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我的功能类似于使用蒙特卡洛积分法确定 pi 值。该函数基本上是在随机位置插入一个分子并估计能量。我将我的 for 循环转换为一个函数，以使用多处理模块在多个内核上运行。但是我从随机数生成器和所有过程的能量值中得到了相似的值。看起来它多次运行该功能，但报告的结果相似。

用户定义函数列表。为简单起见，没有给出细节。

def createRandomValuesforRotationTranslation(boxSpace):
def rotateTranslateMolec(randomValues,molec):
def makemolgroups(newMolec,peg):
def steric_closmol_clashes_vdw2(boxMolecs,ResID):

运行一个 for 循环；

nReplicates = 100 
count = 0
throws = 0

for i in range(nReplicates):
    throws += 1
    randomvalues = createRandomValuesforRotationTranslation(boxSpace)
    newMolec = rotateTranslateMolec(randomvalues,rna_mol)
    boxMolecs = makemolgroups(newMolec,peg)
    output = steric_closmol_clashes_vdw2(boxMolecs,ResID)
    count += output
    ratio = count /throws
    # binomial distribution method V_free = (number of accepted/total)Vbox 
    V_free = (count/throws)*output_vol
    p = count/throws
    std_binom = sqrt(throws*p*(1-p))
    error_binom = (output_vol/throws)*std_binom
    error_binom_fraction =  error_binom/V_free
    if i % 1 == 0:
        print("STEPS %d: BINOMIAL ERROR T.VOLUME %s: ERROR F.VOLUME %s: ESTIMATED VOLUME %s:" %(i, error_binom,error_binom_fraction,ratio))

然后我将 for 循环转换为 ;

def paralle_distances(nReplicates): 
    count = 0
    throws = 0
    for i in range(nReplicates):
        throws += 1
        randomvalues = createRandomValuesforRotationTranslation(boxSpace)
        newMolec = rotateTranslateMolec(randomvalues,rna_mol)
        boxMolecs = makemolgroups(newMolec,peg)
        output = steric_closmol_clashes_vdw2(boxMolecs,ResID)
        count += output
        ratio = count /throws
        # binomial distribution method V_free = (number of accepted/total)Vbox 
        V_free = (count/throws)*output_vol
        p = count/throws
        std_binom = sqrt(throws*p*(1-p))
        error_binom = (output_vol/throws)*std_binom
        error_binom_fraction =  error_binom/V_free
        if i % 1 == 0:
            print("STEPS %d: BINOMIAL ERROR T.VOLUME %s: ERROR F.VOLUME %s: ESTIMATED VOLUME %s:" %(i, error_binom,error_binom_fraction,ratio))
    return

import multiprocessing as mp
pool = mp.Pool(processes=4)
results = [pool.apply_async(paralle_distances, args=(x,)) for x in range(1,5)]

我只在这里打印了随机位置值。

(( 242.281, -50.4288, -7.54141 ), ( -0.679886, 0.674784, 0.287092 ), 201.097 degree)
(( 242.281, -50.4288, -7.54141 ), ( -0.679886, 0.674784, 0.287092 ), 201.097 degree)
(( 242.281, -50.4288, -7.54141 ), ( -0.679886, 0.674784, 0.287092 ), 201.097 degree)
(( 157.376, 67.453, -132.227 ), ( 0.0216526, 0.765258, 0.64336 ), 16.5297 degree)
(( 157.376, 67.453, -132.227 ), ( 0.0216526, 0.765258, 0.64336 ), 16.5297 degree)
(( 242.281, -50.4288, -7.54141 ), ( -0.679886, 0.674784, 0.287092 ), 201.097 degree)

多谢！

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1 回答 1

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我的直觉反应是建议使用x传递给每个线程的值作为随机数生成器的种子，确保每个线程中都有一个单独的生成器实例。如果您使用的生成器不能保证是线程安全的，这可能是您的问题。

于 2015-02-14T08:54:44.123 回答