Replace NaN with previous values

Hello. I have the following issue and i've spent an entire working on it and i never was able to solve it. Now lets say suppose we have a array such as

A =

     NaN   5     6     7     8
      32   NaN   NaN   21    NaN
      NaN   0    12    NaN   6
      34   NaN   NaN   NaN   NaN
      1    24    52     52    44
     NaN   0      2     4      1
     NaN   NaN    3     1      4

^The above is just an example, the data array which i got is large (118x17) to be precise and it has NaN's filled everywhere (like the example of 'A')

Okay so i basically want to copy the value below 'NaN' and replace that value for NaN. And where NaN is at the bottom on the column, i need to copy the above value and replace it with 'NaN'. In short the end result should look like this

A =

      32   5    6     7     8
      32   0    12    21    6
      34   0    12    52    6
      34   24   52    52    44
      1    24   52    52    44
      1    0    2     4     1
      1    0    3     1     4

the code also should be able to work with any kind of array given, not only the example given.

Please please suggest some code, im desperate here.

Thanks in advance,

Answer 1

[Chad Greene]

Use repnan on each column:

A = [NaN   5     6     7     8;
      32   NaN   NaN   21    NaN;
      NaN   0    12    NaN   6;
      34   NaN   NaN   NaN   NaN;
      1    24    52     52    44;
     NaN   0      2     4      1;
     NaN   NaN    3     1      4;]

for k = 1:size(A,2)
    A(:,k) = repnan(A(:,k),'next'); 
    A(:,k) = repnan(A(:,k),'previous'); 
end

A =
  32     5     6     7     8
  32     0    12    21     6
  34     0    12    52     6
  34    24    52    52    44
   1    24    52    52    44
   1     0     2     4     1
   1     0     3     1     4

[Chad Greene]

The first call replaces each NaN with the next finite value, and the second call replaces the leftover NaNs with the previous finite value.

[Chad Greene]

You want loops? We got 'em.

for k = 1:size(A,2) % k moves left to right
  LastReal = NaN; 
  for n = size(A,1):-1:1 % n moves bottom to top
      if isfinite(A(n,k))
          LastReal = A(n,k); 
      else
          A(n,k) = LastReal; 
      end
  end

    LastReal = NaN; 
    for m =1:size(A,1)% m moves top to bottom
        if isfinite(A(m,k))
            LastReal = A(m,k); 
        else
            A(m,k) = LastReal; 
        end
    end
end

It's not the most efficient, but for 118x17 it should be fine.