bundles / scipy 1.17.1 / scipy / interpolate / _bsplines / make_interp_spline

function

`scipy.interpolate._bsplines:make_interp_spline`

source: /scipy/interpolate/_bsplines.py :1416

Signature

   def     make_interp_spline (    x    ,    y    ,    k    =  3   ,    t    =  None   ,    bc_type    =  None   ,    axis    =  0   ,    check_finite    =  True     )

Summary

Create an interpolating B-spline with specified degree and boundary conditions.

Parameters

x : array_like, shape (n,)

Abscissas.

y : array_like, shape (n, ...)

Ordinates.

k : int, optional

B-spline degree. Default is cubic, k = 3.

t : array_like, shape (nt + k + 1,), optional.

Knots. The number of knots needs to agree with the number of data points and the number of derivatives at the edges. Specifically, nt - n must equal len(deriv_l) + len(deriv_r).

bc_type : 2-tuple or None

Boundary conditions. Default is None, which means choosing the boundary conditions automatically. Otherwise, it must be a length-two tuple where the first element (deriv_l) sets the boundary conditions at x[0] and the second element (deriv_r) sets the boundary conditions at x[-1]. Each of these must be an iterable of pairs (order, value) which gives the values of derivatives of specified orders at the given edge of the interpolation interval. Alternatively, the following string aliases are recognized:

"clamped": The first derivatives at the ends are zero. This is equivalent to bc_type=([(1, 0.0)], [(1, 0.0)]).
"natural": The second derivatives at ends are zero. This is equivalent to bc_type=([(2, 0.0)], [(2, 0.0)]).
"not-a-knot" (default): The first and second segments are the same polynomial. This is equivalent to having bc_type=None.
"periodic": The values and the first k-1 derivatives at the ends are equivalent.

axis : int, optional

Interpolation axis. Default is 0.

check_finite : bool, optional

Whether to check that the input arrays contain only finite numbers. Disabling may give a performance gain, but may result in problems (crashes, non-termination) if the inputs do contain infinities or NaNs. Default is True.

Returns

b : `BSpline` object: A BSpline object of the degree k and with knots t.

Notes

Array API Standard Support

make_interp_spline has experimental support for Python Array API Standard compatible backends in addition to NumPy. Please consider testing these features by setting an environment variable SCIPY_ARRAY_API=1 and providing CuPy, PyTorch, JAX, or Dask arrays as array arguments. The following combinations of backend and device (or other capability) are supported.

====================  ====================  ====================
Library               CPU                   GPU
====================  ====================  ====================
NumPy                 ✅                     n/a                 
CuPy                  n/a                   ⛔                   
PyTorch               ✅                     ⛔                   
JAX                   ⚠️ no JIT             ⛔                   
Dask                  ⚠️ computes graph     n/a                 
====================  ====================  ====================

See dev-arrayapi for more information.

Examples

Use cubic interpolation on Chebyshev nodes:

import numpy as np
import matplotlib.pyplot as plt
def cheb_nodes(N):
    jj = 2.*np.arange(N) + 1
    x = np.cos(np.pi * jj / 2 / N)[::-1]
    return x

✓

x = cheb_nodes(20)
y = np.sqrt(1 - x**2)

✓

from scipy.interpolate import BSpline, make_interp_spline
b = make_interp_spline(x, y)
np.allclose(b(x), y)

✓

Note that the default is a cubic spline with a not-a-knot boundary condition

b.k

✓

Here we use a 'natural' spline, with zero 2nd derivatives at edges:

l, r = [(2, 0.0)], [(2, 0.0)]
b_n = make_interp_spline(x, y, bc_type=(l, r))  # or, bc_type="natural"
np.allclose(b_n(x), y)
x0, x1 = x[0], x[-1]
np.allclose([b_n(x0, 2), b_n(x1, 2)], [0, 0])

✓

Interpolation of parametric curves is also supported. As an example, we compute a discretization of a snail curve in polar coordinates

phi = np.linspace(0, 2.*np.pi, 40)
r = 0.3 + np.cos(phi)
x, y = r*np.cos(phi), r*np.sin(phi)  # convert to Cartesian coordinates

✓

Build an interpolating curve, parameterizing it by the angle

spl = make_interp_spline(phi, np.c_[x, y])

✓

Evaluate the interpolant on a finer grid (note that we transpose the result to unpack it into a pair of x- and y-arrays)

phi_new = np.linspace(0, 2.*np.pi, 100)
x_new, y_new = spl(phi_new).T

✓

Plot the result

plt.plot(x, y, 'o')
plt.plot(x_new, y_new, '-')

✗

plt.show()

✓

Build a B-spline curve with 2 dimensional y

x = np.linspace(0, 2*np.pi, 10)
y = np.array([np.sin(x), np.cos(x)])

✓

Periodic condition is satisfied because y coordinates of points on the ends are equivalent

ax = plt.axes(projection='3d')
xx = np.linspace(0, 2*np.pi, 100)
bspl = make_interp_spline(x, y, k=5, bc_type='periodic', axis=1)

✓

ax.plot3D(xx, *bspl(xx))
ax.scatter3D(x, *y, color='red')

✗

plt.show()

✓

Aliases

scipy.interpolate.make_interp_spline