I am new to python, and would like to mimic using the matlab ode15s in python instead of the built-in odeint from scipy.
The code originally is written like this:
newRphi = odeint(PSP,Rphi,t,(b,k,F))[-1,:]
where PSP is defined as:
def PSP(xx,t,b,k,F):
R = xx[0]
phi = xx[1]
Rdot = sum([b[i]*R**(i+1) for i in xrange(len(b))]) + F(t) #indexing from zero
phiDot = 2*pi * k[2]*((R/k[1])**k[0])
yy = hstack((Rdot,phiDot))
return(yy)
from reading the instructions on scipy.integrate.odeint(), this function takes arguments in the following format:
scipy.integrate.odeint(func, y0, t, args=())
which means that func=PSP, y0=Rphi, t=t, args=(b,k,f)
So Rphi goes into PSP function, and gets integrated and becomes yy and comes out, and this function does it repeatedly for every element of t.
Now I want to translate it into something that would mimic ode15s from matlab. From reading some other treads, I found out that I can do that using
ode.set_integrator('vode', method='bdf', order=15)
Now the question becomes, how do I pass the original arguments to this integrator?
I am thinking it would probably look something like this:
ode15s = scipy.integrate.ode(f)
ode15s.set_integrator('vode', method='bdf', order=15)
ode15s.set_initial_value(y0, t0)
I know that the f is my PSP function, y0 is still the same: Rphi,
Here are my questions:
what is my initial value for t0, is it just t[0]?
how do I pass the variables (b,k,f) to the function f=PSP?
when I call this ode15s, how do I integrate through the vector size of t and collect the final values for yy?
Any help would be greatly appreciated. Thank you.
Related
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.
I am trying to approximate the Gauss Linking integral for two straight lines in R^3 using dblquad. I've created this pair of lines as an object.
I have a form for the integrand in parametrisation variables s and t generated by a function gaussint(self,s,t) and this is working. I'm then just trying to define a function which returns the double integral over the two intervals [0,1].
Edit - the code for the function looks like this:
def gaussint(self,s,t):
formnum = self.newlens()[0]*self.newlens()[1]*np.sin(test.angle())*np.cos(test.angle())
formdenone = (np.cos(test.angle())**2)*(t*(self.newlens()[0]) - s*(self.newlens()[1]) + self.adists()[0] - self.adists()[1])**2
formdentwo = (np.sin(test.angle())**2)*(t*(self.newlens()[0]) + s*(self.newlens()[1]) + self.adists()[0] + self.adists()[1])**2
fullden = (4 + formdenone + formdentwo)**(3/2)
fullform = formnum/fullden
return fullform
The various other function calls here are just bits of linear algebra - lengths of lines, angle between them and so forth. s and t have been defined as symbols upstream, if they need to be.
The code for the integration then just looks like this (I've separated it out just to try and work out what was going on:
def approxint(self, s, t):
from scipy.integrate import dblquad
return dblquad(self.gaussint(s,t),0,1, lambda t:0,lambda t:1)
Running it gets me a lengthy bit of somewhat impenetrable process messages, followed by
ValueError: invalid callable given
Any idea where I'm going wrong?
Cheers.
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.
Im trying to speed up my python code by porting a bunch of my nested loops over to fortran and calling them as subroutines.
But alot of my loops call numpy, and special functions from scipy like bessel functions.
Before I try and use fortran I was wondering if it was possible to import scipy and numpy to my fortran subroutine and call the modules for bessel functions?
Else would I have to create the bessel function in fortran in order to use it?
Ideally, I would create some sort of subroutine that would optimize this code below. This is just a snippet of my entire project to give you an idea of what I'm trying to accomplish.
I understand that there are other practices I should implement to improve the speed, but for now I was investigating the benefits of calling fortran subroutines in my main python program.
for m in range(self.MaxNum_Eigen):
#looping throught the eigenvalues for the given maximum number of eigenvalues allotted
bm = self.beta[m]
#not sure
#*note: rprime = r. BUT tprime ~= t.
#K is a list of 31 elements for this particular case
K = (bm / math.sqrt( (self.H2**2) + (bm**2) ))*(math.sqrt(2) / self.b)*((scipy.special.jv(0, bm * self.r))/ (scipy.special.jv(0, bm * self.b))) # Kernel, K0(bm, r).
#initial condition
F = [37] * (self.n1)
# Integral transform of the initial condition
#Fbar = (np.trapz(self.r,self.r*K*F))
'''
matlab syntax trapz(X,Y), x ethier spacing or vector
matlab: trapz(r,r.*K.*F) trapz(X,Y)
python: np.trapz(self.r*K*F, self.r) trapz(Y,X)
'''
#*(np.trapz(self.r,self.r*K*F))
Fbar = np.ones((self.n1,self.n2))*(np.trapz(self.r*K*F, self.r))
#steady state condition: integral is in steady state
SS = np.zeros((sz[0],sz[1]))
coeff = 5000000*math.exp(-(10**3)) #defining value outside of loop with higher precision
for i in range(sz[0]):
for j in range(sz[1]):
'''
matlab reshape(Array, size1, size2) takes multiple arguments the item its resizeing and the new desired shape
create self variables and so we are not re-initializing them over and over agaian?
using generators? How to use generators
'''
s = np.reshape(tau[i,j,:],(1,n3))
# will be used for rprime and tprime in Ozisik solution.
[RR,TT] = np.meshgrid(self.r,s)
'''
##### ERROR DUE TO ROUNDING OF HEAT SOURCE ####
error in rounding 5000000*math.exp(-(10**3)) becomes zero
#log10(e−10000)=−10000∗(0.4342944819)=−4342.944819
#e−1000=10−4342.944819=10−4343100.05518=1.13548386531×10−4343
'''
#g = 5000000*math.exp(-(10**3)) #*(RR - self.c*TT)**2) #[W / m^2] heat source.
g = coeff * (RR - self.c*TT)**2
K = (bm/math.sqrt(self.H2**2 + bm**2))*(math.sqrt(2)/self.b)*((scipy.special.jv(0,bm*RR))/(scipy.special.jv(0,bm*self.b)))
#integral transform of heat source
gbar = np.trapz(RR*K*g, self.r, 2) #trapz(Y,X,dx (spacing) )
gbar = gbar.transpose()
#boundary condition. BE SURE TO WRITE IN TERMS OF s!!!
f2 = self.h2 * 37
A = (self.alpha/self.k)*gbar + ((self.alpha*self.b)/self.k2)*((bm/math.sqrt(self.H2**2 + bm**2))*(math.sqrt(2)/self.b)*((scipy.special.jv(0,bm*self.b))/(scipy.special.jv(0,bm*self.b))))*f2
#A is essentially a constant is this correct all the time?
#What does A represent
SS[i, j] = np.trapz(np.exp( (-self.alpha*bm**2)*(T[i,j] - s) )*A, s)
#INSIDE M loop
K = (bm / math.sqrt((self.H2 ** 2) + (bm ** 2)))*(math.sqrt(2) /self.b)*((scipy.special.jv(0, bm * R))/ (scipy.special.jv(0, bm * self.b)))
U[:,:, m] = np.exp(-self.alpha * bm ** 2 * T)* K* Fbar + K* SS
#print(['Eigenvalue ' num2str(m) ', found at time ' num2str(toc) ' seconds'])
Compilation of answers given in the comments
Answers specific to my code:
As vorticity mentioned my code in itself was not using the numpy, and scipy packages to the fullest extent.
In regards to Bessel, function 'royvib' mentions using using .jo from scipy rather than .jv. Calling the special Bessel function jv. is much more computationally expensive, especially since I knew that I would be using a zeroth order bessel function for many of my declarations the minor change from jv -> j0 solved speed up the process.
In addition, I declared variables outside the loop to prevent expensive calls to searching for my appropriate functions. Example below.
Before
for i in range(SomeLength):
some_var = scipy.special.jv(1,coeff)
After
Bessel = scipy.special.jv
for i in range(SomeLength):
some_var = Bessel(1,coeff)
Storing the function saved time by not using the dot ('.') the command to look through the libraries every single loop. However keep in mind this does make python less readable, which is the main reason I choose to do this project in python. I do not have an exact amount of time this step cut from my process.
Fortran specific:
Since I was able to improve my python code I did not go this route an lack of specifics, but the general answer as stated by 'High Performance Mark' is that yes there are libraries that have been made to handle Bessel functions in Fortran.
If I do port my code over to Fortran or use f2py to mix Fortran and python I will update this answer accordingly.
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.