Replace values in matrix by first non-zero value in previous row

Hi, how can i replace zero values in a matrix by the first occurring non-zero value in that matrix if you would loop backwards in each column? I.e. how can i carry forward a value as long as its next value in the same column is zero (without looping through each value in the matrix!)

Example: m_start = [3;0;0;4;0;5] replace_non_zero_values (m_start) = [3;3;3;4;4;5]

Hope you can help, thanks very much! Steven

Answer 1

[Stephen23]

>> M = [3;0;0;4;0;5];
>> idx = M~=0;
>> tmp = M(idx);
>> tmp(cumsum(idx))
ans =
   3
   3
   3
   4
   4
   5

Note this assumes that the first value is nonzero. You will need to think of how to deal with leading zeros!

[Steven Niggebrugge]

brilliant! thanks

[Steven Niggebrugge]

Could you help me with implementing this for a matrix? i.e. for M = [3 4;0 2;0 0;5 0] which should result in M_adj = [3 4; 3 2; 3 2;5 2]

thanks!

[Steven Niggebrugge]

have to come up with a solution if a value in first row is 0, in that case we need to override this with 1 in the idx matrix...otherwise M(idx) gives error message. 0 in first row also needs to be carried forward until a non-zero value appears.

[Guillaume]

idx = M(:) ~= 0
tmp = M(idx);
reshape(tmp(cumsum(idx)), size(M))

As with the original answer, this will go badly wrong if any column starts with a 0.

[Arturo Camacho Lozano]

Stephen's solution is very clever. Thanks!