CODEWITHSUNDEEP
python-3.xscipyinterpolationboundarycubic-spline
Karl Montalban
Karl Montalban

Reputation: 77

Scipy Interpolation CubicSpline Boundaries

I have an issue using The scipy.interpolate.CubicSpline function. Here is my code :

CS1 = CubicSpline(T,A,bc_type='not-a-knot',extrapolate=bool, axis=1)

Result : CS1 = 

[-8.34442117e+03 -6.94866126e+03 -5.71682333e+03 -4.63872647e+03
 -3.70418976e+03 -2.90303229e+03 -2.22507315e+03 -1.66013142e+03
 -1.19802617e+03 -8.28576513e+02 -5.41601516e+02 -3.26920268e+02
 -1.74351855e+02 -7.37153621e+01 -1.48298738e+01  1.24855245e+01
  1.84117475e+01  1.31297102e+01  6.82032749e+00  9.66451413e+00
  3.15397607e+01  7.05279383e+01  1.09387991e+02  1.32530056e+02
  1.36799756e+02  1.22858734e+02  9.60947464e+01  6.66210660e+01
  4.28224903e+01  2.64229282e+01  1.75832317e+01  1.45176021e+01
  1.39435432e+01  1.33609464e+01  1.23801442e+01  1.09650786e+01
  9.27738095e+00  7.59606003e+00  6.29249366e+00  5.91452686e+00
  6.79882387e+00  7.57144653e+00  6.13515774e+00  2.70590543e+00
  9.34668162e-01  3.86336659e+00  9.73615276e+00  1.52487556e+01
  1.90469811e+01  2.20000000e+01]

There are negative values, which i find odd because the original data is only positive :

[7.0,
 12.0,
 20.0,
 111.0,
 132.0,
 68.0,
 22.0,
 14.0,
 12.0,
 8.0,
 6.0,
 7.0,
 1.0,
 13.0,
 22.0,
 23.0,
 5.0,
 3.0,
 5.0,
 65.0,
 236.0,
 234.0,
 105.0,
 152.0,
 466.0,
 401.0,
 157.0,
 51.0,
 21.0,
 13.0,
 11.0,
 19.0,
 15.0,
 11.0,
 9.0,
 15.0,
 86.0,
 276.0,
 423.0,
 291.0,
 108.0,
 36.0,
 22.0,
 21.0,
 16.0,
 16.0,
 13.0,
 9.0]

And T is only a list that goes one by one from 1 to 48 (48 is the length of A and T) I feel that the issue is from a boundary issue but the problem is only in the beginning...

enter image description here

Any ideas ?

Upvotes: 0

Views: 1063

Answers (1)

user6655984
user6655984

Reputation:

Nothing odd here: a cubic spline on positive data can attain negative values, no matter what the boundary conditions are. If it's necessary to maintain positivity, piecewise linear interpolation (degree 1 spline) is an option. Other options are discussed in How can I find a non-negative interpolation function?

Here is an illustration of why this happens: spl = CubicSpline([-2, -1, 1, 2], [10, 1, 1, 10])

spline

This spline fits a parabola to the given points. The parabola dips into negative territory in the middle, between the points.

That in your example this happened near the boundary is not really important; it can happen anywhere.

Upvotes: 1

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