filling nans between two values

Hi,

I have matrices in which I want to fill the nans with the last available number. for example: A=[ 1 nan;nan 3;nan 2;4 nan;nan nan]; I want it to become A=[ 1 nan;1 3;1 2;4 2;4 2]; Is there an easy way of replacing these nans without running a loop. I have large matrices and I need to do this frequently. So,I am wondering if an easy command exists.

Answer 1

[Andrei Bobrov]

A=[ 1 nan;nan 3;nan 2;4 nan;nan nan]
A(A==0)=inf;
A0 = A;
A0(isnan(A0)) = 0;
n0 = A0~=0;
A1 = A0(n0);
idn = cumsum(n0);
idn(idn==0)=nan;
idx = bsxfun(@plus,idn,[0 idn(end,1:end-1)]);
idx2 = idx;
tnn = isnan(idx2);
idx2(tnn)=1;
out = A1(idx2);
out(tnn)=nan;
out(isinf(out))=0;

small corrections

Aoz = A==A;
A1 = A(Aoz);
i1 = cumsum(Aoz);
im = bsxfun(@plus,[0 i1(end,1:end-1)]);
out = A1(im);
out(i1==0)=nan;

last variant

Aoz = A==A;
A1 = A(Aoz);
i1 = cumsum(Aoz);
im = bsxfun(@plus,i1,[0 cumsum(i1(end,1:end-1))]); 
out = A1(im+(im==0));
out(i1==0)=nan;

more variant with use reshape

>> A
A =
   NaN     0   NaN     0
     6   NaN     6     2
     0     3     6   NaN
   NaN     6     3   NaN
   NaN     6     5     6
>> Aoz = A==A;
A1 = A(Aoz);
i2 = reshape(cumsum(Aoz(:)),size(Aoz));
out = A1(i2+(i2==0));
out(cumsum(Aoz)==0)=nan
out =
   NaN     0   NaN     0
     6     0     6     2
     0     3     6     2
     0     6     3     2
     0     6     5     6
>>

[Oleg Komarov]

Vectorized! +1

Answer 2

[Oleg Komarov]

EDITED

Polished and robustified version of Andrei's algorithm:

A = [NaN   NaN
       1   NaN
     NaN     2
       4     0
     NaN   NaN];

% Index non NaNs and store elements

idxEl = ~isnan(A);
nnNaN = [0; A(idxEl)];

% Positions to non NaN elements

pos = cumsum(idxEl);

% Detect initial NaNs

idxNaN = pos == 0;

% Adust pos with offset for number of non NaN elements in each column

offs = cumsum([1 sum(idxEl(:,1:end-1))]);
pos  = bsxfun(@plus,pos,offs);

% Output

B = nnNaN(pos);

% Place back initial NaNs

B(idxNaN) = NaN

[Andrei Bobrov]

Hi Oleg! Thanks a lot for your vote and cod.

I too corrected my last variant

+1.

[Oleg Komarov]

Thanks but all the merit goes to you. No idea, no risk for small corrections :).