如何快速平滑1000x1000阵列

Question

我们正在模拟法国湾（圣米歇尔山）的沉积，为此，我们将沉积物（以数字表示）放入一个数组中。 沉积物在我们的1000 X1000阵列中是随机的，中间有一个岛。你知道吗

在节目结束时，我们想把我们的海湾除数字等于零的岛屿以外的所有地方都平整。你知道吗

我们做到了，但是跑起来太长了。。有人能帮我们找到一个方法来加速我们的计划吗？你知道吗

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

关于这些循环的正确性，我不能说什么。我刚刚把它们翻译成Numpy数组，所以它的可读性更好一些。你知道吗

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)