如何在Python中並行化Ising模型（多處理程序包）？

Question

我有一個Python腳本，可以對一個2D晶格的Ising模型進行Monte Carlo模擬。 MC模擬是令人尷尬的平行，每個溫度的採樣可以分配到不同的線程。我想爲此使用多處理模塊，但是對於此包來說是新的。我怎樣才能做到這一點？

from __future__ import division 
import numpy as np 
from numpy.random import rand 
import matplotlib.pyplot as plt 
import multiprocessing as mp 

nt = 5 
N  = 16 

Energy = np.zeros(nt) 
Magnetization = np.zeros(nt) 

T = np.linspace(1, 5, nt) 

## monte carlo moves 
def mcmove(config, beta): 
    for i in range(N): 
     for j in range(N): 
       a = np.random.randint(0, N) 
       b = np.random.randint(0, N) 
       s = config[a, b] 
       nb = config[(a+1)%N,b] + config[a,(b+1)%N] + config[(a-1)%N,b] + config[a,(b-1)%N] 
       cost = 2*s*nb 
       if cost < 0: 
        s *= -1 
       elif rand() < np.exp(-cost*beta): 
        s *= -1 
       config[a, b] = s 
    return config 

# calculate thermodynamic variables 
def calcEnergy(config): 
    energy = 0 
    for i in range(len(config)): 
     for j in range(len(config)): 
      S = config[i,j] 
      nb = config[(i+1)%N, j] + config[i,(j+1)%N] + config[(i-1)%N, j] + config[i,(j-1)%N] 
      energy += -nb*S 
    return energy/4. 

def calcMag(config): 
    mag = np.sum(config) 
    return mag 

def simulate(): 
# -->: parallelize T 
    for m in range(len(T)): 
     E1 = M1 = 0 
     def initialstate(N): 
      state = 2*np.random.randint(2, size=(N,N))-1 
      return state 

     eqSteps = 2000 
     config = initialstate(N) 
     for i in range(eqSteps): 
      mcmove(config, 1.0/T[m]) 

     mcSteps = 2000 
     for i in range(mcSteps): 
      mcmove(config, 1.0/T[m]) 
      Ene = calcEnergy(config)   
      Mag = calcMag(config)   

      E1 = E1 + Ene 
      M1 = M1 + Mag 

      Energy[m]   = E1/(mcSteps*N*N) 
      Magnetization[m] = M1/(mcSteps*N*N) 

    return Magnetization, Energy 

Magnetization, Energy = simulate() 
print Magnetization, Energy 

Answer 1

在這裏你去：

from __future__ import division 
import numpy as np 
from numpy.random import rand 
import matplotlib.pyplot as plt 
import multiprocessing as mp 

nt = 5 
N  = 16 

Energy = np.zeros(nt) 
Magnetization = np.zeros(nt) 

T = np.linspace(1, 5, nt) 

## monte carlo moves 
def mcmove(config, beta): 
    for i in range(N): 
     for j in range(N): 
       a = np.random.randint(0, N) 
       b = np.random.randint(0, N) 
       s = config[a, b] 
       nb = config[(a+1)%N,b] + config[a,(b+1)%N] + config[(a-1)%N,b] + config[a,(b-1)%N] 
       cost = 2*s*nb 
       if cost < 0: 
        s *= -1 
       elif rand() < np.exp(-cost*beta): 
        s *= -1 
       config[a, b] = s 
    return config 

# calculate thermodynamic variables 
def calcEnergy(config): 
    energy = 0 
    for i in range(len(config)): 
     for j in range(len(config)): 
      S = config[i,j] 
      nb = config[(i+1)%N, j] + config[i,(j+1)%N] + config[(i-1)%N, j] + config[i,(j-1)%N] 
      energy += -nb*S 
    return energy/4. 

def calcMag(config): 
    mag = np.sum(config) 
    return mag 


def run_one(m): 
    E1 = M1 = _Energy = _Magnetization = 0 

    def initialstate(N): 
     state = 2 * np.random.randint(2, size=(N, N)) - 1 
     return state 

    eqSteps = 2000 
    config = initialstate(N) 
    for i in range(eqSteps): 
     mcmove(config, 1.0/m) 

    mcSteps = 2000 
    for i in range(mcSteps): 
     mcmove(config, 1.0/m) 
     Ene = calcEnergy(config) 
     Mag = calcMag(config) 

     E1 = E1 + Ene 
     M1 = M1 + Mag 

     _Energy= E1/(mcSteps * N * N) 
     _Magnetization = M1/(mcSteps * N * N) 

    return _Energy, _Magnetization 


def parallel_sim(): 
    p = mp.Pool() 
    results = p.map(run_one, T) 
    p.close() 
    p.join() 
    _e = [] 
    _m = [] 
    for _r in results: 
     _e.append(_r[0]) 
     _m.append(_r[1]) 
    return _e, _m 

print parallel_sim() 

我不知道該算法任何事情，請確認我沒有打破它。但是，現在它將T的每個元素映射到並行進程。