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

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

[Torsten]

What if there is another 0 left to it ? Or if the 0 is in the first column ?

[Wietze Zijpp]

Good Question, One way to deal with this could be to search for a value which is not and then replace the zeros with that value

Answer 1

[Star Strider]

One approach using fillmissing —

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.

.

[DGM]

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