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Fill in missing NaNs

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Cuong Nguyen
Cuong Nguyen il 30 Mar 2020
Commentato: Cuong Nguyen il 31 Mar 2020
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 Commenti
Image Analyst
Image Analyst il 30 Mar 2020
Why not use interp1() or regionfill() to linearly interpolate from one side to the other?
Cuong Nguyen
Cuong Nguyen il 30 Mar 2020
I am quite a newbie to matlab, can you make it more specific? Thanks.

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Risposta accettata

Ameer Hamza
Ameer Hamza il 30 Mar 2020
Modificato: Ameer Hamza il 30 Mar 2020
Try this:
x = [1;2;3;4;nan;nan;nan;nan;nan;5;7;8;nan;nan;nan;nan;11;11;12;nan;nan;nan;15];
bb = regionprops(isnan(x));
idx_nnan = find(~isnan(x));
idx_nan = find(isnan(x));
[~,idx] = min(abs(idx_nan - idx_nnan'), [], 2);
x(idx_nan) = x(idx_nnan(idx));
a = [bb.Centroid];
a(1:2:end) = [];
bb = bb(a==fix(a));
for i=1:numel(bb)
idx_center = bb(i).Centroid(2);
idx = cumsum(bb(i).BoundingBox([2 4])) + [-0.5 0.5];
x(idx_center) = mean(x(idx));
end
Original x:
x =
1
2
3
4
NaN
NaN
NaN
NaN
NaN
5
7
8
NaN
NaN
NaN
NaN
11
11
12
NaN
NaN
NaN
15
New x:
x =
1.0000
2.0000
3.0000
4.0000
4.0000 % <--- nearesr value
4.0000 % <--- nearesr value
4.5000 % average
5.0000 % <--- nearesr value
5.0000 % <--- nearesr value
5.0000
7.0000
8.0000
8.0000 % <--- nearesr value
8.0000 % <--- nearesr value
11.0000 % <--- nearesr value
11.0000 % <--- nearesr value
11.0000
11.0000
12.0000
12.0000 % <--- nearesr value
13.5000 % average
15.0000 % <--- nearesr value
15.0000
  4 Commenti
Cuong Nguyen
Cuong Nguyen il 31 Mar 2020
Oh nice nice. Thank you, I got something to learn from this!
Ameer Hamza
Ameer Hamza il 31 Mar 2020
Glad to be of help.

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Più risposte (2)

darova
darova il 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 Commento
Cuong Nguyen
Cuong Nguyen il 31 Mar 2020
I tried yours and I got this message "Array indices must be positive integers or logical values". It also shows that the error is in "A(ix) = mean(A(ix([-1 1])))" line.

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Andrei Bobrov
Andrei Bobrov il 31 Mar 2020
This question is a repeat of this question:
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 Commenti
Andrei Bobrov
Andrei Bobrov il 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
Cuong Nguyen il 31 Mar 2020
Ahh ok. Can you explain more about this line? It's something about the distance to the edges of NaN cluster I reckon, isn't it?
d = [0;diff(bwdist(~isnan(x)),2);0]==-2;

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