并行化：使用python并行化比使用monte-carlo模拟的并行化时间长

Question

嘿

我正在进行散射介质中光子输运的蒙特卡罗模拟。我正在尝试将其并行化，但与串行模拟相比，很难观察到运行时间方面的任何性能改进

montecarlo代码可以在下面找到。Photon类包含各种计算单光子输运和散射的方法，而RunPhotonPackage类运行给定厚度L的散射介质的光子系列N。这些是目前我唯一的输入参数：

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

我运行下面的序列代码来记录不同层厚度长度和给定光子数的位置直方图。在

^{pr2}$

那么这次磨合：

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

所以时间加倍了！！！我在Ubuntu32位的四核上运行这个程序，并使用ipcluster start -n 4在本地主机上启动一个控制器和四个引擎。我原以为并行化的代码将在运行串行代码所需时间的1/4内运行，但显然不是这样。在

为什么会这样？如何纠正？在

谢谢你的建议。在

格雷格

Answer 1

我做了一些修改来简化你的例子。在我的Mac电脑上，串行版本的运行时间大约为18秒，而带有4个引擎的并行版本的运行时间大约为一半。鉴于任务的持续时间不均衡，这似乎是合理的。在

按照之前的设置方式，发动机出现错误，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，我只是黑客攻击系统路径对于本例，但在生产环境中，您可能会适当地处理此问题。在

我想你不想在循环里“等”一下。另外，async_map（）方法似乎比async_apply（）更方便。在

要运行此操作，请创建一个目录，将以下代码复制到名为“的文件中”光子.py，并在那里创建一个空的“init.py”。修改插入到的代码中的行系统路径以引用您的新目录。在那里更改目录并运行“python光子.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)