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Am I correct in my approach regarding interpolation?

1 vue (au cours des 30 derniers jours)
tzaloupas
tzaloupas le 14 Fév 2013
Clôturé : MATLAB Answer Bot le 20 Août 2021
Dear all
I have the vector
A={
[ NaN]
[ NaN]
[1.0877]
[1.0909]
[ NaN]
[ NaN]
[ NaN]
[ NaN]
[ NaN]
[ NaN]
[ NaN]
[ NaN]
[ NaN]
[ NaN]
[ NaN]
[ NaN]
[ NaN]
[ NaN]
};
and the date vector
D={'10/2008'
'11/2008'
'12/2008'
'1/2009'
'2/2009'
'3/2009'
'4/2009'
'5/2009'
'6/2009'
'7/2009'
'8/2009'
'9/2009'
'10/2009'
'11/2009'
'12/2009'
'1/2010'
'2/2010'
'3/2010'
'4/2010'
'5/2010'
'6/2010'
'7/2010'
'8/2010'
'9/2010'
'10/2010'
'11/2010'
'12/2010'
'1/2011'
'2/2011'
'3/2011'
'4/2011'
'5/2011'
'6/2011'
'7/2011'
'8/2011'
'9/2011'
};
and I am doing interpolation
xi = datenum(D, 'mm/yyyy');
z = interp1(xi(1:2:end),cell2mat(A(:,1)),xi);
Equivalently,
idx = ~isnan(cell2mat(A(:,1)));
z= interp1(xi(idx),cell2mat(A(idx,1)),xi);
Could you please verify that I am correct in my approach ?
thanks

Réponses (1)

Azzi Abdelmalek
Azzi Abdelmalek le 14 Fév 2013
v=cell2mat(A)
xi = datenum(D, 'mm/yyyy');
z = interp1(xi(3:4),v(3:4),xi,'spline');
  2 commentaires
tzaloupas
tzaloupas le 14 Fév 2013
Hi Azzi. thanks What is the difference between your approach and mine?Is this linear interpolation?
Satyam Gaba
Satyam Gaba le 24 Août 2018
No it's not a linear function. Spline interpolation can be used for polynomial interpolation for low degree polynomials as well as higher degrees.

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