Question

给出二维网格上的函数的值z。网格坐标称为x和y。

现在，我想插入数据，这样我就可以得到任何（x，y）的z值。

为确保我的示例清晰，我不想缩短我的代码：

import numpy as np
import pylab as pl
from scipy.interpolate import SmoothBivariateSpline
from scipy.interpolate import interp1d

def Func(x, y): # Define a test-function.
    return y**2 # A parabola.

SparseGridX, SparseGridY = np.mgrid[0:1:5j, 0:1:5j] # Generate a sparsely meshed grid.

SparseGridZ = Func(SparseGridX, SparseGridY) # A function evaluated for np-arrays of arbitrary shape, again yields np-arrays of the same shape. Hence, this yields again an array.

pl.figure(num='Sparsely sampled function.')
pl.pcolor(SparseGridX, SparseGridY, SparseGridZ) # Plot the sparsely sampled function on the grid.
pl.colorbar();pl.title("Sparsely sampled function.")

FineListX = np.linspace(0,1,20) # Generate a fine-mesh list along x...
FineListY = np.linspace(0,1,20) # ... as well as along y...
FineGridX, FineGridY = np.mgrid[0:1:20j, 0:1:20j] # ... and the corresponding fine-mesh grid.

ListedSparseGridX = SparseGridX.flatten() # Attain the list of x-coordinates in the fine-mesh grid.
ListedSparseGridY = SparseGridY.flatten() # Attain the list of y-coordinates in the fine-mesh grid.

ListedSparseGridZ = Func(ListedSparseGridX,ListedSparseGridY) # This yields a list, in which an element is the result of Func for the corresponding combination of elements of FineListX and FineListY.

IntObj = SmoothBivariateSpline(ListedSparseGridX, ListedSparseGridY, ListedSparseGridZ, kx=1, ky=1) # This yields an interpolated object of the sparsely meshed points. 
Interpolation = IntObj(FineListX,FineListX) # This evaluates the interpolated object at the points of the fine-mesh grid and returns the corresponding array.

pl.figure(num='Fine-meshed interpolation.')
pl.pcolor(FineGridX, FineGridY, Interpolation) # Plot the interpolation in a fine-mesh grid.
pl.colorbar();pl.title("Fine-meshed interpolation.")

IntObj1Dim = interp1d(SparseGridY[3], SparseGridZ[3]) # This yields a one-dimensional interpolated object. The index 3 is arbitrary. 
Interpolation1Dim = IntObj1Dim(FineListY) # This evaluates the interpolated object along the fine-meshed y-coordinate.

pl.figure(num="Plot only along y-coordinate.")
pl.plot(SparseGridY[3],SparseGridZ[3], label="Points to interpolate", marker="o", ls="None") # Plot the points (along y) that were used for the interpolation.
pl.plot(FineListY,Interpolation[5], label="Interpolation via SmoothBivariateSpline") # Plot the interpolation (along y).
pl.plot(FineListY,Interpolation1Dim, label="Interpolation via interp1d") # Plot the one-dimensional interpolation.
pl.legend(loc="best")
pl.title("Plot only along y-coordinate.")

如您所见，interp1d提供了一个完全符合要点的分段线性函数。相比之下，SmoothBivariateSpline只是产生一条直线。我的目标是使用SmoothBivariateSpline生成插值，它也是与interp1d生成的一样的分段线性。

我尝试使用SmoothBivariateSpline的参数s，但我没有成功。特别是，我试过s=0。在这种情况下，插值应该完全符合数据点，但事实并非如此。

提前致谢！

Answer 1

现在我发现LinearNDInterpolator在这里更有用。

ListedSparseGrid = np.asarray([(i,j) for i, j in zip(ListedSparseGridX,ListedSparseGridY)]) # Attain a list of all pairs of the elements of ListedSparseGridX and ListedSparseGridY.
IntObjLNDI = LinearNDInterpolator(ListedSparseGrid, ListedSparseGridZ) # This yields an interpolated object of the sparsely meshed points. 
InterpolationLNDI = np.asarray([[IntObjLNDI(i,j) for j in FineListY] for i in FineListX]) # This evaluates the interpolated object at the points of the fine-mesh grid and returns the corresponding array.

pl.figure(num='Fine-meshed interpolation using LinearNDInterpolator.')
pl.pcolor(FineGridX, FineGridY, InterpolationLNDI) # Plot the interpolation in a fine-mesh grid.
pl.colorbar();pl.title("Fine-meshed interpolation using LinearNDInterpolator.")

并且1-D结果等于interp1d的结果，通过添加

可以看到

pl.plot(FineListY,InterpolationLNDI[40], label="Interpolation via LinearNDInterpolator") # Plot the interpolation (along y).

到名为＆＃34的图;仅沿y坐标绘图。＆＃34;上方。

通过SmoothBivariateSpline进行分段线性插值

1 个答案: