My question is if there's a way to take some values in a function that are not
integrated in odeint.
Exemple: if I have a derivative dy(x)/dt = A*x+ln(x) and before to get this equation I computed A throught of a intermediate equation like A = B*D . I would like to take the A's value during the process.
More detailed (only exemple):
def func(y,t)
K = y[0]
B = 3
A = cos(t**2) + B
dy/dt = A*t+ln(t)
return [dy/dt]
Can I take A's values of function?
The answer for Josh Karpel
The code is like that:
def Reaction(state,t):
# Integrate Results
p = state[0]
T = state[1]
# function determine enthalpy of system
f1(T,p) = enthalpy
# function determine specific volume of system
f2(T,p) = specific volume
# function determine heat release by reactions
f3(T,p,t) = heat release by reactions
# Derivatives
dp/dt = f(T,p,enthalpy,specific volume,heat release by reactions)
dT/dt = f(T,p,enthalpy,specific volume,heat release by reactions)
The real code is bigger than that. But, I would like to know if there is a way to store the values of f1 (enthalpy), f2 (specific volume), f3 (heat release) as a vector or tuple during the process of solution of odeint with the same size of p and T.
It's not entirely clear what you want, but it sounds like you need to pass another value to the function you're integrating over. There are two options I can think of:
scipy.integrate.odeint takes an args argument which contains extra arguments to be passed to the integrand function, which could then have signature y(t, A).
You could use functools.partial to construct a new function which has the argument A for the integrand function y(t, A) already set.
Related
It is now one week that i am stuck with this and i don't know why it is happening. So i want to present to you my problem to see if you have a solution.
I have this code. My purpose here is to update this variables A and B using a loop. So, i first execute all calcs using A and B that i give when i call the class, then the code compute gamma and xi and finally i want to compute new A and B.
I create in my class first the functions that i use to find the values that i need to use for computing gamma and xi. Then i create functions that i use to find gamma and xi. In this function i call the previous functions with self.fun.
Then i create the function that i show here to compute new A and B values and i insert this in a loop because i want to iterate this function until i reach a convergence.
But...
This method works correctly to compute A, but, when it needs to compute B, it uses the new A before computing B. It implies that, when it computes A, old A and B are used to compute gamma and xi and then A. Where it computes B, it uses new A and B to compute gamma and xi but i want that it uses the same that it uses to compute first A.
def update(self):
for n in range(self.iter):
gamma = self.gamma()
xi = self.xi()
# new trans matrix
new_A = []
for i in range(len(self.A)):
temp = []
for j in range(len(self.A[i])):
numerator = 0
denominator = 0
for r in range(len(self.seq)):
for t in range(len(xi[r])):
numerator += xi[r][t][i][j]
denominator += gamma[r][t][i]
aij = numerator / denominator
temp.append(aij)
new_A.append(temp)
self.A = new_A
emission_signals = unique(self.seq[0])
emission_matrix = []
for k in range(len(emission_signals)):
emission_matrix.append([])
for i in range(len(self.A)):
gamma_vec = []
gamma_num = []
for r in range(len(self.seq)):
for t in range(len(self.seq[r])):
gamma_append = gamma[r][t][i]
gamma_vec.append(gamma_append)
if self.seq[r][t] == emission_signals[k]:
gamma_num_append = gamma[r][t][i]
gamma_num.append(gamma_num_append)
bik = sum(gamma_num) / sum(gamma_vec)
emission_matrix[k].append(bik)
new_B = {}
keys = emission_signals
for i in range(len(keys)):
new_B[keys[i]] = emission_matrix[i]
self.A = new_A
self.B = new_B
return {'A': self.A, 'B': self.B}
I don't know if i'm explaining well but this is my problem.
Hope that you can help me!
Thank you !
It's difficult without seeing the rest of your code, but it looks like you are referencing mutable objects A and B (list and dict) multiple times, and editing them from different places.
Just take into account, in python, every time you assign a mutable object to a variable, you are only creating a new reference to same object and not copying it. So every edit you do on that object, will be reflected on all its references.
Ok. the problem was this:
print(self.update().get('A'))
print(self.update().get('B'))
I change it in this way and now it works but i do not understand the logic behind this.
a = self.update()
print('\n---New transition matrix... ---')
print(a.get('A'))
print('\n---New Emission matrix... ---')
print(a.get('B'))
Maybe in the previous way i call update 2 times and then calculations go wrong. i don't know.
I am using sympy to help automate the process of finding equations of motion for some systems using the Euler Lagrange method. What would really make this easy is if I could define a function q and specify its time derivative qd --> d/dt(q) = qd. Likewise I'd like to specify d/dt(qd) = qdd. This is helpful because as part of the process of finding the equations of motion, I need to take derivatives (with respect to time, q, and qd) of expressions that are functions of q and qd. In the end I'll end up with an equation in terms of q, qd, and qdd and I'd like to be able to either print this neatly or use lambdify to convert this to a neat numpy function for use in a simulation.
Currently I've accomplished this is a very roundabout and annoying way by defining q as a function and qd as the derivative of that function:
q = sympy.Function('q', real=True)(t)
q_diff = diff(q,t)
This is fine for most of the process but then I end up with a messy expression filled with "Derivative (q, t)" and "Derivative (Derivative(q, t), t)" which is hard to wrangle into a neat printing format and difficult to turn into a numpy function using lambdify. My current solution is thus to use the subs function and replace q_diff and diff(q_diff, t) with sympy symbols qd and qdd respectively, which cleans things up and makes manipulating the expression much easier. This seems like a bad hack though and takes tons of time to do for more complicated equations with lots of state variables.
What I'd like is to define a function, q, with a specific value for the time derivative. I'd pass in that value when creating the function, and sympy could treat it like a generic Function object but would use whatever I'd given it for the time derivative instead of just saying "Derivative(q, t)". I'd like something like this:
qdd = sympy.symbols('qdd')
qd = my_func(name='qd', time_deriv=qdd)
q = my_func(name='q', time_deriv=qd)
diff(q, t)
>>> qd
diff(q**2, t)
>>> 2*q*qd
diff(diff(q**2, t))
>>> 2*q*qdd + 2*qd**2
expr = q*qd**2
expr.subs(q, 5)
>>> 5*qd**2
Something like that, where I could still use the subs command and lambdify command to substitute numeric values for q and qd, would be extremely helpful. I've been trying to do this but I don't understand enough of how the base sympy.Function class works to get this going. This is what I have right now:
class func(sp.Function):
def __init__(self, name, deriv):
self.deriv = deriv
self.name = name
def diff(self, *args, **kwargs):
return self.deriv
def fdiff(self, argindex=1):
assert argindex == 1
return self.deriv
This code so far does not really work, I don't know how to specify that specifically the time derivative of q is qd. Right now all derivatives of q are returning q?
I don't know if this is just a really bad solution, if I should be avoiding this issue entirely, or if there's already a clean way to solve this. Any advice would be very appreciated.
SciPy can solve ode equations by scipy.integrate.odeint or other packages, but it gives result after the function has been solved completely. However, if the ode function is very complex, the program will take a lot of time(one or two days) to give the whole result. So how can I mointor the step it solve the equations(print out result when the equation hasn't been solved completely)?
When I was googling for an answer, I couldn't find a satisfactory one. So I made a simple gist with a proof-of-concept solution using the tqdm project. Hope that helps you.
Edit: Moderators asked me to give an explanation of what is going on in the link above.
First of all, I am using scipy's OOP version of odeint (solve_ivp) but you could adapt it back to odeint. Say you want to integrate from time T0 to T1 and you want to show progress for every 0.1% of progress. You can modify your ode function to take two extra parameters, a pbar (progress bar) and a state (current state of integration). Like so:
def fun(t, y, omega, pbar, state):
# state is a list containing last updated time t:
# state = [last_t, dt]
# I used a list because its values can be carried between function
# calls throughout the ODE integration
last_t, dt = state
# let's subdivide t_span into 1000 parts
# call update(n) here where n = (t - last_t) / dt
time.sleep(0.1)
n = int((t - last_t)/dt)
pbar.update(n)
# we need this to take into account that n is a rounded number.
state[0] = last_t + dt * n
# YOUR CODE HERE
dydt = 1j * y * omega
return dydt
This is necessary because the function itself must know where it is located, but scipy's odeint doesn't really give this context to the function. Then, you can integrate fun with the following code:
T0 = 0
T1 = 1
t_span = (T0, T1)
omega = 20
y0 = np.array([1], dtype=np.complex)
t_eval = np.arange(*t_span, 0.25/omega)
with tqdm(total=1000, unit="‰") as pbar:
sol = solve_ivp(
fun,
t_span,
y0,
t_eval=t_eval,
args=[omega, pbar, [T0, (T1-T0)/1000]],
)
Note that anything mutable (like a list) in the args is instantiated once and can be changed from within the function. I recommend doing this rather than using a global variable.
This will show a progress bar that looks like this:
100%|█████████▉| 999/1000 [00:13<00:00, 71.69‰/s]
You could split the integration domain and integrate the segments, taking the last value of the previous as initial condition of the next segment. In-between, print out whatever you want. Use numpy.concatenate to assemble the pieces if necessary.
In a standard example of a 3-body solar system simulation, replacing the code
u0 = solsys.getState0();
t = np.arange(0, 100*365.242*day, 0.5*day);
%timeit u_res = odeint(lambda u,t: solsys.getDerivs(u), u0, t, atol = 1e11*1e-8, rtol = 1e-9)
output: 1 loop, best of 3: 5.53 s per loop
with a progress-reporting code
def progressive(t,N):
nk = [ int(n+0.5) for n in np.linspace(0,len(t),N+1) ]
u0 = solsys.getState0();
u_seg = [ np.array([u0]) ];
for k in range(N):
u_seg.append( odeint(lambda u,t: solsys.getDerivs(u), u0, t[nk[k]:nk[k+1]], atol = 1e11*1e-8, rtol = 1e-9)[1:] )
print t[nk[k]]/day
for b in solsys.bodies: print("%10s %s"%(b.name,b.x))
return np.concatenate(u_seg)
%timeit u_res = progressive(t,20)
output: 1 loop, best of 3: 5.96 s per loop
shows only a slight 8% overhead for console printing. With a more substantive ODE function, the fraction of the reporting overhead will reduce significantly.
That said, python, at least with its standard packages, is not the tool for industrial-scale number-crunching. Always use compiled versions with strong typing of variables to reduce interpretative overhead as much as possible.
Use some heavily developed and tested package like Sundials or the julia-lang framework differentialequations.jl directly coding the ODE function in an appropriate compiled language. Use the higher-order methods for larger step sizes, thus smaller steps. Test if using implicit or exponential/Rosenbrock methods reduces the number of steps or ODE function evaluations per fixed interval further. The difference can be a factor of 10 to 100 in speedup.
Use a python wrapper of the above with some acceleration-friendly implementation of your ODE function.
Use the quasi-source-translating tool JITcode to translate your python ODE function to a spaghetti list of C instruction that then give a compiled function that can be (almost) directly called from the compiled FORTRAN kernel of odeint.
Simple and Clear.
If you want to integrate an ODE from T0 to T1:
In the last line of the code, before return, you can use print((t/T1)*100,end='')
Then use a sys.stdout.flush() to keep the same line of printing.
Here is an example. My integrating time [0 0.2]
ddt[-2]=(beta/(Ap2*(L-x)))*(-Qgap+Ap*u)
ddt[-1]=(beta/(Ap2*(L+x)))*(Qgap-Ap*u)
print("\rCompletion percentage "+str(format(((t/0.2)*100),".4f")),end='')
sys.stdout.flush()
return ddt
It slows a bit the solving process by fraction of seconds, but it serves perfectly the purpose rather than creating new functions.
To preface this question, I understand that it could be done better. But this is a question in a class of mine and I must approach it this way. We cannot use any built in functions or packages.
I need to write a function to approximate the numerical value of the second derivative of a given function using finite difference. The function is below we are using.
2nd Derivative Formula (I lost the login info to my old account so pardon my lack of points and not being able to include images).
My question is this:
I don't understand how to make the python function accept the input function it is to be deriving. If someone puts in the input 2nd_deriv(2x**2 + 4, 6) I dont understand how to evaluate 2x^2 at 6.
If this is unclear, let me know and I can try again to describe. Python is new to me so I am just getting my feet wet.
Thanks
you can pass the function as any other "variable":
def f(x):
return 2*x*x + 4
def d2(fn, x0, h):
return (fn(x0+h) - 2*fn(x0) + fn(x0-h))/(h*h)
print(d2(f, 6, 0.1))
you can't pass a literal expression, you need a function (or a lambda).
def d2(f, x0, h = 1e-9):
func = f
if isinstance(f, str):
# quite insecure, use only with controlled input
func = eval ("lambda x:%s" % (f,))
return (func(x0+h) - 2*func(x0) + func(x0-h))/(2*h)
Then to use it
def g(x):
return 2*x**2 + 4
# using explicit function, forcing h value
print d2(g, 6, 1e-10)
Or directly:
# using lambda and default value for h
print d2(lambda x:2x**2+4, 6)
EDIT
updated to take into account that f can be a string or a function
I want to use the newton function loaded as
from scipy.optimize import newton
in order to find the zeros of a function enetered by the user. I write a script that first ask to the user to specify a function together with its first derivative, and also the starting point of the algorithm. First of all typing help(newton) I saw which parameters takes the function and the relative explanation:
newton(func, x0, fprime=None, args=(), tol=1.48e-08, maxiter=50)
func : function
The function whose zero is wanted. It must be a function of a
single variable of the form f(x,a,b,c...), where a,b,c... are extra
arguments that can be passed in the `args` parameter.
In which way I have to pass my function? If I use for func e.g. x**3 (and its first derivative) the response is NameError: name 'x' is not defined. On the internet I find that first I have to define my function and its first derivative and pass the names as parameters. So I made the following
fie = raw_input('Enter function in terms of x (e.g. x**2 - 2*x). F= ')
dfie = raw_input('Enter first derivative of function above DF = ')
x0 = input('Enter starting point x0 = ')
def F(x,fie):
y = eval(fie)
return y
def DF(x, dfie):
dy = eval(dfie)
return dy
print newton(F,x0,DF)
But I get the output
102 for iter in range(maxiter):
103 myargs = (p0,) + args
--> 104 fder = fprime(*myargs)
105 if fder == 0:
106 msg = "derivative was zero."
TypeError: DF() takes exactly 2 arguments (1 given)
and the same issue for F if I omit DF. Looking at the code in /usr/local/share/src/scipy/scipy/optimize/zeros.py I see that it evaluates the first derivative with fder=fprime(*myargs) so maybe I have to put in args something that make it working. I was thinking about it but no solution comes to me.
First, be aware that using eval makes your program vulnerable to malicious users. If that concern does not apply, you can create F and DF like this:
F = eval('lambda x :'+fie)
DF = eval('lambda x :'+dfie)
Then both functions expect only a single argument, and you can leave the args argument empty.
EDIT. If you really want to stick to your code as closely as possible, this should also work, but it does not look very nice to me. newton will send the same args to both functions.
def F(x,fie,dfie):
y = eval(fie)
return y
def DF(x,fie,dfie):
dy = eval(dfie)
return dy
print newton(F,x0,DF,(fie,dfie))