如何快速平滑 1000x1000 的数组

Question

我们正在模拟法国海湾（圣米歇尔山）的沉积物，为此，我们将沉积物（以数字表示）放入数组中。 存款在我们的 1000x1000 阵列中是随机的，中间有一个岛。

在程序结束时，我们希望将我们的海湾除数字为零的岛屿之外的所有地方都进行平滑处理。

我们做到了，但是运行时间太长了。有人可以帮助我们找到一种方法来加速我们的程序吗？

for j in range (0,375):
    for decale in range (250): 

        diff=M[749+decale][j]-M[749+decale+1][j] 
        if diff>=3: 
            M[749+decale][j]-=floor(19*diff/20) 
            M[749+decale+1][j]+=floor(19*diff/20)

for j in range (625,1000):  #for the S-W of the bay
    for decale in range (250): 

        diff=M[749+decale][j]-M[749+decale+1][j] 
        if diff>=3:
            M[749+decale][j]-=floor(19*diff/20) 
            M[749+decale+1][j]+=floor(19*diff/20) 

for i in range (800,1000):


    for decale in range (799): For the West of the bay

        diff=M[i][200+decale]-M[i][200+decale+1]
        if diff>=3: 
            M[i][200+decale]-=floor(19*diff/20) 
            M[i][200+decale+1]+=floor(19*diff/20)

    for decale in range (799): #Idem

        diff=M[i][799-decale]-M[i][799-decale-1] 
        if diff>=3: 
            M[i][799-decale]-=floor(19*diff/20) 
            M[i][799-decale-1]+=floor(19*diff/20) 

for j in range (0,375): #Along the island
    for decale in range (350):
        diff=M[850-decale][j]-M[850-decale-1][j] 
        if diff>=3: 
            M[850-decale][j]-=floor(19*diff/20)
            M[850-decale-1][j]+=floor(19*diff/20) 

for j in range(625,1000): #Idem
    for decale in range (350):
        diff=M[850-decale][j]-M[850-decale-1][j] 
        if diff>=3: 
            M[850-decale][j]-=floor(19*diff/20) 
            M[850-decale-1][j]+=floor(19*diff/20)

for i in range (500,549): #Idem
    for decale in range (500): 

        diff=M[i][499+decale]-M[i][499+decale+1] 
        if diff>=3:
            M[i][499+decale]-=floor(19*diff/20) 
            M[i][499+decale+1]+=floor(19*diff/20) 

    for decale in range (500):  #Idem

        diff=M[i][500-decale]-M[i][500-decale-1] 
        if diff>=3: 
            M[i][500-decale]-=floor(19*diff/20)   
            M[i][500-decale-1]+=floor(19*diff/20) 

return M

Answer 1

我不能说这些循环的正确性。我只是将它们翻译成 Numpy 数组，所以它可以更具可读性。

import numpy as np

M = np.array(M)

diff = M[749:-1,:375]-M[750:,:375]
M[749:-1,:375][diff>=3] -= np.floor(19*diff[diff>=3]/20)
M[750:  ,:375][diff>=3] += np.floor(19*diff[diff>=3]/20)

diff = M[749:-1,625:]-M[750:,625:] # for the S-W of the bay
M[749:-1,625:][diff>=3] -= np.floor(19*diff[diff>=3]/20)
M[750:  ,625:][diff>=3] += np.floor(19*diff[diff>=3]/20)

diff = M[800:,200:-1]-M[800:,201:] # for the West of the bay
M[800:,200:-1][diff>=3] -= np.floor(19*diff[diff>=3]/20)
M[800:,201:  ][diff>=3] += np.floor(19*diff[diff>=3]/20)

diff = M[800:,1:800]-M[800:,:799] #Idem 
M[800:,1:800][diff>=3] -= np.floor(19*diff[diff>=3]/20)
M[800:, :799][diff>=3] += np.floor(19*diff[diff>=3]/20)

diff = M[501:851,:375]-M[500:850,:375] #Along the island
M[501:851,:375][diff>=3] -= np.floor(19*diff[diff>=3]/20)
M[500:850,:375][diff>=3] += np.floor(19*diff[diff>=3]/20)

diff = M[501:851,625:]-M[500:850,625:] #Idem
M[501:851,625:][diff>=3] -= np.floor(19*diff[diff>=3]/20)
M[500:850,625:][diff>=3] += np.floor(19*diff[diff>=3]/20)

diff = M[500:549,499:-2]-M[500:549,500:-1] #Idem
M[500:549,499:-2][diff>=3] -= np.floor(19*diff[diff>=3]/20)
M[500:549,500:-1][diff>=3] += np.floor(19*diff[diff>=3]/20)

diff = M[500:549,1:501]-M[500:549,:500] #Idem
M[500:549,1:501][diff>=3] -= np.floor(19*diff[diff>=3]/20)
M[500:549,:500][diff>=3] += np.floor(19*diff[diff>=3]/20)