)) = 0

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Walter Roberson
Walter Roberson le 3 Mar 2013
Suppose you set Newmatrix to 0 because matrix matched vector(1). Now what happens when you go on to the next j to test if matrix matched vector(2) ?

Image Analyst
Image Analyst le 3 Mar 2013
Try it this way:
newMatrix = matrix % Initialize
for k = 1 : length(vector)
newMatrix(matrix==vector(k)) = 0
end
  1 commentaire
Ionut  Anghel
Ionut Anghel le 3 Mar 2013
Thanks It is working as well

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