Fill in missing NaNs
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I am trying to fill these NaNs following this rule: If there is a single NAN, I want the NAN to be filled in with the average of the numbers before and after. If there is more than one NAN. I want the NAN to be filled in with the nearest number. For example, row 2305 should be the average of 16.3 and 14.8, from 2298 to 2304 should be 16.3, and from 2306 to 2312 should be 14.8.

4 commentaires
darova
le 30 Mar 2020
Do you have any code? What problem are you facing?
Cuong Nguyen
le 30 Mar 2020
Image Analyst
le 30 Mar 2020
Why not use interp1() or regionfill() to linearly interpolate from one side to the other?
Cuong Nguyen
le 30 Mar 2020
Réponse acceptée
Plus de réponses (2)
darova
le 30 Mar 2020
Use bwlabel
A1 = isnan(A); % find NaN
[L,n] = bwlabel(A1); % label each region
xx = 1:length(A);
for i = 1:n % loop through each region
BW = L==i; % select region
if sum(BW(:))>1 % if more than one 'NaN'
ix1 = find(BW,1,'first'); % first index of region
ix2 = floor(mean(BW.*xx)); % find mean index in region
ix3 = find(B2,1,'last'); % last index of region
A(ix1:ix2) = A(ix1-1); % fill first part
A(ix2+1:ix3) = A(ix3+1); % fill second part
else
ix = find(BW);
A(ix) = mean(A(ix([-1 1]))); % average
end
end
1 commentaire
Cuong Nguyen
le 31 Mar 2020
Andrei Bobrov
le 31 Mar 2020
x = [1;2;3;4;nan;nan;nan;nan;nan;5;7;8;nan;nan;nan;nan;11;11;12;nan;nan;nan;15];
out = [x,f1(x)]
x = [16.3;nan(15,1);14.8];
out = [x, f1(x)]
Here f1:
function out = f1(x)
b1 = fillmissing(x,'linear');
b2 = fillmissing(x,'nearest');
d = [0;diff(bwdist(~isnan(x)),2);0]==-2;
out = b2;
out(d) = b1(d);
end
3 commentaires
Cuong Nguyen
le 31 Mar 2020
Andrei Bobrov
le 31 Mar 2020
No! It only says that this version is available to me :) and this code will work with R2016b and later.
Cuong Nguyen
le 31 Mar 2020
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