Replacing specific values in each matrix row in all values in the matrix

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I have a matrix as follows A=[8 66 92 101 0; 14 52 80 76 0; 16 66 33 51 7]. I want to read the rows column by column and assign all values in the matrix the first value of the row, my code so far works (I have also made an if loop to ignore the zeros), and the output is B=[8 8 8 8 0; 14 14 14 14 0; 16 16 16 16 16], however this isn't exactly what I need. For example, the matrix I need is B=[8 8 8 8 0: 14 14 14 14 0; 8 8 8 8 8] since A(1,2)=66 and this should have already been assigned to the value 8. So what I am looking for is to iterate over each row and column, assign all numbers in the row to the same value throughout the matrix, and whenever the value appears in a row later on the whole row again be assigned to the value, 8, in this example. I hope that this is clear enough. So far, the code is:
F=Locations;
indices = find(F(:,1)==0);
F(indices,:) = [];
for l=1:size(F,1);
for k=1:size(F,2);
if F(l,k)~=0;
[R,L]=find(F==F(l,k));
F(R,L)=F(l,1);
end
end
end

Akzeptierte Antwort

Stephen23
Stephen23 am 1 Feb. 2016
Bearbeitet: Stephen23 am 1 Feb. 2016
Try this:
A = [8 66 92 101 0; 14 52 80 76 0; 16 66 33 51 7; 1 76 66 0 0]
% get unique elements, row-wise:
[~,X,Y] = unique(A.','first');
% indices where elements first appear:
S = size(A.');
[C,R] = ind2sub(S,X(Y));
R = reshape(R,S)
% for each row find "first" appearance:
R(0==A.') = Inf;
D = A(min(R,[],1));
% create output array using "first" values:
out = bsxfun(@times,D(:),cumprod(+(A>0),2))
creates this:
out =
8 8 8 8 0
14 14 14 14 0
8 8 8 8 8
8 8 8 0 0
  4 Kommentare
Stephen23
Stephen23 am 1 Feb. 2016
Bearbeitet: Stephen23 am 1 Feb. 2016
Please see my edited question.
Also note that in case a row contains more than one value from other lines, the highest row wins (not the first value). If you want to match the first element, then replace the line that defines D with these three lines:
E = ~Z & 0~=diff([1:S(2);R],1,1);
E(1,:) = all(E==0,1);
D = A(R(E & cumsum(E,1)==1))
creates:
out =
8 8 8 8 0
14 14 14 14 0
8 8 8 8 8
14 14 14 0 0
Matlab User
Matlab User am 1 Feb. 2016
Thank you very much, it is very much appreciated!

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