Question

我正在scipy.interpolate.RegularGridInterpolator使用method='linear'。获取插值非常简单（参见https://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.interpolate.RegularGridInterpolator.html处的示例）。获得渐变和插值的好方法是什么？

一种可能性是多次调用内插器并使用有限差分“手动”计算梯度。考虑到对内插器的每次调用可能已经在计算引擎盖下的梯度，这感觉很浪费。那是对的吗？如果是这样，我如何修改RegularGridInterpolator以返回插值函数值及其渐变？

为了清楚起见，我对我正在插值的函数的“真实”梯度不感兴趣 - 只是线性逼近的梯度，例如， https://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.interpolate.RegularGridInterpolator.html示例中my_interpolating_function的渐变。

这是一个例子。我有一个函数f，我构建了一个线性插值器f_interp，我对f_interp的渐变感兴趣（而不是f的渐变）。我可以使用有限差分来计算它，但有更好的方法吗？我假设RegularGridInterpolator已经在计算引擎盖下的渐变 - 并快速完成。如何修改它以返回梯度以及插值？

import matplotlib.pyplot as plt
import numpy as np
import scipy.interpolate as interp

def f(x, y, z):
    return 0.01*x**2 + 0.05*x**3 + 5*x*y + 3*x*y*z + 0.1*x*(y**2)*z + 9*x*z**2 - 2*y

x_grid = np.linspace(0.0, 10.0, 20)
y_grid = np.linspace(-10.0, 10.0, 20)
z_grid = np.linspace(0.0, 20.0, 20)

mesh = np.meshgrid(x_grid, y_grid, z_grid, indexing="ij")

f_on_grid = f(mesh[0], mesh[1], mesh[2])
assert np.isclose(f_on_grid[0, 0, 0], f(x_grid[0], y_grid[0], z_grid[0]))  # Sanity check

grid = (x_grid, y_grid, z_grid)
f_interp = interp.RegularGridInterpolator(grid, f_on_grid, method="linear",
                                          bounds_error=False, fill_value=None)

dense_x = np.linspace(0.0, 20.0, 400)
plt.plot(dense_x, [f_interp([x, 1.0, 2.0])[0] for x in dense_x], label="y=1.0, z=2.0")
plt.plot(dense_x, [f_interp([x, 1.0, 4.0])[0] for x in dense_x], label="y=1.0, z=4.0")
plt.legend()
plt.show()

f_interp([0.05, 1.0, 2.0])  # Linearly interpolated value, distinct from f(0.05, 1.0, 2.0)

## Suppose I want to compute both f_interp and its gradient at point_of_interest
point_of_interest = np.array([0.23, 1.67, 5.88])
f_interp(point_of_interest)  # Function value -- how can I get f_interp to also return gradient?

## First gradient calculation using np.gradient and a mesh around point_of_interest +- delta
delta = 0.10
delta_mesh = np.meshgrid(*([-delta, 0.0, delta], ) * 3, indexing="ij")
delta_mesh_long = np.column_stack((delta_mesh[0].flatten(),
                                   delta_mesh[1].flatten(),
                                   delta_mesh[2].flatten()))
assert delta_mesh_long.shape[1] == 3
point_plus_delta_mesh = delta_mesh_long + point_of_interest.reshape((1, 3))
values_for_gradient = f_interp(point_plus_delta_mesh).reshape(delta_mesh[0].shape)
gradient = [x[1, 1, 1] for x in np.gradient(values_for_gradient, delta)]
gradient  # Roughly [353.1, 3.8, 25.2]

## Second gradient calculation using finite differences, should give similar result
gradient = np.zeros((3, ))
for idx in [0, 1, 2]:
    point_right = np.copy(point_of_interest)
    point_right[idx] += delta
    point_left = np.copy(point_of_interest)
    point_left[idx] -= delta
    gradient[idx] = (f_interp(point_right)[0] - f_interp(point_left)[0]) / (2*delta)

gradient  # Roughly [353.1, 3.8, 25.2]

这是f和f_interp的图片。我对f_interp（实线）的渐变感兴趣：

Answer 1

没有

以下是scipy.interpolate.RegularGridInterpolator幕后的内容：

class CartesianGrid(object):
    """
    Linear Multivariate Cartesian Grid interpolation in arbitrary dimensions
    This is a regular grid with equal spacing.
    """
    def __init__(self, limits, values):
        self.values = values
        self.limits = limits

    def __call__(self, *coords):
        # transform coords into pixel values
        coords = numpy.asarray(coords)
        coords = [(c - lo) * (n - 1) / (hi - lo) for (lo, hi), c, n in zip(self.limits, coords, self.values.shape)]

        return scipy.ndimage.map_coordinates(self.values, coords, 
            cval=numpy.nan, order=1)

https://github.com/JohannesBuchner/regulargrid/blob/master/regulargrid/cartesiangrid.py

使用scipy.ndimage.map_coordinates进行线性插值。 coords包含像素坐标中的位置。您应该能够使用这些权重，以及每个维度的下限值和上限值来确定插值的上升程度。

但是，渐变还取决于角点的值。

您可以在此处找到数学：https://en.wikipedia.org/wiki/Trilinear_interpolation

scipy的RegularGridInterpolator可以通过一次调用返回值和渐变吗？

1 个答案: