python - scipy 的 RegularGridInterpolator 能否通过一次调用同时返回值和梯度？

Question

我正在使用 scipy.interpolate.RegularGridInterpolator 和 method='linear'。获取内插值很简单(参见 https://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.interpolate.RegularGridInterpolator.html 中的示例)。获取梯度和插值的好方法是什么？

一种可能性是多次调用插值器并使用有限差分“手动”计算梯度。鉴于对插值器的每次调用可能已经在后台计算梯度，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梯度？

明确地说，我对我插值的函数的“真实”梯度不感兴趣——只是线性近似的梯度，例如https://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.interpolate.RegularGridInterpolator.html 示例中 my_interpolating_function 的梯度.

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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 的梯度感兴趣(与f)。我可以使用有限差分来计算它，但是有更好的方法吗？我假设 RegularGridInterpolator 已经在后台计算梯度——而且速度很快。我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]

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这是 f 和 f_interp 的图片。我对 f_interp(实线)的梯度感兴趣:

Answer 1

没有。

下面是 scipy.interpolate.RegularGridInterpolator 在幕后所做的事情:

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

它使用 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计算插值上升的陡峭程度。

但是，梯度还取决于角点的值。

你可以在这里找到数学:https://en.wikipedia.org/wiki/Trilinear_interpolation