使用scipy.interpolate.griddata进行2d插值的不同结果

问题描述

我需要在2d numpy数组中对某些Nan值进行插值，例如，请参见下图:

I need to perform an interpolation of some Nan values in a 2d numpy array, see for example the following picture:

在我目前的方法中，我使用scipy.interpolate.griddata用于插值过程.但是我注意到在两个轴上镜像数组，即d2 = d[::-1, ::-1]插值得出不同的结果.这是一个完整的示例:

In my current approach I use scipy.interpolate.griddatafor the interpolation procedure. However I noticed that whenmirroring the array on both axis i.e. d2 = d[::-1, ::-1]the interpolation gives different results.Here is a complete example :

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

def replace_outliers(f):
    mask = np.isnan(f)
    lx, ly = f.shape
    x, y = np.mgrid[0:lx, 0:ly]
    z = interp.griddata(np.array([x[~mask].ravel(),y[~mask].ravel()]).T,
                                          f[~mask].ravel(),
                                          (x,y), method='linear', fill_value=0)
    return z

def main():
    d = np.load('test.npy')
    d2 = d[::-1, ::-1]

    dn = replace_outliers(d)
    dn2 = replace_outliers(d2)

    print np.sum(dn - dn2[::-1, ::-1])

    plt.imshow(dn-dn2[::-1, ::-1], interpolation='nearest')
    plt.colorbar()
    plt.show()


if __name__=='__main__':
    main()

这给出了两个插值之间的差异:

This gives the difference between the two interpolations:

或np.sum评估为-62.7

那么如何简单地对数组进行镜像在插值过程中给出不同的结果?我使用的坐标可能有问题吗?

So how can it be that a simple mirroring of the arraygives different results in the interpolation procedure?Is there maybe something wrong with the coordinates I use ?

推荐答案

原因很可能是线性插值基于基于三角形.但是，这种正方形网格对于Delaunay三角剖分来说是退化的情况，并且三角剖分并不是唯一的.我可以想象结果取决于数据点的顺序.

The reason likely is that the linear interpolation is triangle-based.However, such a square grid is a degenerate case for Delaunay triangulation, and the triangulation is not unique. I can imagine the outcome depends on the order of the data points.

对于丢失的数据点，我猜这两种情况对应于空白区域的不同三角剖分:

For a missing data point, I would guess the two cases correspond to different triangulations of the empty space:

                           A                 A
*   *   *              *---*---*         *---*---*
                       | /   \ |         | / | \ |
*       *        =>   D*-------*B   or  D*   |   *B
                       | \   / |         | \ | / |
*   *   *              *---*---*         *---*---*
                           C                 C

如果现在计算中心的值，则从一个三角获得(B + D)/2，从另一个三角获得(A + C)/2.

If you now compute the value at the center, you get (B+D)/2 from one triangulation and (A+C)/2 from the other.

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