Change the value of a matrix according to the indexes stored in another one.

조회 수: 6 (최근 30일)
Hey, goodies, let's pretend I have the next matrix.
Z = zeros(5,6);
And in another matrix called idx I have the information of which rows and columns of Z I want to change the value, for example:
idx = [1 1; 2 5; 4 6]
Being the first column the position of the columns I want to change and the value of the second idx column the position of the rows I want to change, and if the value i want to obtein in Z in idx position is 5, what i expect to have is:
This is just an example, in the realization I have many values that I want to change, therefore the option:
Z(1,1) = 5;
Z(2,5) = 5;
Z(4,6) = 5;
Is not an option
I'm looking for something that allows me to do it automatically, without the need for loops, if anyone knows it would help me a lot, thanks.

채택된 답변

KSSV
KSSV 2020년 6월 2일
편집: KSSV 2020년 6월 2일
Read about sub2ind
Z =zeros(5,6) ;
idx = [1 1; 2 5; 4 6] ;
idx = sub2ind(size(Z),idx(:,1),idx(:,2)) ;
Z(idx) = 5

추가 답변 (1개)

Chris Angeloni
Chris Angeloni 2020년 6월 2일
편집: Chris Angeloni 2020년 6월 2일
You probably want to get the index as a linear subscript instead of row,column, then index a vectorized version of your original matrix.
Z = zeros(5,6);
idx = [1 1; 2 5; 4 6];
% save size of Z
sz = size(Z);
% vectorize Z
Z = Z(:);
% make row,col index to linear index
lInd = sub2ind(sz,idx(:,1),idx(:,2))
Z(lInd) = 5;
% resize Z to original size
Z = reshape(Z,sz(1),sz(2));
  댓글 수: 2
Alejandro Fernández
Alejandro Fernández 2020년 6월 2일
Thank you very much for the answer, it works, however KSSV's answer is much smaller in terms of code.
Chris Angeloni
Chris Angeloni 2020년 6월 2일
Yes, I saw it after! I forgot you can use the linear index to index the original matrix

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