Filling the gaps in a vector

I have a vector of nans

A=[nan;nan;2;nan;4;nan;nan;nan;7;nan;nan;nan;nan] how can I fill the nan gaps by the closest number (for beginning and mid values it is the closest upper;for the last values it is the closest lower value). i,e how can reproduce the vector to become A=[2;2;2;4;4;7;7;7;7;7;7;7;7]

Answer 1

[Andrei Bobrov]

A=[nan;nan;2;nan;4;nan;nan;nan;7;nan;nan;nan;nan];
b = ~isnan(A);
k = cumsum(flipud(b));
k(k==0) = 1;
n = flipud(A(b));
s = n(k);
out = flipud(s);

or

t = ~isnan(A);
k = find(t) + 1;
z = zeros(size(A));
z(k(k <= numel(A))) = 1;
q = cumsum(z) + 1;
q(q > nnz(t)) =  max(q) - 1;
p = A(t);
out = p(q);

Answer 2

[Image Analyst]

Do you have the Image Processing Toolbox? If so, you can use imdilate, if you're clever about it.

[joseph Frank]

Currently I don't but I will check with the university I think they have it

Answer 3

[Azzi Abdelmalek]

A=[nan;nan;2;nan;4;nan;nan;nan;7;nan;nan;nan;nan]
B=A;
idx=find(isnan(A));
idx1=fliplr(find(~isnan(A)));
for k=1:numel(idx)
  a=idx(k);
  [~,ii]=min(abs(a-idx1));
  B(idx(k))=A(idx1(ii));
end

[joseph Frank]

there is one additional 4

Answer 4

[Walter Roberson]

You might also be interested in John D'Errico's FEX contribution inpaint_nans