Replace elements of matrix

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

[Mike Lynch]

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 ...

Answer 1

[Youssef Khmou]

hi, try ;

 F=matrix;
 for i=1:length(vector)
   F(F==vector(i))=0;
 end

[Ionut Anghel]

Thank you. Very efficient

[BJ Anderson]

See the comments below about changem, which is the most versatile and efficient tool for such problems.

In general, you should be avoiding loops in MATLAB, and changem is the thought-out implementation that does just that.

Another option that avoids a loop is 'ismember':

matrix(ismember(matrix,vector))=0;

The ismember function in this case returns a logical array the same size as matrix, with 1s wherever it finds a member of vector, and zero otherwise. It is implemented here as an indexing array, so now all elements in matrix where the above condition is true is set to 0--or whatever arbitrary value you need.

[Stephen23]

"In general, you should be avoiding loops in MATLAB"

Why ?

[Joel Bly]

@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]

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)

Answer 2

[Metin Ozturk]

The more vectorized and easier way to do this could be as follows:

new_matrix = changem(matrix,zeros(length(vector),1),vector);

[BJ Anderson]

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]

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.

Answer 3

[Bruno Luong]

F = matrix .* ~ismember(matrix,vector)

[Bruno Luong]

A variant is:

matrix(ismember(matrix,vector)) = 0

Answer 4

[Walter Roberson]

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) ?

Answer 5

[Image Analyst]

Try it this way:

newMatrix = matrix % Initialize
for k = 1 : length(vector)
  newMatrix(matrix==vector(k)) = 0
end

[Ionut Anghel]

Thanks It is working as well