Jak szybko wygładzić tablicę 1000x1000 - python, tablice, python-2.7

Symulujemy sedymentację we francuskiej zatoce(Mont Saint Michel), w tym celu umieszczamy osady (przedstawione jako liczby) w tablicy. Depozyt jest losowy w naszej tablicy 1000x1000 z wyspą pośrodku.

Pod koniec programu chcielibyśmy wygładzić naszą zatokę wszędzie poza wyspą, gdzie liczby są równe zero.

Zrobiliśmy to, ale jest o wiele za długo. Czy ktoś może nam pomóc znaleźć sposób na przyspieszenie naszego programu?

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

Odpowiedzi:

Nie mogę powiedzieć nic o poprawności tych pętli. Właśnie przetłumaczyłem je na tablicę Numpy, więc może być nieco bardziej czytelna.

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)