how can i use "variable" constants in scipy.optimize functions? I am trying to create an iterative optimisation algorithm, which updates certain parameters in the objective function after each optimisation run.
to use a very simple example of what i want to do:
from scipy import optimize as opt
def f(x, R):
return R * (x[0]**2 + x[1]**3)
R = 0.1 # initial R value
y = []
y.append([2,2]) # initial point
for i in range(0,10):
y.append(opt.fmin(f, y[i])) # how can i include 'R' in this line??
R = some_function_to_update_R(R)
any help would be appreciated
EDIT:
would it help to re-declare the objective function each time i optimise? so make the loop look like this instead?
for i in range(0,10):
def f_temp(x_temp):
return f(x_temp,R)
y.append(opt.fmin(f_temp, y[i]))
R = some_function_to_update_R(R)
or is there some better way?
fmin supports an optional args argument, representing a tuple of additional arguments to be passed to the function you're trying to optimize:
y.append(opt.fmin(f, y[i], args=(R,)))
This is explained in the documentation for fmin; you should get into the habit of checking the documentation when you want to figure out how to do something.
Related
I'm working on a project where I'm writing a code that hopefully works similarly to scipy.integrate.solve_ivp. I'm supposed to allow for there to be extra parameters by using *args. Basically one of the functions I'm supposed to run is what I've defined in my code as f2a and I'm supposed to assign Vin,D,k as extra parameters. It won't let me assign all three of them. Also I was never given a value for k (don't know if that's important or if assigning it as *k takes care of it). I've put my entire code so far for this project below. F2a is at the top.
import numpy as np
import math
tspan=np.array([0,20])
#defining functions
Vin=150 #m^3/min
H0=7 #initial height, meters
D=7 #diameter, m**2
#they never told us what k is...k must be args thingy?
def f2a(t,H,Vin,D,*k):
Vin=150 #m^3/min
D=7 #diameter, m**2
#they never told us what k is...k must be args
dhdt=4/(math.pi*D**2)*(Vin-k*np.sqrt(H))
return(dhdt)
x0=1 #initial cond.
y0=1
def fb2(J,t):
x=J[0]
y=J[1]
dxdt=0.25*y-x
dydt=3*x-y
return([dxdt,dydt])
#x0 and y0 are initial conditions
def odeRK4(function,tspan,R,h,*args):
#args allows for the possibility of extra parameters to be added?
#R is vector of inital conditions
x0=R[0]
y0=R[1]
#writing statement for what to do if h isnt given/other thing
if h==None:
h=.01*(tspan[1]-tspan[0])
elif h> tspan[1]-tspan[0]:
h=.01*(tspan[1]-tspan[0])
else:
h=h
#defining the 2-element array (i hope)
#pretty sure tspan is range of t values
x0=tspan[0] #probably 0 if this is meant for time
xn=tspan[1] #whatever time we want it to end at?
#xn is final x value-t
#x0 is initial
def f(x,y):
f= eval(function)
return f
#eval is used to evaluate whatever I put in the function place when I recall ode
#this won't work without eval to run the functions
t_values=np.arange(0,20,21)
for i in t_values:
#rk4 method
k1=h*(f(x0,y0))
k2=h*(f((x0+h/2), (y0+k1/2)))
k3=h*(f((x0+h/2), (y0+k2/2)))
k4=h*(f((x0+h), (y0+k3)))
n=(k1+2*k2+2*k3+k4)/6
#new y value
yn=y0+n
#makes it so new y and x are used for next # in range
y0=yn
xn=x0+h
print ('Y=')
print(yn)
print('When t=')
print(xn)
print('For 3A:')
odeRK4("f2a(x,y,*k)", [0,20],[0,H0], None)
I'm currently trying to solve this integral using SciPy:
I was first advised to use interpolation, which I tried but cannot figure out for some reason, but would probably be a good approach. I found this post about using np.vectorize and I think it might still work, but I am getting an error. Here is the code that I have written thus far (also note that n and n,eq are not indices, they're just variable names):
import numpy as np
from scipy import integrate
def K(x): #This is a function in the integral.
b = 0.252
return b*(((4/(x**3))+(3/(x**2))+1/x) + (4/(x**3) + 1/(x**2))*np.exp(-x))
def Xntot_integrand(x,z): #Defining the integrand
Xneq_x = (1+np.exp(x))**(-1) #This is the term outside the integral and squared within it.
return Xneq_x(x)**2 * np.exp(K(z) - K(x)) * np.exp(x)
Xntot_integrand = np.vectorize(Xntot_integrand)
def Xntot_integrated(x,z):
return quad(Xntot_integrand, 0, z)
Xntot_integrated=np.vectorize(Xntot_integrated)
T_narrow = np.linspace(1,0.01,100) #Narrow T range from 1 to 0.01 MeV
z_narrow = Q/T_narrow
final_integrated_Xneq = Xntot_integrated(z_narrow)
I am getting the error that I am missing a positional argument when I call Xntot_integrated (which makes sense, I think it is still in the two variables x and z).
So I suppose the issue is stemming from where I use quad() because after it is integrated, x should go away. Any advice? Should I use tabulation/interpolation instead?
You need to be using the args keyword argument of integrate.quad to pass additional inputs to the function, so it would look like this:
def Xntot_integrated(z):
return integrate.quad(Xntot_integrand, 0, z, args=(z,))
Note here x is not an input to the integrated function, only z, the first input to the integrand is the integration variable and any extra information is passed via args=(z,) tuple.
alternatively you can define a wrapper that knows z from context and only takes the integration variable as input:
def Xntot_integrated(z):
def integrand(x):return Xntot_integrand(x,z)
return integrate.quad(integrand, 0, z)
but most API's that take a function typically have a keyword argument to specify those inputs. (threading.Thread comes to mind.)
also your Xneq_x should probably be a function itself since you accidentally use it as such inside your integrand (it is just a value there right now) and you will need to use it outside the integration anyway :)
As I'm really struggleing to get from R-code, to Python code, I would like to ask some help. The code I want to use has been provided to my from withing the mathematics forum of stackexchange.
https://math.stackexchange.com/questions/2205573/curve-fitting-on-dataset
I do understand what is going on. But I'm really having a hard time trying to solve the R-code, as I have never seen anything of it. I have written the function to return the sum of squares. But I'm stuck at how I could use a function similar to the optim function. And also I don't really like the guesswork at the initial values. I would like it better to run and re-run a type of optim function untill I get the wanted result, because my needs for a nearly perfect curve fit are really high.
def model (par,x):
n = len(x)
res = []
for i in range(1,n):
A0 = par[3] + (par[4]-par[1])*par[6] + (par[5]-par[2])*par[6]**2
if(x[i] == par[6]):
res[i] = A0 + par[1]*x[i] + par[2]*x[i]**2
else:
res[i] = par[3] + par[4]*x[i] + par[5]*x[i]**2
return res
This is my model function...
def sum_squares (par, x, y):
ss = sum((y-model(par,x))^2)
return ss
And this is the sum of squares
But I have no idea on how to convert this:
#I found these initial values with a few minutes of guess and check.
par0 <- c(7,-1,-395,70,-2.3,10)
sol <- optim(par= par0, fn=sqerror, x=x, y=y)$par
To Python code...
I wrote an open source Python package (BSD license) that has a genetic algorithm (Differential Evolution) front end to the scipy Levenberg-Marquardt solver, it functions similarly to what you describe in your question. The github URL is:
https://github.com/zunzun/pyeq3
It comes with a "user-defined function" example that's fairly easy to use:
https://github.com/zunzun/pyeq3/blob/master/Examples/Simple/FitUserDefinedFunction_2D.py
along with command-line, GUI, cluster, parallel, and web-based examples. You can install the package with "pip3 install pyeq3" to see if it might suit your needs.
Seems like I have been able to fix the problem.
def model (par,x):
n = len(x)
res = np.array([])
for i in range(0,n):
A0 = par[2] + (par[3]-par[0])*par[5] + (par[4]-par[1])*par[5]**2
if(x[i] <= par[5]):
res = np.append(res, A0 + par[0]*x[i] + par[1]*x[i]**2)
else:
res = np.append(res,par[2] + par[3]*x[i] + par[4]*x[i]**2)
return res
def sum_squares (par, x, y):
ss = sum((y-model(par,x))**2)
print('Sum of squares = {0}'.format(ss))
return ss
And then I used the functions as follow:
parameter = sy.array([0.0,-8.0,0.0018,0.0018,0,200])
res = least_squares(sum_squares, parameter, bounds=(-360,360), args=(x1,y1),verbose = 1)
The only problem is that it doesn't produce the results I'm looking for... And that is mainly because my x values are [0,360] and the Y values only vary by about 0.2, so it's a hard nut to crack for this function, and it produces this (poor) result:
Result
I think that the range of x values [0, 360] and y values (which you say is ~0.2) is probably not the problem. Getting good initial values for the parameters is probably much more important.
In Python with numpy / scipy, you would definitely want to not loop over values of x but do something more like
def model(par,x):
res = par[2] + par[3]*x + par[4]*x**2
A0 = par[2] + (par[3]-par[0])*par[5] + (par[4]-par[1])*par[5]**2
res[np.where(x <= par[5])] = A0 + par[0]*x + par[1]*x**2
return res
It's not clear to me that that form is really what you want: why should A0 (a value independent of x added to a portion of the model) be so complicated and interdependent on the other parameters?
More importantly, your sum_of_squares() function is actually not what least_squares() wants: you should return the residual array, you should not do the sum of squares yourself. So, that should be
def sum_of_squares(par, x, y):
return (y - model(par, x))
But most importantly, there is a conceptual problem that is probably going to plague this model: Your par[5] is meant to represent a breakpoint where the model changes form. This is going to be very hard for these optimization routines to find. These routines generally make a very small change to each parameter value to estimate to derivative of the residual array with respect to that variable in order to figure out how to change that variable. With a parameter that is essentially used as an integer, the small change in the initial value will have no effect at all, and the algorithm will not be able to determine the value for this parameter. With some of the scipy.optimize algorithms (notably, leastsq) you can specify a scale for the relative change to make. With leastsq that is called epsfcn. You may need to set this as high as 0.3 or 1.0 for fitting the breakpoint to work. Unfortunately, this cannot be set per variable, only per fit. You might need to experiment with this and other options to least_squares or leastsq.
I wrote a function in Python 2.7:
# Python #
def function_py(par):
#something happens
return(value)
and I want to use this function as an argument for another function in R. More precisely, I want to perform to compute the Sobol' indices using the following function:
# R #
library('sensitivity')
sobol(function_py_translated, X1,X2)
where function_py_translated would b the R equivalent of function_py.
I'm trying to use the rpy2 module, and for a simple function, I could make a working case:
import rpy2.rinterface as ri
import rpy2.robjects.numpy2ri
sensitivity = importr('sensitivity')
radd = ri.baseenv.get('+')
def costfun(X):
a = X[0]
b = X[1]
return(radd(a,b))
costfunr=ri.rternalize(costfun)
X1 = robjects.r('data.frame(matrix(rnorm(2*1000), nrow = 1000))')
X2 = robjects.r('data.frame(matrix(rnorm(2*1000), nrow = 1000))')
sobinde = sensitivity.sobol(costfunr,X1,X2)
print(sobinde.__getitem__(11))
The main problem is that I had to redefine the "+". Is there a way to work around this ? Being able to pass an arbitrary function without prior transformation ? The function I want to analyze is much more complicated.
Thank you very much for your time
I have the following problem: I have two sets of data (set T and set F). And the following functions:
x(T) = arctan(T-c0), A(x(T)) = arctan(x(T) -c1),
B(x(T)) = arctan(x(T) -c2)
and Y(x(t),F) = ((A(x(t)) - B(x(t)))/2 - A(x(t))arctan(F-c3) + B(x(t))arctan(F-c4))
# where c0,c1,c2,c3,c4 are constants
Now I want to create a surface plot of Y. And for that I would like to implement Y as a python (numpy) function what turns out to be quite complicated, because Y takes other functions as input.
Another idea of mine was to evaluate x, B and A on the data separately and store the results in numpy arrays. With those I also could get the output of the function Y , but I don't know which way is better in order to plot the data and I really would like to know how to write Y as a python function.
Thank you very much for your help
It is absolutely possible to use functions as input parameters to other functions. A use case could look like:
def plus_one(standard_input_parameter_like_int):
return standard_input_parameter_like_int + 1
def apply_function(function_as_input, standard_input_parameter):
return function_as_input(standard_input_parameter)
if(__name__ == '__main__'):
print(apply_function(plus_one, 1))
I hope that helps to solve your specific problem.
[...] somethin like def s(x,y,z,*args,*args2): will yield an
error.
This is perfectly normal as (at least as far as I know) there is only one variable length non-keyword argument list allowed per function (that has to be exactly labeled as *args). So if you remove the asterisks (*) you should actually be able to run s properly.
Regarding your initial question you could do something like:
c = [0.2,-0.2,0,0,0,0]
def x(T):
return np.arctan(T-c[0])
def A(xfunc,T):
return np.arctan(xfunc(T) - c[1])
def B(xfunc,T):
return np.arctan(xfunc(T) - c[2])
def Y(xfunc,Afunc,Bfunc,t,f):
return (Afunc(xfunc,t) - Bfunc(xfunc,t))/2.0 - Afunc(xfunc,t) * np.arctan(f - c[3]) + Bfunc(xfunc,t)*np.arctan(f-c[4])
_tSet = np.linspace(-1,1,20)
_fSet = np.arange(-1,1,20)
print Y(x,A,B,_tSet,_fSet)
As you can see (and probably already tested by yourself judging from your comment) you can use functions as arguments. And as long as you don't use any 'if' conditions or other non-vectorized functions in your 'sub'-functions the top-level function should already be vectorized.