一个问题。
使用 multiprocessing 是为了获得更快的代码。
但是在使用以下框架之后，获取输出所花费的时间与普通代码所花费的时间相同或更多。
import multiprocessing
def code() :
my code
if __name__ == '__main__' :
p = multiprocessing.Process(target = code)
p.start()
p.join()
因为是 2 个处理器的笔记本电脑，运行此代码后程序要我导入数据两次。 问题是时间以这种方式没有任何意义。我遇到了长时间只要没有并行的正常代码。
import numpy as np
from scipy.integrate import odeint
from math import *
from scipy.integrate import quad
import pandas as pd
import os
import warnings
warnings.filterwarnings("ignore")
#you need add the following 3 lines
from multiprocessing import Pool
from multiprocessing import Process
import multiprocessing
print("Model 4, Equation 11")
print("")
###################### STEP NUMBER #######################
N = int(input("PLEASE ENTER NUMBER OF STEP WALKS: ")) # Step walk by user
dec=int(input("NUMBER OF DECIMAL PLACES OF OUTPUTS (RECOMENDED 10-15)?"))
print("")
print("PLEASE WAIT, METROPOLIS HASTINGS IS RUNNING ... ")
print("")
def FIT():
##########################################################
od0o = np.zeros((N,))
od0o[0]=0.72
od0n = np.zeros((N,))
Mo = np.zeros((N,))
Mo[0]= 0
Mn = np.zeros((N,))
co = np.zeros((N,))
co[0]= 0.84
cn = np.zeros((N,))
bo = np.zeros((N,))
bo[0]= 0.02
bn = np.zeros((N,))
H0o = np.zeros((N,))
H0o[0]= 70
H0n = np.zeros((N,))
Orco = np.zeros((N,))
Orco[0]= 0.0003
Orcn = np.zeros((N,))
temp=1e10 # a big number
##########################################################
CovCMB=[[3.182,18.253,-1.429],
[18.253,11887.879,-193.808],
[-1.429,-193.808,4.556]] # CMB DATA
##########################################################
def OD_H(U,z,c,b,Orc):
od, H = U
Omegai = 3 * b * ((1- od - 2*(Orc)**0.5) + (1- od - 2*(Orc)**0.5)**2/(1-2*(Orc)**0.5)) #equation 10
div1=np.divide((c**2-od),(1+z),where=(1+z)!=0)
div2=np.divide(H ,(1+z),where=(1+z)!=0)
dMdt = (div1)*(6*od+6-9*(od/c**2)+ Omegai)*(1+c**2+od*(1-3/c**2))**(-1)
dHdt = (div2)*(6*od+6-9*(od/c**2)+ Omegai)*(1+c**2+od*(1-3/c**2))**(-1)
return [dMdt, dHdt]
def solution(H0,z1,z,od0,c,b,Orc):
U = odeint(OD_H,[od0,H0],[z1,z], args=(c,b,Orc))[-1]
od, H = U
return H
##########################################################
def DMCMB1(H0,z1,z,od0,c,b,Orc):
dm = 1090 * 1/solution(H0,z1,z,od0,c,b,Orc)
return dm
def R1(H0,z1,z,od0,c,b,Orc):
#r=sqrt(Om)*(70/299000)*rszstar(z,Om,Od)
r = sqrt(1-od0-2*(Orc)**0.5)*DMCMB1(H0,z1,z,od0,c,b,Orc)
return r
def lA1(H0,z1,z,od0,c,b,Orc):
la=((3.14*299000/H0)*DMCMB1(H0,z1,z,od0,c,b,Orc))/(153)
return la
def CMBMATRIX1(H0,z1,z,od0,c,b,Orc,M):
hmCMB1=[lA1(H0,z1,z,od0,c,b,Orc)-301.57, R1(H0,z1,z,od0,c,b,Orc)-1.7382+M, 0.0222-0.02262]
vmCMB1=[[lA1(H0,z1,z,od0,c,b,Orc)-301.57], [R1(H0,z1,z,od0,c,b,Orc)-1.7382], [0.0222-0.02262]]
fmCMB1=np.dot(hmCMB1,CovCMB)
smCMB1=np.dot(fmCMB1,vmCMB1)[0]
return smCMB1
######################################################
def TOTAL(H0, od0, c, b,Orc, M) :
total = CMBMATRIX1(H0,0,1090,od0,c,b,Orc,M)
return total
######################################################
################## MCMC - MH #########################
highest=0
pat='C:/Users/21/Desktop/MHT/Models/outputs'
file_path = os.path.join(pat,'Model3.txt')
file_path2 = os.path.join(pat,'Model3min.txt')
with open(file_path, 'w') as f: # DATA WILL BE SAVED IN THIS FILE, PLEASE BECAREFUL TO CHANGE THE NAME IN EACH RUN TO AVOIDE REWRITING.
with open(file_path2, 'w') as d:
for i in range (1,N):
num = 0
R = np.random.uniform(0,1)
while True:
num += 1
od0n[i] = od0o[i-1] + 0.001 * np.random.normal()
H0n[i] = H0o[i-1] + 0.01 * np.random.normal()
bn[i] = bo[i-1] + 0.001 * np.random.normal()
cn[i] = co[i-1] + 0.001 * np.random.normal()
Mn[i] = Mo[i-1] + 0.01 * np.random.normal()
Orcn[i] = Orco[i-1] + 0.00001 * np.random.normal()
L = np.exp(-0.5 * (TOTAL(H0n[i], od0n[i], cn[i], bn[i],Orcn[i], Mn[i]) - TOTAL(H0o[i-1], od0o[i-1], co[i-1], bo[i-1],Orco[i-1], Mo[i-1]))) # L = exp(-( x^2 )/2)
LL=min(1,max(L,0))
if LL>R:
od0o[i]= od0n[i]
H0o[i] = H0n[i]
bo[i] = bn[i]
co[i] = cn[i]
Mo[i] = Mn[i]
Orco[i] = Orcn[i]
chi = TOTAL(H0o[i], od0o[i], co[i], bo[i],Orco[i], Mo[i])
else:
od0o[i]= od0o[i-1]
H0o[i] = H0o[i-1]
bo[i] = bo[i-1]
co[i] = co[i-1]
Mo[i] = Mo[i-1]
Orco[i] = Orco[i-1]
chi = TOTAL(H0o[i], od0o[i], co[i], bo[i],Orco[i], Mo[i])
if (Mo[i]>0 and 0<bo[i]<0.09 and Orco[i]>0) or num>100: # constraining the value to stay in positive area
highest = max(num, highest)
break
f.write("{0}\t{1}\t{2}\t{3}\t{4}\t{5}\t{6}\t{7}\t{8}\t{9}\t{10}\t{11}\t{12}\n".format(round(chi,dec),' ',round(H0o[i],dec),' ',round(od0o[i],dec),' ',
round(co[i],dec),' ',round(bo[i],dec),' ',
round(Orco[i],dec),' ',round(Mo[i],dec)))
if chi<temp:
temp=chi
aa = H0o[i]
bb = od0o[i]
cc = co[i]
dd = bo[i]
ee = Mo[i]
ff=Orco[i]
Om=1-2*sqrt(Orco[i])-od0o[i]
# minimum of chi and its parameters
d.write("{0}\t{1}\t{2}\t{3}\t{4}\t{5}\t{6}\t{7}\t{8}\t{9}\t{10}\t{11}\t{12},\t{13}\t{14}\n".format(round(temp,dec), "H =", round(aa,dec), "Orc=",
round(ff,dec), "OD =",round(bb,dec),"c =",
round(cc,dec),"b =", round(dd,dec),
"M =",round(ee,dec),"Om =",round(Om,dec)))
print(round(temp,dec), "H =", round(aa,dec), "Orc=",round(ff,dec), "OD =",round(bb,dec),"c =", round(cc,dec),"b =", round(dd,dec), "M =",round(ee,dec),"Om =",round(Om,dec))
#print(highest)
print("")
#test = input("Press the enter to exit...")
#print(test)
if __name__ == '__main__':
p = multiprocessing.Process(target=FIT)
p.start()
p.join()
最佳答案
我认为您错过了 multiprocessing 的主要概念。它不会更快地运行你的代码，它只是让你在另一个进程中运行一些东西来绕过 GIL ( https://wiki.python.org/moin/GlobalInterpreterLock )。
它可用于对不同输入值的某些函数进行并行计算，例如 docs 中的示例
from multiprocessing import Pool
def f(x):
return x*x
if __name__ == '__main__':
p = Pool(5)
print(p.map(f, [1, 2, 3]))
这导致在不同的进程中计算 f，并且每个进程都返回单独的值
关于python - 使用多处理不会减少计算时间，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50722291/