Can the function variables be dynamically updated with each call?

Question

Can the function variables be dynamically updated with each call?

I am currently tackling an Ordinary Differential Equation (ODE) system using the solve_ivp function and I aim to modify the local variables within the function each time it is called by the solver. Specifically, I wish to update the Lagrange multipliers (lambdas_i) so that the subsequent step in solve_ivp utilizes the values from the previous one. (The ''reconstruct'' function comes from a Python module that employs a method for reconstructing a size distribution based on given moments). Is there a way to achieve this? Below is the code snippet:

import time 
import numpy as np
import scipy.integrate as integrate
from numpy import sqrt, sin, cos, pi
import math 
import pylab as pp
from pymaxent import reconstruct

start = time.time()

'''Initializing variables'''

t=np.linspace(0, 60,31)
tspan=(0,60)

initial_m=[]
for i in range(4):
    def distr(L,i=i):
        return (L**i)*0.0399*np.exp(-((L-50)**2)/200)
   
    m, err=integrate.quad(distr, 0, np.inf)
    print('m(',i,')=',m)
    initial_m.append(m)

'''Solving the ODE system using Maximum Entropy with G(L)=1+0.002*L'''

def moments(t,y):
    m0 = y[0]
    m1 = y[1]
    m2 = y[2]
    m3 = y[3]
    Lmean=m1
    σ=(m2-Lmean**2)**(1/2)
    Lmin=Lmean-3*σ
    Lmax=Lmean+4*σ
    bnds=[Lmin,Lmax]
    L=np.linspace(Lmin,Lmax)
    
    sol, lambdas_i= reconstruct(mu=y ,bnds=bnds)
    print(lambdas_i)
    
    dm0dt=0
        
    def moment1(L):
        return(sol(L)+0.002*sol(L)*L)                
    dm1dt, err1=integrate.quad(moment1,Lmin,Lmax)
    
    def moment2(L):
        return(2*L*(sol(L)+0.002*sol(L)*L))
    
    dm2dt, err2=integrate.quad(moment2,Lmin,Lmax)
    
    def moment3(L):
        return(3*L**2*(sol(L)+0.002*sol(L)*L))
    
    dm3dt, err3=integrate.quad(moment3,Lmin,Lmax)
    
    return(dm0dt,dm1dt,dm2dt,dm3dt)

'''Using the Backward Differentiation Formula (BDF) algorithm step by step'''

r=integrate.solve_ivp(moments,tspan,initial_m,method='BDF',jac=None,t_eval=t,rtol=10**(-3))

end = time.time()

print('Total time taken:', {end-start})

python iteration ode timestep

Answer 1

Answer №1

If you're looking for a solution, here is one approach you could take. I'll provide a simplified example problem to demonstrate the strategy that can be applied to your specific case.

def sequence():
  list = [1, 0]
  while True:
    previous = list[1]
    current = list[0] + previous
    list = [previous, current]
    print(current)
    input()

The function above showcases how you can generate the next term of the fibonacci sequence, where each term is the sum of the two preceding terms. The input command serves to create breaks between iterations, given the infinite nature of this sequence. To enhance flexibility and utility, consider transforming the function into a generator like so:

def sequence():
  list = [1, 0]
  while True:
    previous = list[1]
    current = list[0] + previous
    list = [previous, current]
    yield current

To achieve similar results, you can iterate over the generator as follows:

for item in sequence():
  print(item)
  input()

This method may seem redundant at first glance. However, utilizing a generator becomes advantageous when you require access to the next number of the sequence at any point within your code, rather than being confined to a loop structure. For instance:

gen = sequence()
next(gen) # returns 1
next(gen) # returns 1
next(gen) # returns 2
next(gen) # returns 3
next(gen) # returns 5

Following this pattern...

An alternative approach involves using a global variable instead of a local variable:

list = [1, 0]

def sequence():
  previous = list[1]
  current = list[0] + previous
  list[0] = previous
  list[1] = current
  return current

By defining the variable list outside the function scope, it maintains its state beyond individual function calls. Thus, running the updated code remains consistent with the previous implementation:

while True:
  print(sequence())
  input()

Both methods discussed can be integrated into your own code base effectively.

Answer 2

If you're looking for a solution, here is one approach you could take. I'll provide a simplified example problem to demonstrate the strategy that can be applied to your specific case.

def sequence():
  list = [1, 0]
  while True:
    previous = list[1]
    current = list[0] + previous
    list = [previous, current]
    print(current)
    input()

The function above showcases how you can generate the next term of the fibonacci sequence, where each term is the sum of the two preceding terms. The input command serves to create breaks between iterations, given the infinite nature of this sequence. To enhance flexibility and utility, consider transforming the function into a generator like so:

def sequence():
  list = [1, 0]
  while True:
    previous = list[1]
    current = list[0] + previous
    list = [previous, current]
    yield current

To achieve similar results, you can iterate over the generator as follows:

for item in sequence():
  print(item)
  input()

This method may seem redundant at first glance. However, utilizing a generator becomes advantageous when you require access to the next number of the sequence at any point within your code, rather than being confined to a loop structure. For instance:

gen = sequence()
next(gen) # returns 1
next(gen) # returns 1
next(gen) # returns 2
next(gen) # returns 3
next(gen) # returns 5

Following this pattern...

An alternative approach involves using a global variable instead of a local variable:

list = [1, 0]

def sequence():
  previous = list[1]
  current = list[0] + previous
  list[0] = previous
  list[1] = current
  return current

By defining the variable list outside the function scope, it maintains its state beyond individual function calls. Thus, running the updated code remains consistent with the previous implementation:

while True:
  print(sequence())
  input()

Both methods discussed can be integrated into your own code base effectively.

Can the function variables be dynamically updated with each call?

Answer №1

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