# Filling the gaps in a vector

1 view (last 30 days)
joseph Frank on 23 Jan 2013
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]

Andrei Bobrov on 23 Jan 2013
Edited: Andrei Bobrov on 23 Jan 2013
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);

Image Analyst on 23 Jan 2013
Do you have the Image Processing Toolbox? If so, you can use imdilate, if you're clever about it.
joseph Frank on 23 Jan 2013
Currently I don't but I will check with the university I think they have it

Azzi Abdelmalek on 23 Jan 2013
Edited: Azzi Abdelmalek on 23 Jan 2013
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 on 24 Jan 2013

Walter Roberson on 23 Jan 2013
You might also be interested in John D'Errico's FEX contribution inpaint_nans