Replacing a zero value of a matrix with the value left to it
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Wietze Zijpp
el 6 de Abr. de 2022
Suppose I have a matrix
A =
8 1 6
3 0 7
4 9 2
Now I would like the 0 value to be replaced by the value 3 (the value left of it).
This is an illustrative example for a small matrix. I would like to apply this approach to a much larger matrix
2 comentarios
Torsten
el 6 de Abr. de 2022
What if there is another 0 left to it ? Or if the 0 is in the first column ?
Respuesta aceptada
Star Strider
el 6 de Abr. de 2022
A = [8 1 6
3 0 7
4 9 2];
A(A==0) = NaN;
B = fillmissing(A,'previous',2) % With Original Matrix
A = [8 1 6
0 0 7
4 9 2];
A(A==0) = NaN;
B = fillmissing(A,'nearest',2) % With Matrix With First Two Rows Being Zero
It would be necessary to use an if block to test for the second condition and then choose the 'nearest' method for that condition. That could go something like this:
A = [8 1 6
0 0 7
4 9 2
5 0 0];
B = A; % Create Result MAtrix
B(B==0) = NaN;
for k = 1:size(A,1)
ixr = find(A(k,:)==0);
if ~isempty(ixr) & any(ixr==1)
B(k,:) = fillmissing(B(k,:),'nearest',2);
elseif ~isempty(ixr) & ~any(ixr==1)
B(k,:) = fillmissing(B(k,:),'previous',2);
end
end
A
B
The loop, and testing each row, seems to me to be the only way to code this, considering that two different interpolation methods are required, depending on the position of the zeros in each row.
.
1 comentario
DGM
el 6 de Abr. de 2022
Editada: DGM
el 6 de Abr. de 2022
Well this is way neater than what I came up with, but I can offer this bit of a modification from what I was doing:
A = [8 1 6
0 3 7 % this row uses 'nearest'
4 9 2
5 0 0];
B = [fliplr(A) A]; % book-matched array
B(B==0) = NaN;
B = fillmissing(B,'previous',2);
B = B(:,size(A,2)+1:end) % trim off excess
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