Replace elements of matrix
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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 个评论
Mike Lynch
2020-11-24
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 ...
采纳的回答
Youssef Khmou
2013-3-3
hi, try ;
F=matrix;
for i=1:length(vector)
F(F==vector(i))=0;
end
5 个评论
Joel Bly
2019-4-11
@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
2019-4-12
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)
更多回答(4 个)
Metin Ozturk
2018-8-1
The more vectorized and easier way to do this could be as follows:
new_matrix = changem(matrix,zeros(length(vector),1),vector);
2 个评论
BJ Anderson
2019-3-12
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
2019-3-12
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.
Walter Roberson
2013-3-3
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) ?
0 个评论
Image Analyst
2013-3-3
Try it this way:
newMatrix = matrix % Initialize
for k = 1 : length(vector)
newMatrix(matrix==vector(k)) = 0
end
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