Question

我正在尝试写一个方程来执行以下操作：

1）积分方程

2）存储该公式供以后使用

3）在100个不同的间隔上对第一个方程进行数值积分并求出第二个方程，每次增加一个固定量

import math
from sympy import *
import kvalues
import time
import random
import pandas as pd
import matplotlib.pyplot as plt

第一个任务很简单，我是这样完成的：

def integration_gas(number,Fa_0,Fb_0,Fc_0,v_0,a,b,c,d,e):    
  Ca_0 = Fa_0/v_0
  Cb_0 = Fb_0/v_0
  Cc_0 = Fc_0/v_0
  Ft_0 = Fb_0 + Fa_0 + Fc_0
  theta1 = Cb_0/Ca_0
  stoic1 = b/a
  theta2 = Cc_0/Ca_0
  stoic2 = c/a
  stoic3 = d/a
  stoic4 = e/a
  Cd = stoic3*x
  Ce = stoic4*x
  sigma = e+d-c-b-1
  epsilon = (Fa_0/Ft_0)*sigma
  Ca_eq = Ca_0*((1-x)/(1+epsilon*x))
  Cb_eq = Ca_0*((1*theta1-stoic1*x)/(1+epsilon*x))
  Cc_eq = Ca_0*((1*theta2-stoic2*x)/(1+epsilon*x))                 
  ra = 1*(Ca_eq**a)*(Cb_eq**b)*(Cc_eq**c)*final_k[number-1]    
  equation = Fa_0/ra 
  int1 = Integral(equation,x)    
  pprint(int1)
  evaluate = int1.doit()     
  pprint(evaluate)
  return equation

这部分代码可以很好地工作，直到第二部分。

def Ra_gas(number,Fa_0,Fb_0,Fc_0,v_0,a,b,c,d,e): 
    Ca_0 = Fa_0/v_0
    Cb_0 = Fb_0/v_0
    Cc_0 = Fc_0/v_0
    Ft_0 = Fb_0 + Fa_0 + Fc_0
    theta1 = Cb_0/Ca_0
    stoic1 = b/a
    theta2 = Cc_0/Ca_0
    stoic2 = c/a    
    sigma = e+d-c-b-1
    epsilon = (Fa_0/Ft_0)*sigma
    Ca_eq = Ca_0*((1-x)/(1+epsilon*x))
    Cb_eq = Ca_0*((1*theta1-stoic1*x)/(1+epsilon*x))
    Cc_eq = Ca_0*((1*theta2-stoic2*x)/(1+epsilon*x))                 
    ra = 1*(Ca_eq**a)*(Cb_eq**b)*(Cc_eq**c)*final_k[number-1]
    pprint(ra)
    return ra

这部分代码也可以正常工作。所以对于最后一部分，我有以下代码：

Number = 4
FA0 = 10
FB0 = 25
FC0 = 5
V0 = 2
A = 1
B = 2
C = 0.5
D = 1
E = 1

Ra = []
volume = []
Xff = []
eq1 = integration_gas(Number,FA0,FB0,FC0,V0,A,B,C,D,E)
Ra1 = Ra_gas(Number,FA0,FB0,FC0,V0,A,B,C,D,E)
#print(Ra1)
Xf = 0.01

# Calculates the reaction rate and volume for every interval of conversion
while Xf <=1:
    int2 = Integral(eq1,(x,0,Xf))
    volume.append(int2.doit())
    f = lambdify(x,Ra1,"math")
    f(Xf)
    Ra.append(f(Xf))    
    Xff.append(Xf)
    Xf += 0.01

然后我将结果绘制出来。我写的所有内容在某些情况下都可以正常工作，并且大约需要10到15秒即可完成。但是，特别是在这种情况下，我已经将这个代码运行了5个多小时，没有任何解决方案。如何优化此代码？

Answer 1

看看sympy，它可以符号积分原始方程式，然后可以通过numpy对其求值。对于“真实”数学Python有点慢，scipy Stack（numpy，matplotlib，sympy ...）要快得多。

虽然5个多小时有点长，但是您确定它确实可以执行吗？

编辑：一个简单的尝试抱歉，刚才注意到您要羊羔化，您可能想包括您的进口货，以便人们看到您在使用什么。

开头：

import numpy as np

让我们看一下这段代码：

Xf = 0.01
while Xf <=1: 
    int2 = Integral(eq1,(x,0,Xf)) 
    volume.append(int2.doit()) 
    f = lambdify(x,Ra1,"math") #you're lambdifying each iteration that takes time
    f(Xf) # no assignment here, unless you're doing something in place this line does nothing
    Ra.append(f(Xf)) 
    Xff.append(Xf) 
    Xf += 0.01

与此类似：

Xf = np.arange(0.01, 1.01, 0.01) #vector with values from 0.01 to 1 in steps of 0.01
f = np.vectorize(lambdify(x,Ra1,"math")) # you anonymous function but able to take np vectors/np arrays
Ra = f(Xf)
#Xff would be Xf

极慢的计算

1 个答案: