如何设置 B 样条结果的上限和下限以获得合理的插值

Question

我有插值功能：

from scipy import interpolate
def f(x):
  x_points = [38508,38510,38512]
  y_points = [0.249267578125,0.181396484375,0.1912841796875]

  tck = interpolate.splrep(x_points, y_points,k=2,xb=38508,xe=38512)
  return interpolate.splev(x, tck)

当我评估 f(38503) 时，输出是 0.75，这与 y_points.

完全不同

关于如何使用这种或其他插值方法减少此错误的任何建议？

Answer 1

正如 RishiG 在评论中指出的那样，您要做的是外推。

object oriented approach 有一个额外的参数：ext。

from scipy import interpolate

def f(x):
    x_points = [38508, 38510, 38512]
    y_points = [0.249267578125, 0.181396484375, 0.1912841796875]

    tck = interpolate.splrep(x_points, y_points,k=2,xb=38508,xe=38512)
    return interpolate.splev(x, tck)

def g(x):
    x_points = [38508, 38510, 38512]
    y_points = [0.249267578125, 0.181396484375, 0.1912841796875]

    spl = interpolate.UnivariateSpline(x_points, y_points, k=2, ext=3)

    return spl(x)

if __name__=='__main__':

    print(f(38503))
    print(g(38503))

输出：

0.7591400146484374
0.249267578125

编辑：

This similar question 可能也很有趣。

如何设置 B 样条结果的上限和下限以获得合理的插值

How to set upper and lower bound on B-Spline result to get a reasonable interpolation

python

math

interpolation