如何在Python(多处理包)中并行化Ising模型?
我有一个Python脚本,可以对二维晶格的伊辛模型进行蒙特卡罗模拟。MC模拟是令人尴尬的并行,每个温度的采样可以分配到不同的线程。我想为此使用多处理模块,但这是一个新的软件包。我该怎么做如何在Python(多处理包)中并行化Ising模型?,python,parallel-processing,python-multiprocessing,Python,Parallel Processing,Python Multiprocessing,我有一个Python脚本,可以对二维晶格的伊辛模型进行蒙特卡罗模拟。MC模拟是令人尴尬的并行,每个温度的采样可以分配到不同的线程。我想为此使用多处理模块,但这是一个新的软件包。我该怎么做 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
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
来自未来进口部的
将numpy作为np导入
从numpy.random导入rand
将matplotlib.pyplot作为plt导入
将多处理作为mp导入
nt=5
N=16
能量=np.零(nt)
磁化强度=np.零(nt)
T=np.linspace(1,5,nt)
##蒙特卡罗移动
def mcmove(配置,测试版):
对于范围(N)中的i:
对于范围(N)内的j:
a=np.random.randint(0,N)
b=np.random.randint(0,N)
s=配置[a,b]
nb=配置[(a+1)%N,b]+配置[a,(b+1)%N]+配置[(a-1)%N,b]+配置[a,(b-1)%N]
成本=2*s*nb
如果成本<0:
s*=-1
elif rand():并行化T
对于范围内的m(len(T)):
E1=M1=0
def初始状态(N):
state=2*np.random.randint(2,size=(N,N))-1
返回状态
方程步骤=2000
配置=初始状态(N)
对于范围内的i(eqSteps):
mcmove(配置,1.0/T[m])
mcSteps=2000
对于范围内的i(mcSteps):
mcmove(配置,1.0/T[m])
Ene=calcEnergy(配置)
Mag=calcMag(配置)
E1=E1+Ene
M1=M1+Mag
能量[m]=E1/(微秒级*N*N)
磁化强度[m]=M1/(mcSteps*N*N)
回磁
磁化,能量=模拟()
打印磁化、能量
给你:
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()
来自未来进口部的
将numpy作为np导入
从numpy.random导入rand
将matplotlib.pyplot作为plt导入
将多处理作为mp导入
nt=5
N=16
能量=np.零(nt)
磁化强度=np.零(nt)
T=np.linspace(1,5,nt)
##蒙特卡罗移动
def mcmove(配置,测试版):
对于范围(N)中的i:
对于范围(N)内的j:
a=np.random.randint(0,N)
b=np.random.randint(0,N)
s=配置[a,b]
nb=配置[(a+1)%N,b]+配置[a,(b+1)%N]+配置[(a-1)%N,b]+配置[a,(b-1)%N]
成本=2*s*nb
如果成本<0:
s*=-1
elif rand()
我对算法一无所知,请确认我没有弄坏它。现在它将T的每个元素映射到并行进程 如果答案能回答你的问题,你可以接受。