Python中的离散Laplacian（del2等价）

Question

我需要与Matlab（八进制）离散Laplacian运算符（函数）del2（）等价的Python/Numpy。我尝试了一些Python解决方案，但没有一个与del2的输出相匹配。我有八度音阶

image = [3 4 6 7; 8 9 10 11; 12 13 14 15;16 17 18 19]
del2(image)

这就产生了结果

   0.25000  -0.25000  -0.25000  -0.75000
  -0.25000  -0.25000   0.00000   0.00000
   0.00000   0.00000   0.00000   0.00000
   0.25000   0.25000   0.00000   0.00000

在Python上我试过

import numpy as np
from scipy import ndimage
import scipy.ndimage.filters

image =  np.array([[3, 4, 6, 7],[8, 9, 10, 11],[12, 13, 14, 15],[16, 17, 18, 19]])
stencil = np.array([[0, 1, 0],[1, -4, 1], [0, 1, 0]])
print ndimage.convolve(image, stencil, mode='wrap')

结果是

[[ 23  19  15  11]
 [  3  -1   0  -4]
 [  4   0   0  -4]
 [-13 -17 -16 -20]]

我也试过了

scipy.ndimage.filters.laplace(image)

结果就是这样

[[ 6  6  3  3]
 [ 0 -1  0 -1]
 [ 1  0  0 -1]
 [-3 -4 -4 -5]]

所以所有的输出似乎都不匹配。八度码del2.m表明它是拉普拉斯算子。我遗漏了什么吗？

Answer 1

通过使用适当的stencil对数组进行卷积，可以使用卷积来计算laplacian：

from scipy.ndimage import convolve
stencil= (1.0/(12.0*dL*dL))*np.array(
        [[0,0,-1,0,0], 
         [0,0,16,0,0], 
         [-1,16,-60,16,-1], 
         [0,0,16,0,0], 
         [0,0,-1,0,0]])
convolve(e2, stencil, mode='wrap')

Answer 2

根据这里的代码

http://cns.bu.edu/~tanc/pub/matlab_octave_compliance/datafun/del2.m

我试图编写一个与Python等价的程序。这似乎是工作，任何反馈将不胜感激。

import numpy as np

def del2(M):
    dx = 1
    dy = 1
    rows, cols = M.shape
    dx = dx * np.ones ((1, cols - 1))
    dy = dy * np.ones ((rows-1, 1))

    mr, mc = M.shape
    D = np.zeros ((mr, mc))

    if (mr >= 3):
        ## x direction
        ## left and right boundary
        D[:, 0] = (M[:, 0] - 2 * M[:, 1] + M[:, 2]) / (dx[:,0] * dx[:,1])
        D[:, mc-1] = (M[:, mc - 3] - 2 * M[:, mc - 2] + M[:, mc-1]) \
            / (dx[:,mc - 3] * dx[:,mc - 2])

        ## interior points
        tmp1 = D[:, 1:mc - 1] 
        tmp2 = (M[:, 2:mc] - 2 * M[:, 1:mc - 1] + M[:, 0:mc - 2])
        tmp3 = np.kron (dx[:,0:mc -2] * dx[:,1:mc - 1], np.ones ((mr, 1)))
        D[:, 1:mc - 1] = tmp1 + tmp2 / tmp3

    if (mr >= 3):
        ## y direction
        ## top and bottom boundary
        D[0, :] = D[0,:]  + \
            (M[0, :] - 2 * M[1, :] + M[2, :] ) / (dy[0,:] * dy[1,:])

        D[mr-1, :] = D[mr-1, :] \
            + (M[mr-3,:] - 2 * M[mr-2, :] + M[mr-1, :]) \
            / (dy[mr-3,:] * dx[:,mr-2])

        ## interior points
        tmp1 = D[1:mr-1, :] 
        tmp2 = (M[2:mr, :] - 2 * M[1:mr - 1, :] + M[0:mr-2, :])
        tmp3 = np.kron (dy[0:mr-2,:] * dy[1:mr-1,:], np.ones ((1, mc)))
        D[1:mr-1, :] = tmp1 + tmp2 / tmp3

    return D / 4

Answer 3

也许你在找^{}。