python中的monte carlo模拟：使用ipyparallel的并行化比序列化花费更长的时间

Question

hei，

我正在制作散射介质中光子传输的蒙特卡罗模拟。我试图对其进行并列化，但与串行模拟相比，难以观察到运行时间方面的任何性能提升

montecarlo代码可以在下面找到。 Photon类包含用于计算单光子的传输和散射的各种方法，而类RunPhotonPackage对于给定厚度L的散射介质运行一系列N光子。那些是我目前唯一的输入参数：

import matplotlib.pyplot as plt
import numpy as np
from numpy.random import random as rand


NPHOTONS = 100000 # Nb photons
PI  = np.pi
EPS = 1.e-6
L = 100. # scattering layer thickness

class Photon():

    mut = 0.02 
    k = [0,0,1]

    def __init__(self,ko,pos):
        Photon.k = ko
        self.x = pos[0]
        self.y = pos[1]
        self.z = pos[2]

    def move(self):
        ksi = rand(1)
        s = -np.log(1-ksi)/Photon.mut

        self.x = self.x + s*Photon.k[0]
        self.y = self.y + s*Photon.k[1]
        self.z = self.z + s*Photon.k[2]       
        zPos = self.z
        return zPos 

    def exittop(self):

        newZpos = 0


    def exitbase(self):
        newZpos = 0


    def HG(self,g):
        rand_teta = rand(1)
        costeta = 0.5*(1+g**2-((1-g**2)/(1-g + 2.*g*rand_teta))**2)/g

        return costeta

    def scatter(self):
        # calculate new angle of scattering
        phi = 2*PI*rand(1)                
        costeta = self.HG(0.85)
        sinteta = (1-costeta**2)**0.5 


        sinphi = np.sin(phi) 
        cosphi = np.cos(phi)

        temp = (1-Photon.k[2]**2)**0.5

        if np.abs(temp) > EPS:        

            mux = sinteta*(Photon.k[0]*Photon.k[2]*cosphi-Photon.k[1]*sinphi)/temp + Photon.k[0]*costeta 
            muy = sinteta*(Photon.k[1]*Photon.k[2]*cosphi+Photon.k[0]*sinphi)/temp + Photon.k[1]*costeta
            muz = -sinteta*cosphi*temp + Photon.k[2]*costeta

        else:
            mux = sinteta*cosphi 
            muy = sinteta*sinphi
            if Photon.k[2]>=0:
                muz = costeta
            else:
                muz = -costeta


        # update the new direction of the photon 
        Photon.k[0] = mux
        Photon.k[1] = muy
        Photon.k[2] = muz        


class RunPhotonPackage():

    def __init__(self,L,NPHOTONS):
        self.L = L
        self.NPHOTONS = NPHOTONS

    def RunPhoton(self):
        Dist_Pos = np.zeros((3,self.NPHOTONS))
        # loop over number of photon
        for i in range(self.NPHOTONS):

            # inititate initial photon direction
            k_init = [0,0,1]
            k_init_norm = k_init/np.linalg.norm(k_init) # initial photon direction.
            # initiate new photon with initial direction   
            pos_init = [0,0,0]
            newPhoton = Photon(k_init_norm,pos_init)
            newZpos = 0.

            # while the photon is still in the layer, move it and scatter it
            while ((newZpos >= 0.) and (newZpos <= self.L)):

                newZpos = newPhoton.move()
                newscatter = newPhoton.scatter()

            Dist_Pos[0,i] = newPhoton.x
            Dist_Pos[1,i] = newPhoton.y
            Dist_Pos[2,i] = newPhoton.z


        return Dist_Pos

我运行以下序列代码来记录各种层厚度长度和给定光子数的位置直方图。

import time
tic = time.time()
dictresult = {}
for L in np.arange(10,100,10):
    print('L={0} m'.format(L))
    Dist_Pos = RunPhotonPackage(L,10000).RunPhoton()
    dictresult['{0}'.format(L)]=Dist_Pos
toc = time.time()
print('sec Elapsed: {0}s'.format(toc-tic))

然后进入：

sec Elapsed: 26.425330162s

当我尝试使用ipyparallel并行化代码时：

import ipyparallel
clients = ipyparallel.Client()
clients.ids
dview = clients[:]

dview.execute('import numpy as np')
dview.execute('from numpy.random import random as rand')
dview['PI'] = np.pi
dview['EPS']= 1.e-6

dview.push({"Photon": Photon, "RunPhotonPackage": RunPhotonPackage})

def RunPhotonPara(L):
    LayerL = RunPhotonPackage(L,10000)
    dPos = LayerL.RunPhoton()
    return dPos

tic = time.time()
dictresultpara = []
for L in np.arange(10,100,10):
    print('L={0}'.format(L))
    value = dview.apply_async(RunPhotonPara,L)
    dictresultpara.append(value)
    clients.wait(dictresultpara)
toc = time.time()
print('sec Elapsed: {0}s'.format(toc-tic))

它运行于：

sec Elapsed: 55.4289810658s

超过一倍的时间!!!我在带有四个内核的ubuntu 32位上运行它，并使用ipcluster start -n 4在localhost上启动一个控制器和4个引擎。我期待这段参数的代码运行时间约为运行序列号的1/4，但显然不会。

为什么这样以及如何纠正它？

感谢您的任何建议。

格雷格

Answer 1

我做了一些更改以简化您的示例。串行版本在我的Mac上运行大约18秒，并且具有4个引擎的并行版本在大约一半的时间内运行。鉴于任务的持续时间不均匀，这似乎是合理的。

以前设置的方式，引擎中发生错误，因此快速返回。似乎通过字典传递类是不够的。相反，代码现在导入定义每个引擎上的类的模块。 请注意，我只是为此示例攻击了sys.path，但可能是在生产环境中，您可以适当地处理此问题。

我认为你不想在循环中“等待”。此外，async_map（）方法似乎比async_apply（）更方便。

要运行此操作，请创建一个目录，将以下代码复制到该目录中名为“photon.py”的文件中，并在其中创建一个空的“ init .py”。修改插入sys.path的代码中的行以引用新目录。在那里更改目录并运行“python photon.py”：

# photon.py

import ipyparallel
import numpy as np
from numpy.random import random as rand
import time

NPHOTONS = 100000 # Nb photons
PI  = np.pi
EPS = 1.e-6
L = 100. # scattering layer thickness

class Photon():

    mut = 0.02 
    k = [0,0,1]

    def __init__(self,ko,pos):
        Photon.k = ko
        self.x = pos[0]
        self.y = pos[1]
        self.z = pos[2]

    def move(self):
        ksi = rand(1)
        s = -np.log(1-ksi)/Photon.mut

        self.x = self.x + s*Photon.k[0]
        self.y = self.y + s*Photon.k[1]
        self.z = self.z + s*Photon.k[2]       
        zPos = self.z
        return zPos 

    def exittop(self):

        newZpos = 0


    def exitbase(self):
        newZpos = 0


    def HG(self,g):
        rand_teta = rand(1)
        costeta = 0.5*(1+g**2-((1-g**2)/(1-g + 2.*g*rand_teta))**2)/g

        return costeta

    def scatter(self):
        # calculate new angle of scattering
        phi = 2*PI*rand(1)                
        costeta = self.HG(0.85)
        sinteta = (1-costeta**2)**0.5 


        sinphi = np.sin(phi) 
        cosphi = np.cos(phi)

        temp = (1-Photon.k[2]**2)**0.5

        if np.abs(temp) > EPS:        

            mux = sinteta*(Photon.k[0]*Photon.k[2]*cosphi-Photon.k[1]*sinphi)/temp + Photon.k[0]*costeta 
            muy = sinteta*(Photon.k[1]*Photon.k[2]*cosphi+Photon.k[0]*sinphi)/temp + Photon.k[1]*costeta
            muz = -sinteta*cosphi*temp + Photon.k[2]*costeta

        else:
            mux = sinteta*cosphi 
            muy = sinteta*sinphi
            if Photon.k[2]>=0:
                muz = costeta
            else:
                muz = -costeta


        # update the new direction of the photon 
        Photon.k[0] = mux
        Photon.k[1] = muy
        Photon.k[2] = muz        


class RunPhotonPackage():

    def __init__(self,L,NPHOTONS):
        self.L = L
        self.NPHOTONS = NPHOTONS

    def RunPhoton(self):
        Dist_Pos = np.zeros((3,self.NPHOTONS))
        # loop over number of photon
        for i in range(self.NPHOTONS):

            # inititate initial photon direction
            k_init = [0,0,1]
            k_init_norm = k_init/np.linalg.norm(k_init) # initial photon direction.
            # initiate new photon with initial direction   
            pos_init = [0,0,0]
            newPhoton = Photon(k_init_norm,pos_init)
            newZpos = 0.

            # while the photon is still in the layer, move it and scatter it
            while ((newZpos >= 0.) and (newZpos <= self.L)):

                newZpos = newPhoton.move()
                newscatter = newPhoton.scatter()

            Dist_Pos[0,i] = newPhoton.x
            Dist_Pos[1,i] = newPhoton.y
            Dist_Pos[2,i] = newPhoton.z

        return Dist_Pos

def RunPhoton(L):
    print('L={0}'.format(L))
    return RunPhotonPackage(L, 10000).RunPhoton()

def serialTest(values):
    print "Running serially..."
    tic = time.time()
    results = map(RunPhoton, values)
    print results
    toc = time.time()
    print('sec Elapsed: {0}s'.format(toc-tic))

def parallelTest(values):
    print "Running in parallel..."
    client = ipyparallel.Client()
    view = client[:]

    view.execute('import sys')

    # CHANGE THIS PATH TO REFER TO WHEREVER YOU PUT THIS CODE
    view.execute('sys.path.insert(0, "/Users/rjp/ipp")')
    view.execute('from photon import *')

    tic = time.time()
    asyncResults = view.map_async(RunPhoton, values)
    print asyncResults.get()
    toc = time.time()
    print('sec Elapsed: {0}s'.format(toc-tic))    


if __name__ == "__main__":
    values = np.arange(10, 100, 10)

    serialTest(values)
    parallelTest(values)