How to interpolate at NaN values?

I have a vector [1 2 3 NaN 4 4.5 5.5 NaN NaN 6 6 7 NaN NaN NaN 8] and I want [1 2 3 3.5 4 4.5 5.5 5.75 5.75 6 6 7 7.5 7.5 7.5 8]. Is it possible to do this without a for cycle?

[jonas]

Yes, see previous question here ( link )

[Paolo]

Is it really interpolation? Whenever you have three consecutive NaN they all have the same value.

Answer 1

[Stephen23]

No loops, no third-party functions, and works for all MATLAB versions:

>> V = [1 2 3 NaN 4 4.5 5.5 NaN NaN 6 6 7 NaN NaN NaN 8];
>> X = ~isnan(V);
>> Y = cumsum(X-diff([1,X])/2);
>> Z = interp1(1:nnz(X),V(X),Y)
Z =      1.00    2.00    3.00    3.50    4.00    4.50    5.50    5.75    5.75    6.00    6.00    7.00    7.50    7.50    7.50    8.00

[Fernand ASSENE]

Hello Dear Stephen,

First thank for sharing.

I have the problem with your code when matrix begins with nan values, it does not work.

Please do you how can i fix this issue???

[Stephen23]

@Fernand ASSENE: in the original question all NaN were replaced by the average of the two non-NaN values on either side. If you have a NaN at the start then there are no such two values and so you will have to be more specific about what you expect to be returned, e.g.:

1. the first non-NaN value
2. a value extrapolated from the first two or more non-NaN values.
3. something else...

Note that none of these are the same as the original rule, nor is any of them "better" than the others, so I cannot guess which output you want. Consider the example in the original question appended with NaN i.e. [NaN,1,2,3,NaN,4...], what would you expect the first output value to be?:

1. 1
2. 0
3. something else...

Answer 2

[Jan]

a = [1 2 3 NaN 4 4.5 5.5 NaN NaN 6 6 7 NaN NaN NaN 8];
b = 0.5 * (fillmissing(a, 'previous') + fillmissing(a, 'next'))

Or

v        = ~isnan(a);
G        = griddedInterpolant(find(v), a(v), 'previous');
idx      = find(~v);
bp       = G(idx);
G.Method = 'next';
bn       = G(idx);
b        = a;
b(idx)   = (bp + bn) / 2;

Some timings:

n = 20;
a = rand(1, 1e6);
a(rand(1, 1e6) < 0.3) = NaN;

tic
for k = 1:n
  b = 0.5 * (fillmissing(a, 'previous') + fillmissing(a, 'next'));
end
toc

tic;
for k = 1:n
  v        = ~isnan(a);
  G        = griddedInterpolant(find(v), a(v), 'previous');
  idx      = find(~v);
  bp       = G(idx);
  G.Method = 'next';
  bn       = G(idx);
  b        = a;
  b(idx)   = (bp + bn) / 2;
end
toc

tic;
for k = 1:n
  X = ~isnan(a);
  Y = cumsum(X-diff([1,X])/2);
  b = interp1(1:nnz(X),a(X),Y);
end
toc

R2016b, i7:

 Elapsed time is 2.901105 seconds.
 Elapsed time is 0.598722 seconds.
 Elapsed time is 1.032346 seconds.

Answer 3

[Syed Abdul Salam]

Why not use smooth function with 2 levels, it will result the same as mentioned in accepted answer.

V = [1 2 3 NaN 4 4.5 5.5 NaN NaN 6 6 7 NaN NaN NaN 8];

Z = smooth(V,2)

Z = [1 2 3 3.500 4 4.500 5.500 5.500 6 6 6 7 7 7.500 8 8]