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Am I correct in my approach regarding interpolation?

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tzaloupas
tzaloupas il 14 Feb 2013
Chiuso: MATLAB Answer Bot il 20 Ago 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

Risposte (1)

Azzi Abdelmalek
Azzi Abdelmalek il 14 Feb 2013
v=cell2mat(A)
xi = datenum(D, 'mm/yyyy');
z = interp1(xi(3:4),v(3:4),xi,'spline');
  2 Commenti
tzaloupas
tzaloupas il 14 Feb 2013
Hi Azzi. thanks What is the difference between your approach and mine?Is this linear interpolation?
Satyam Gaba
Satyam Gaba il 24 Ago 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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