在Python中将较大的不规则网格插值到另一个不规则网格 [英] Interpolate large irregular grid onto another irregular grid in Python

问题描述

我正在尝试使用Python将复杂值从一个不规则网格插值到另一个不规则网格.网格是二维的，有103,113个数据点.我正在使用Python 2.6.6，Scipy 0.7.2，Numpy 1.3.0，Matplotlib 0.99.3

I am trying to interpolate complex values from one irregular grid to another irregular grid using Python. The grids are in 2D and there are 103,113 data points. I am using Python 2.6.6, Scipy 0.7.2, Numpy 1.3.0, Matplotlib 0.99.3

在使用griddata的Matlab中，这大约需要5秒钟.

In Matlab using griddata this is achieved in roughly 5 seconds.

BnGRID2  = griddata(R_GRID1,Z_GRID1,BnGRID1,R_GRID2,Z_GRID2) (MATLAB)

(请注意，所有数组均为201 x 513)

(Note all arrays are 201 x 513)

但是，如果我尝试使用matplotlib.mlab.griddata，即使我尝试仅使用实部，也会出现memoryError:

However, if I try using matplotlib.mlab.griddata I get a memoryError even if I try to work with the real part only:

mlab.griddata(R_GRID1.flatten(),Z_GRID1.flatten(),num.real(BnGRID1.flatten()),R_GRID2.flatten(),Z_GRID2.flatten())

如果我尝试使用interp2d，则会遇到分段错误，并且Python退出:

If I try using interp2d I get a segmentation fault and Python exits:

a = interp.interp2d(R_GRID1,Z_GRID1,num.real(BnGRID1))

我已经尝试过使用KDTree，这似乎可以正常工作，但是，与Matlab的几秒钟相比，它花费了几分钟，但是我对这个选项的探索还不够.

I have tried using KDTree and this seems to work ok, however, it takes a few minutes compared with the few seconds for Matlab, but I haven't explored this option too much yet.

您是否想知道是否有人对如何像Matlab一样快地完成此工作有任何想法?我注意到较新版本的Scipy也具有griddata，有人知道它是否可以处理较大的不规则网格?

Was wondering if anyone has any ideas how I can get this done as quickly as Matlab seems to? I noticed that the newer version of Scipy also has griddata, does anyone know if this can handle large irregular grids?

推荐答案

Scipy的griddata似乎能够处理这种大小的数据集而不会出现问题:

Scipy's griddata seems to be able to deal with data sets of this size without problems:

import numpy as np
import scipy.interpolate

# old grid
x, y = np.mgrid[0:1:201j, 0:1:513j]
z = np.sin(x*20) * (1j + np.cos(y*3))**2   # some data

# new grid
x2, y2 = np.mgrid[0.1:0.9:201j, 0.1:0.9:513j]

# interpolate onto the new grid
z2 = scipy.interpolate.griddata((x.ravel(), y.ravel()), z.ravel(), (x2, y2), method='cubic')

在旧的AMD Athlon上，griddata步骤大约需要5秒钟.

The griddata step takes about 5s on an old AMD Athlon.

如果您的数据在网格上(即，与值z [i，j]对应的坐标为(x [i]，y [j]))，则可以使用scipy.interpolate.RectBivariateSpline

If your data is on a grid (i.e., the coordinates corresponding to value z[i,j] are (x[i], y[j])), you can get more speed by using scipy.interpolate.RectBivariateSpline

z3 = (scipy.interpolate.RectBivariateSpline(x[:,0], y[0,:], z.real)(x2[:,0], y2[0,:])
 + 1j*scipy.interpolate.RectBivariateSpline(x[:,0], y[0,:], z.imag)(x2[:,0], y2[0,:]))

需要0.05秒.它的速度要快得多，因为即使您的网格间距不规则，只要网格是矩形，就可以使用更有效的算法.

which takes 0.05s. It's much faster, because even if your grid spacings are irregular, a more efficient algorithm can be used as long as the grid is rectangular.

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