Replacing a zero value of a matrix with the value left to it

A = [8     1     6
     3     0     7
     4     9     2];
A(A==0) = NaN;
B = fillmissing(A,'previous',2)                     % With Original Matrix
B = 3×3
     8     1     6
     3     3     7
     4     9     2
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
B = 3×3
     8     1     6
     7     7     7
     4     9     2

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
A = 4×3
     8     1     6
     0     0     7
     4     9     2
     5     0     0
B
B = 4×3
     8     1     6
     7     7     7
     4     9     2
     5     5     5

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.

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DGM 2022-4-6

编辑：DGM 2022-4-6

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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
B = 4×3
     8     1     6
     3     3     7
     4     9     2
     5     5     5