SciPy interp2D用于坐标对

问题描述

我正在使用scipy.interpolate.interp2d为曲面创建插值函数.然后，我有两个实际数据数组，我想为其计算插值点.如果将两个数组传递给interp2d函数，我将得到一个包含所有点的数组，而不仅仅是点对.

I'm using scipy.interpolate.interp2d to create an interpolation function for a surface. I then have two arrays of real data that I want to calculate interpolated points for. If I pass the two arrays to the interp2d function I get an array of all the points, not just the pairs of points.

我对此的解决方案是将两个数组压缩到一个坐标对列表中，然后将其循环传递给插值函数:

My solution to this is to zip the two arrays into a list of coordinate pairs and pass this to the interpolation function in a loop:

f_interp = interpolate.interp2d(X_table, Y_table,Z_table, kind='cubic')

co_ords = zip(X,Y)
out = []
for i in range(len(co_ords)):
    X = co_ords[i][0]
    Y = co_ords[i][1]
    value = f_interp(X,Y)
    out.append(float(value))

我的问题是，是否有更好的方法(更优雅，Pythonic?)来实现相同的结果?

My question is, is there a better (more elegant, Pythonic?) way of achieving the same result?

推荐答案

一个就可以了

for Xtmp,Ytmp in zip(X,Y):
    ...

在您的循环中.甚至更好，只是

in your loop. Or even better, just

out = [float(f_interp(XX,YY)) for XX,YY in zip(X,Y)]

替换循环.

换句话说，我建议改为使用interpolate.griddata .它的行为往往比interp2d好得多，并且可以接受任意形状的点作为输入.如您所见，interp2d插值器只会在网格上返回您的值.

On a different note, I suggest using interpolate.griddata instead. It tends to behave much better than interp2d, and it accepts arbitrary-shaped points as input. As you've seen, interp2d interpolators will only return you values on a mesh.

这篇关于SciPy interp2D用于坐标对的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！