沿曲线评估 scipy 2D 插值的输出 [英] Evaluate the output from scipy 2D interpolation along a curve

问题描述

我有从网格点 x, y 处的二维函数 f 采样的数据 z，如 z = f(x, y).

I have data z sampled from a 2D function f at grid points x, y, as in z = f(x, y).

很容易通过 f = interp2d(x, y, z) 用 scipy.interp2d 插入 f.

It is easy to interpolate f with scipy.interp2d via f = interp2d(x, y, z).

然而，评估 f(x, y) 会返回一个完整的 2D 网格，就像我已经完成一样

However, evaluating f(x, y) returns an entire 2D grid as if I had done

xx, yy = np.meshgrid(x, y)
f(xx, yy)

我想要的行为是简单地返回值 [f(x[i], y[i]) for i in range(len(x))]，我认为这是numpy 中几乎任何其他方法的行为.

The behaviour I want is to simply return the values [f(x[i], y[i]) for i in range(len(x))], which I believe is the behaviour for pretty much any other method in numpy.

我想要这种行为的原因是我正在寻找沿着 f 的表面在时间"上由 (t, u(t)).

The reason I want this behaviour is that I'm looking for the path traced out along the surface of f over "time" by the pair (t, u(t)).

同样令人惊讶的是，np.diag(f(t, u(t))) 与 np.array([f(ti, u(ti))] 对于 ti 不同在 t]) 中，所以我不清楚如何从通过 interp2d 返回的内容获得路径 f(t, u(t)).

It is also surprising that np.diag(f(t, u(t))) differs from np.array([f(ti, u(ti)) for ti in t]), so It's not clear to me how to get at the path f(t, u(t)) from what is returned via interp2d.

关于 diag，我只是觉得我们应该有 np.diag(f(t, u(t))) == np.array([f(ti), u(ti)) for ti in t])，但事实并非如此.

About diag, I just thought it seemed we should have np.diag(f(t, u(t))) == np.array([f(ti, u(ti)) for ti in t]), but that isn't the case.

完整示例:

def f(t, u):
    return (t**2) * np.exp((u**2) / (1 + u**2))

x = np.linspace(0, 1, 250)
xx, yy = np.meshgrid(x, x)

z = f(xx, yy)
f = scipy.interpolate.interp2d(x, y, z)

print(f(x, y))
print(np.array([f(xi, yi)[0] for xi, yi in zip(x, y)]))

我希望两个 print 语句的输出相同.

I would like the output of both print statements to be the same.

推荐答案

interp2d 方法返回一个对象，该对象的调用方法期望 x、y 向量是矩形网格的坐标.您没有从返回数组的对角线上获得所需值的原因是它首先对 x, y 进行排序.

The method interp2d returns an object whose call method expects the x, y vectors to be coordinates of a rectangular grid. And the reason you don't get the desired values from the diagonal of the returned array is that it sorts x, y first.

但是有一种解决方法，我也在在 B 样条基础上查询二元样条上的多个点 中使用了它.执行后

But there is a workaround, which I also used in Querying multiple points on a bivariate spline in the B-spline basis. After executing

import scipy.interpolate as si
f = si.interp2d(x, y, z)

评估 f 不是通过调用它，而是通过传递它的 tck 属性，然后是你的 x、y 坐标，到内部的 bispeu 方法.像这样:

evaluate f not by calling it, but by passing its tck properties, followed by your x, y coordinates, to internal bispeu method. Like this:

print(si.dfitpack.bispeu(f.tck[0], f.tck[1], f.tck[2], f.tck[3], f.tck[4], x, y)[0])

以上返回与慢循环相同

print(np.array([f(xi, yi)[0] for xi, yi in zip(x, y)]))

说明

对象 f 秘密地是 1 阶 B 样条.样条参数(节点、系数、阶数)包含在其 tck 属性中，可以使用直接通过低阶例程达到预期的效果.

Explanation

The object f is secretly a B-spline of order 1. The spline parameters (knots, coefficients, order) are contained in its tck property and can be used directly by lower-order routines to the desired effect.

(理想情况下，f 的调用方法将有一个布尔参数 grid，我们将其设置为 False 以使其知道我们不需要网格评估.唉，它没有实现.)

(Ideally, the call method of f would have a Boolean parameter grid which we'd set to False to let it know we don't want grid evaluation. Alas, it's not implemented.)

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