I am trying to minimize a function of a complex (vector) variable using scipy.optimize. My results so far indicate that it may not be possible. To investigate the problem, I have implemented a simple example - minimize the 2-norm of a complex vector with an offset:
import numpy as np
from scipy.optimize import fmin
def fun(x):
return np.linalg.norm(x - 1j * np.ones(2), 2)
sol = fmin(fun, x0=np.ones(2) + 0j)
The output is
Optimization terminated successfully.
Current function value: 2.000000
Iterations: 38
Function evaluations: 69
>>> sol
array([-2.10235293e-05, 2.54845649e-05])
Clearly, the solution should be
array([0.+1.j, 0.+1.j])
Disappointed with this outcome, I have also tried scipy.optimize.minimize:
from scipy.optimize import minimize
def fun(x):
return np.linalg.norm(x - 1j * np.ones(2), 1)
sol = minimize(fun, x0=np.ones(2) + 0j)
The output is
>>> sol
fun: 2.0
hess_inv: array([[ 9.99997339e-01, -2.66135332e-06],
[-2.66135332e-06, 9.99997339e-01]])
jac: array([0., 0.])
message: 'Optimization terminated successfully.'
nfev: 24
nit: 5
njev: 6
status: 0
success: True
x: array([6.18479071e-09+0.j, 6.18479071e-09+0.j])
Not good either. I have tried specifying all of the possible methods for minimize (supplying the Jacobian and Hessian as necessary), but none of them reach the correct result. Most of them cause ComplexWarning: Casting complex values to real discards the imaginary part, indicating that they cannot handle complex numbers correctly.
Is this possible at all using scipy.optimize?
If so, I would very much appreciate if someone can tell me what I am doing wrong.
If not, do you perhaps have suggestions for alternative optimization tools (for Python) that allow this?
The minimization methods of SciPy work with real arguments only. But minimization on the complex space Cn amounts to minimization on R2n, the algebra of complex numbers never enters the consideration. Thus, adding two wrappers for conversion from Cn to R2n and back, you can optimize over complex numbers.
def real_to_complex(z): # real vector of length 2n -> complex of length n
return z[:len(z)//2] + 1j * z[len(z)//2:]
def complex_to_real(z): # complex vector of length n -> real of length 2n
return np.concatenate((np.real(z), np.imag(z)))
sol = minimize(lambda z: fun(real_to_complex(z)), x0=complex_to_real(np.ones(2) + 0j))
print(real_to_complex(sol.x)) # [-7.40376620e-09+1.j -8.77719406e-09+1.j]
You mention Jacobian and Hessian... but minimization only makes sense for real-valued functions, and those are never differentiable with respect to complex variables. The Jacobian and Hessian would have to be computed over R2n anyway, treating the real and imaginary parts as separate variables.
I have needed to minimize the departure of a complex valued model function upon complex valued parameters, over a real domain.
A toy example:
def f(x, a, b):
ab = complex(a,b);
return np.exp(x*ab)
And suppose that I have data DATA for x = np.arange(N). Note that x is real.
What I did was this:
def helper(x, a, b):
return abs(f(x,a,b) - DATA[x])
and then I can use curve_fit():
curve_fit(helper, np.arange(N), np.zeros(N), p0 = [1,0])
What is happening is this: By subtracting the data from the model function, the new "ideal" output is all zeroes, which can be (must be) real in order for curve_fit() to work. The complex parameter ab = a + jb has been broken into its real and imaginary parts. The helper() function returns the absolute value of the difference between the model and the data.
A critical issue is that curve_fit() doesn't evaluate any other x values than those you give it. Otherwise DATA[x] would fail.
Note that by using abs() I'm achieving an L1 fit (more or less). One could just as well use abs()**2 to get an L2 fit ... but why one would use L1 or L2 is a topic for another day.
You could fret, "suppose that the x[] aren't integers (but are real)?" which my code requires. Well, that's doable, simply by putting them into an array, and indexing that. There's probably some clever hack using a dictionary that would address this issue, too.
Sorry about the code formatting; haven't figured out the markup yet.
Given a 2D point p, I'm trying to calculate the smallest distance between that point and a functional curve, i.e., find the point on the curve which gives me the smallest distance to p, and then calculate that distance. The example function that I'm using is
f(x) = 2*sin(x)
My distance function for the distance between some point p and a provided function is
def dist(p, x, func):
x = np.append(x, func(x))
return sum([[i - j]**2 for i,j in zip(x,p)])
It takes as input, the point p, a position x on the function, and the function handle func. Note this is a squared Euclidean distance (since minimizing in Euclidean space is the same as minimizing in squared Euclidean space).
The crucial part of this is that I want to be able to provide bounds for my function so really I'm finding the closest distance to a function segment. For this example my bounds are
bounds = [0, 2*np.pi]
I'm using the scipy.optimize.minimize function to minimize my distance function, using the bounds. A result of the above process is shown in the graph below.
This is a contour plot showing distance from the sin function. Notice how there appears to be a discontinuity in the contours. For convenience, I've plotted a few points around that discontinuity and the "closet" points on the curve that they map to.
What's actually happening here is that the scipy function is finding a local minimum (given some initial guess), but not a global one and that is causing the discontinuity. I know finding the global minimum of any function is impossible, but I'm looking for a more reliable way to find the global minimum.
Possible methods for finding a global minimum would be
Choose a smart initial guess, but this amounts to knowing approximately where the global minimum is to begin with, which is using the solution of the problem to solve it.
Use a multiple initial guesses and choose the answer which gets to the best minimum. This however seems like a poor choice, especially when my functions get more complicated (and higher dimensional).
Find the minimum, then perturb the solution and find the minimum again, hoping that I may have knocked it into a better minimum. I'm hoping that maybe there is some way to do this simply without evoking some complicated MCMC algorithm or something like that. Speed counts for this process.
Any suggestions about the best way to go about this, or possibly directions to useful functions that may tackle this problem would be great!
As suggest in a comment, you could try a global optimization algorithm such as scipy.optimize.differential_evolution. However, in this case, where you have a well-defined and analytically tractable objective function, you could employ a semi-analytical approach, taking advantage of the first-order necessary conditions for a minimum.
In the following, the first function is the distance metric and the second function is (the numerator of) its derivative w.r.t. x, that should be zero if a minimum occurs at some 0<x<2*np.pi.
import numpy as np
def d(x, p):
return np.sum((p-np.array([x,2*np.sin(x)]))**2)
def diff_d(x, p):
return -2 * p[0] + 2 * x - 4 * p[1] * np.cos(x) + 4 * np.sin(2*x)
Now, given a point p, the only potential minimizers of d(x,p) are the roots of diff_d(x,p) (if any), as well as the boundary points x=0 and x=2*np.pi. It turns out that diff_d may have more than one root. Noting that the derivative is a continuous function, the pychebfun library offers a very efficient method for finding all the roots, avoiding cumbersome approaches based on the scipy root-finding algorithms.
The following function provides the minimum of d(x, p) for a given point p:
import pychebfun
def min_dist(p):
f_cheb = pychebfun.Chebfun.from_function(lambda x: diff_d(x, p), domain = (0,2*np.pi))
potential_minimizers = np.r_[0, f_cheb.roots(), 2*np.pi]
return np.min([d(x, p) for x in potential_minimizers])
Here is the result:
suppose I have the following Problem:
I have a complex function A(x) and a complex function B(y). I know these functions cross in the complex plane. I would like to find out the corresponding x and y of this intersection point, numerically ( and/or graphically). What is the most clever way of doing that?
This is my starting point:
import matplotlib.pyplot as plt
import numpy as np
from numpy import sqrt, pi
x = np.linspace(1, 10, 10000)
y = np.linspace(1, 60, 10000)
def A_(x):
return -1/( 8/(pi*x)*sqrt(1-(1/x)**2) - 1j*(8/(pi*x**2)) )
A = np.vectorize(A_)
def B_(y):
return 3/(1j*y*(1+1j*y))
B = np.vectorize(B_)
real_A = np.real(A(x))
imag_A = np.imag(A(x))
real_B = np.real(B(y))
imag_B = np.imag(B(y))
plt.plot(real_A, imag_A, color='blue')
plt.plot(real_B, imag_B, color='red')
plt.show()
I don't have to plot it necessarily. I just need x_intersection and y_intersection (with some error that depends on x and y).
Thanks a lot in advance!
EDIT:
I should have used different variable names. To clarify what i need:
x and y are numpy arrays and i need the index of the intersection point of each array plus the corresponding x and y value (which again is not the intersection point itself, but some value of the arrays x and y ).
Here I find the minimum of the distance between the two curves. Also, I cleaned up your code a bit (eg, vectorize wasn't doing anything useful).
import matplotlib.pyplot as plt
import numpy as np
from numpy import sqrt, pi
from scipy import optimize
def A(x):
return -1/( 8/(pi*x)*sqrt(1-(1/x)**2) - 1j*(8/(pi*x**2)) )
def B(y):
return 3/(1j*y*(1+1j*y))
# The next three lines find the intersection
def dist(x):
return abs(A(x[0])-B(x[1]))
sln = optimize.minimize(dist, [1, 1])
# plotting everything....
a0, b0 = A(sln.x[0]), B(sln.x[1])
x = np.linspace(1, 10, 10000)
y = np.linspace(1, 60, 10000)
a, b = A(x), B(y)
plt.plot(a.real, a.imag, color='blue')
plt.plot(b.real, b.imag, color='red')
plt.plot(a0.real, a0.imag, "ob")
plt.plot(b0.real, b0.imag, "xr")
plt.show()
The specific x and y values at the intersection point are sln.x[0] and sln.x[1], since A(sln.x[0])=B(sln.x[1]). If you need the index, as you also mention in your edit, I'd use, for example, numpy.searchsorted(x, sln.x[0]), to find where the values from the fit would insert into your x and y arrays.
I think what's a bit tricky with this problem is that the space for graphing where the intersection is (ie, the complex plane) does not show the input space, but one has to optimize over the input space. It's useful for visualizing the solution, then, to plot the distance between the curves over the input space. That can be done like this:
data = dist((X, Y))
fig, ax = plt.subplots()
im = ax.imshow(data, cmap=plt.cm.afmhot, interpolation='none',
extent=[min(x), max(x), min(y), max(y)], origin="lower")
cbar = fig.colorbar(im)
plt.plot(sln.x[0], sln.x[1], "xw")
plt.title("abs(A(x)-B(y))")
From this it seems much more clear how optimize.minimum is working -- it just rolls down the slope to find the minimum distance, which is zero in this case. But still, there's no obvious single visualization that one can use to see the whole problem.
For other intersections one has to dig a bit more. That is, #emma asked about other roots in the comments, and there I mentioned that there's no generally reliable way to find all roots to arbitrary equations, but here's how I'd go about looking for other roots. Here I won't lay out the complete program, but just list the changes and plots as I go along.
First, it's obvious that for the domain shown in my first plot that there's only one intersection, and that there are no intersection in the region to the left. The only place there could be another intersection is to the right, but for that I'll need to allow the sqrt in the def of B to get a negative argument without throwing an exception. An easy way to do this is to add 0j to the argument of the sqrt, like this, sqrt(1+0j-(1/x)**2). Then the plot with the intersection becomes
I plotted this over a broader range (x=np.linspace(-10, 10, 10000) and y=np.linspace(-400, 400, 10000)) and the above is the zoom of the only place where anything interesting is going on. This shows the intersection found above, plus the point where it looks like the two curves might touch (where the red curve, B, comes to a point nearly meeting the blue curve A going upward), so that's the new interesting thing, and the thing I'll look for.
A bit of playing around with limits, etc, show that B is coming to a point asymptotically, and the equation of B is obvious that it will go to 0 + 0j for large +/- y, so that's about all there is to say for B.
It's difficult to understand A from the above plot, so I'll look at the real and imaginary parts independently:
So it's not a crazy looking function, and the jumping between Re=const and Im=const is just the nature of sqrt(1-x-2), which is pure complex for abs(x)<1 and pure real for abs(x)>1.
It's pretty clear now that the other time the curves are equal is at y= +/-inf and x=0. And, quick look at the equations show that A(0)=0+0j and B(+/- inf)=0+0j, so this is another intersection point (though since it occurs at B(+/- inf), it's sort-of ambiguous on whether or not it would be called an intersection).
So that's about it. One other point to mention is that if these didn't have such an easy analytic solution, like it wasn't clear what B was at inf, etc, one could also graph/minimize, etc, by looking at B(1/y), and then go from there, using the same tools as above to deal with the infinity. So using:
def dist2(x):
return abs(A(x[0])-B(1./x[1]))
Where the min on the right is the one initially found, and the zero, now at x=-0 and 1./y=0 is the other one (which, again, isn't interesting enough to apply an optimizer here, but it could be interesting in other equations).
Of course, it's also possible to estimate this by just finding the minimum of the data that goes into the above graph, like this:
X, Y = np.meshgrid(x, y)
data = dist((X, Y))
r = np.unravel_index(data.argmin(), data.shape)
print x[r[1]], y[r[0]]
# 2.06306306306 1.8008008008 # min approach gave 2.05973231 1.80069353
But this is only approximate (to the resolution of data) and involved many more calculations (1M compared to a few hundred). I only post this because I think it might be what the OP originally had in mind.
Briefly, two analytic solutions are derived for the roots of the problem. The first solution removes the parametric representation of x and solves for the roots directly in the (u, v) plane, where for example A(x): u(x) + i v(y) gives v(u) = f(u). The second solution uses a polar representation, e.g. A(x) is given by r(x) exp(i theta(x)), and offers a better understanding of the behavior of the square root as x passes through unity towards zero. Possible solutions occurring at the singular points are explored. Finally, a bisection root finding algorithm is constructed as a Python iterator to invert certain solutions. Summarizing, the one real root can be found as a solution to either of the following equations:
and gives:
x0 = -2.059732
y0 = +1.800694
A(x0) = B(y0) = (-0.707131, -i 0.392670)
As in most problems there are a number of ways to proceed. One can use a "black box" and hopefully find the root they are looking for. Sometimes an answer is all that is desired, and with a little understanding of the functions this may be an adequate way forward. Unfortunately, it is often true that such an approach will provide less insight about the problem then others.
For example, algorithms find it difficult locating roots in the global space. Local roots may be found with other roots lying close by and yet undiscovered. Consequently, the question arises: "Are all the roots accounted for?" A more complete understanding of the functions, e.g. asymptotic behaviors, branch cuts, singular points, can provide the global perspective to better answer this, as well as other important questions.
So another possible solution would be building one's own "black box." A simple bisection routine might be a starting point. Robust if the root lies in the initial interval and fairly efficient. This encourages us to look at the global behavior of the functions. As the code is structured and debugged the various functions are explored, new insights are gained, and the algorithm has become a tool towards a more complete solution to the problem. Perhaps, with some patience, a closed-form solution can be found. A Python iterator is constructed and listed below implementing a bisection root finding algorithm.
Begin by putting the functions A(x) and B(x) in a more standard form:
C(x) = u(x) + i v(x)
and here the complex number i is brought out of the denominator and into the numerator, casting the problem into the form of functions of a complex variable. The new representation simplifies the original functions considerably. The real and imaginary parts are now clearly separated. An interesting graph is to plot A(x) and B(x) in the 3-dimensional space (u, v, x) and then visualize the projection into the u-v plane.
import numpy as np
from numpy import real, imag
import matplotlib.pyplot as plt
ax = fig.gca(projection='3d')
s = np.linspace(a, b, 1000)
ax.plot(f(s).real, f(s).imag, z, color='blue')
ax.plot(g(s).real, g(s).imag, z, color='red')
ax.plot(f(s).real, f(s).imag, 0, color='black')
ax.plot(g(s).real, g(s).imag, 0, color='black')
The question arises: "Can the parametric representation be replaced so that a relationship such as,
A(x): u(x) + i v(x) gives v(u) = f(u)
is obtained?" This will provide A(x) as a function v(u) = f(u) in the u-v plane. Then, if for
B(x): u(x) + i v(x) gives v(u) = g(u)
a similar relationship can be found, the solutions can be set equal to one another,
f(u) = g(u)
and the root(s) computed. In fact, it is convenient to look for a solution in the square of the above equation. The worst case is that an algorithm will have to be built to find the root, but at this point the behavior of our functions are better understood. For example, if f(u) and g(u) are polynomials of degree n then it is known that there are n roots. The best case is that a closed-form solution might be a reward for our determination.
Here is more detail to the solution. For A(x) the following is derived:
and v(u) = f(u) is just v(u) = constant. Similarly for B(x) a slightly more complex form is required:
Look at the function g(u) for B(x). It is imaginary if u > 0, but the root must be real since f(u) is real. This means that u must be less then 0, and there is both a positive and negative real branch to the square root. The sign of f(u) then allows one to pick the negative branch as the solution for the root. So the fact that the solution must be real is determined by the sign of u, and the fact that the real root is negative specifies what branch of the square root to choose.
In the following plot both the real (u < 0) and complex (u > 0) solutions are shown.
The camera looks toward the origin in the back corner, where the red and blue curves meet. The z-axis is the magnitude of f(u) and g(u). The x and y axes are the real/complex values of u respectively. The blue curves are the real solution with (3 - |u|). The red curves are the complex solution with (3 + |u|). The two sets meet at u = 0. The black curve is f(u) equal to (-pi/8).
There is a divergence in g(u) at |u| = 3 and this is associated with x = 0. It is far removed from the solution and will not be considered further.
To obtain the roots to f = g it is easier to square f(u) and equate the two functions. When the function g(u) is squared the branches of the square root are lost, much like squaring the solutions for x**2 = 4. In the end the appropriate root will be chosen by the sign of f(u) and so this is not an issue.
So by looking at the dependence of A and B, with respect to the parametric variable x, a representation for these functions was obtained where v is a function of u and the roots found. A simpler representation can be obtained if the term involving c in the square root is ignored.
The answer gives all the roots to be found. A cubic equation has at most three roots and one is guaranteed to be real. The other two may be imaginary or real. In this case the real root has been found and the other two roots are complex. Interestingly, as c changes these two complex roots may move into the real plane.
In the above figure the x-axis is u and the y axis is the evaluated cubic equation with constant c. The blue curve has c as (pi/8) squared. The red curve uses a larger and negative value for c, and has been translated upwards for purposes of demonstration. For the blue curve there is an inflection point near (0, 0.5), while the red curve has a maximum at (-0.9, 2.5) and a minimum at (0.9, -0.3).
The intersection of the cubic with the black line represents the roots given by: u**3 + c u + 3c = 0. For the blue curve there is one intersection and one real root with two roots in the complex plane. For the red curve there are three intersections, and hence 3 roots. As we change the value of the constant c (blue to red) the one real root undergoes a "pitchfork" bifurcation, and the two roots in the complex plane move into the real space.
Although the root (u_t, v_t) has been located, obtaining the value for x requires that (u, v) be inverted. In the present example this is a trivial matter, but if not, a bisection routine can be used to avoid the algebraic difficulties.
The parametric representation also leads to a solution for the real root, and it rounds out the analysis with an independent verification of the first result. Second, it answers the question about what happens at the singularity in the square root. Third, it gives a greater understanding of the multiplicity of roots.
The steps are: (1) convert A(x) and B(x) into polar form, (2) equate the modulus and argument giving two equations in two unknowns, (3) make a substitution for z = x**2. Converting A(x) to polar form:
Absolute value bars are not indicated, and it should be understood that the moduli r(x) and s(x) are positive definite as their names imply. For B(x):
The two equations in two unknowns:
Finally, the cubic solution is sketched out here where the substitution z = x**2 has been made:
The solution for z = x**2 gives x directly, which allows one to substitute into both A(x) and B(x). This is an exact solution if all terms are maintained in the cubic solution, and there is no error in x0, y0, A(x0), or B(x0). A simpler representation can be found by considering terms proportional to 1/d as small.
Before leaving the polar representation consider the two singular points where: (1) abs(x) = 1, and (2) x = 0. A complicating factor is that the arguments behave something like 1/x instead of x. It is worthwhile to look at a plot of the arctan(a) and then ask yourself how that changes if its argument is replaced by 1/a. The following graphs will then look less foreign.
Consider the polar representation of B(x). As x approaches 0 the modulus and argument tend toward infinity, i.e. the point is infinitely far from the origin and lies along the y-axis. Approaching 0 from the negative direction the point lies along the negative y-axis with varphi = (-pi/2), while approaching from the other direction the point lies along the positive y-axis with varphi = (+pi/2).
A somewhat more complicated behavior is exhibited by A(x). A(x) is even in x since the modulus is positive definite and the argument involves only x**2. There is a symmetry across the y-axis that allows us to only consider the x > 0 plane.
At x = 1 the modulus is just (pi/8), and as x continues to approach 0 so does r(x). The behavior of the argument is more complex. As x approaches unity from large positive values the argument is diverging towards +inf and so theta is approaching (+pi/2). As x passes through 1 the argument becomes complex. At x equals 0 the argument has reached its minimum value of -i. For complex arguments the arctan is given by:
The following are plots of the arguments for A(x) and B(x). The x-axis is the value of x, and the y-axis is the value of the angle in units of pi. In the first plot theta is shown in blue curves, and as x approaches 1 the angle approaches (+pi/2). Theta is real because abs(x) >= 1, and notice it is symmetric across the y-axis. The black curve is varphi and as x approaches 0 it approaches plus or minus (pi/2). Notice it is an odd function in x.
In the second plot A(x) is shown where abs(x) < 1 and the argument becomes complex. Near x = 1 theta is equal to (+pi/2), the blue curve, minus a small imaginary part, the red curve. As x approaches zero theta is equal to (+pi/2) minus a large imaginary part. At x equals 0 the argument is equal to -i and theta = (+pi/2) minus an infinite imaginary part, i.e ln(0) = -inf:
The values for x0 and y0 are determined by the set of equations that equate modulus and argument of A(x) and B(x), and there are no other roots. If x0 = 0 was a root, then it would fall out of these equations. The same holds for x0 = 1. In fact, if one uses approximations in the argument of A(x) about these points, and then substitutes into the equation for the modulus, the equality cannot be maintained there.
Here is another perspective: consider the set of equations where x is assumed large and call it x_inf. The equation for the argument then gives x_inf = y_inf, where 1 is neglected with respect to x_inf squared. Upon substitution into the second equation a cubic is obtained in x_inf. Will this give the correct answer? Yes, if x0 is actually large, and in this case you might get away with it since x0 is approximately 2. The difference between the sqrt(4) and the sqrt(5) is around 10%. But does this mean that x_inf = 100 is a solution? No it does not: x_inf is only a solution if it equals x0.
The initial reason for examining the problem in the first place was to find a context for building a root-finding bisection routine as a Python iterator. This can be used to find any of the roots discussed here, and looks something like this:
class Bisection:
def __init__(self, a, b, func, max_iter):
self.max_iter = max_iter
self.count_iter = 0
self.a = a
self.b = b
self.func = func
fa = func(self.a)
fb = func(self.b)
if fa*fb >= 0.0:
raise ValueError
def __iter__(self):
self.x1 = self.a
self.x2 = self.b
self.xmid = self.x1 + ((self.x2 - self.x1)/2.0)
return self
def __next__(self):
f1 = self.func(self.x1)
f2 = self.func(self.x2)
error = abs(f1 - f2)
fmid = self.func(self.xmid)
if fmid == 0.0:
return self.xmid
if f1*fmid < 0:
self.x2 = self.xmid
else:
self.x1 = self.xmid
self.xmid = self.x1 + ((self.x2 - self.x1)/2.0)
f1 = self.func(self.x1)
fmid = self.func(self.xmid)
self.count_iter += 1
if self.count_iter >= self.max_iter:
raise StopIteration
return self.xmid
The routine does only a minimal amount in the way of catching exceptions and was used to find x for the given solution in the u-v plane. The arguments a and b give the lower and upper brackets for the root to be found. The argument func is the function for the root to be found. This might look like: u0 - B(x).real. The constant max_iterations tells the iterator to stop after a given number of bisections has been attempted.
I am trying to fit a polynomial to my data, e.g.
import scipy as sp
x = [1,6,9,17,23,28]
y = [6.1, 7.52324, 5.71, 5.86105, 6.3, 5.2]
and say I know the degree of polynomial (e.g.: 3), then I just use scipy.polyfit method to get the polynomial of a given degree:
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
fittedModelFunction = sp.polyfit(x, y, 3)
func = sp.poly1d(fittedModelFunction)
++++++++++++++++++++++++++++++
QUESTIONS: ++++++++++++++++++++++++++++++
1) How can I tell in addition that the resulting function func must be always positive (i.e. f(x) >= 0 for any x)?
2) How can I further define a constraint (e.g. number of (local) min and max points, etc.) in order to get a better fitting?
Is there smth like this:
http://mail.scipy.org/pipermail/scipy-user/2007-July/013138.html
but more accurate?
Always Positve
I haven't been able to find a scipy reference that determines if a function is positive-definite, but an indirect way would be to find the all the roots - Scipy Roots - of the function and inspect the limits near those roots. There are a few cases to consider:
No roots at all
Pick any x and evaluate the function. Since the function does not cross the x-axis because of a lack of roots, any positive result will indicate the function is positive!
Finite number of roots
This is probably the most likely case. You would have to inspect the limits before and after each root - Scipy Limits. You would have to specify your own minimum acceptable delta for the limit however. I haven't seen a 2-sided limit method provided by Scipy, but it looks simple enough to make your own.
from sympy import limit
// f: function, v: variable to limit, p: point, d: delta
// returns two limit values
def twoSidedLimit(f, v, p, d):
return limit(f, v, p-d), limit(f, v, p+d)
Infinite roots
I don't think that polyfit would generate an oscillating function, but this is something to consider. I don't know how to handle this with the method I have already offered... Um, hope it does not happen?
Constraints
The only built-in form of constraints seems to be limited to the optimize library of SciPy. A crude way to enforce constraints for polyfit would be to get the function from polyfit, generate a vector of values for various x, and try to select values from the vector that violate the constraint. If you try to use filter, map, or lambda it may be slow with large vectors since python's filter makes a copy of the list/vector being filtered. I can't really help in this regard.
I am trying to integrate a function over a list of point and pass the whole array to an integration function in order ot vectorize the thing. For starters, calling scipy.integrate.quad is way too slow since I have something like 10 000 000 points to integrate. Using scipy.integrate.romberg does the trick much faster, almost instantaneous while quad is slow since you must loop over it or vectorize it.
My function is quite complicated, but for demonstation purpose, let's say I want to integrate x^2 from a to b, but x is an array of scalar to evaluate x. For example
import numpy as np
from scipy.integrate import quad, romberg
def integrand(x, y):
return x**2 + y**2
quad(integrand, 0, 10, args=(10) # this fails since y is not a scalar
romberg(integrand, 0, 10) # y works here, giving the integral over
# the entire range
But this only work for fixed bounds. Is there a way to do something like
z = np.arange(20,30)
romberg(integrand, 0, z) # Fails since the function doesn't seem to
# support variable bounds
Only way I see it is to re-implement the algorithm itself in numpy and use that instead so I can have variable bounds. Any function that supports something like this? There is also romb, where you must supply the values of integrand directly and a dx interval, but that will be too imprecise for my complicated function (the marcum Q function, couldn't find any implementation, that could be another way to dot it).
The best approach when trying to evaluate a special function is to write a function that uses the properties of the function to quickly and accurately evaluate it in all parameter regimes. It is quite unlikely that a single approach will give accurate (or even stable) results for all ranges of parameters. Direct evaluation of an integral, as in this case, will almost certainly break down in many cases.
That being said, the general problem of evaluating an integral over many ranges can be solved by turning the integral into a differential equation and solving that. Roughly, the steps would be
Given an integral I(t) which I will assume is an integral of a function f(x) from 0 to t [this can be generalized to an arbitrary lower limit], write it as the differential equation dI/dt = f(x).
Solve this differential equation using scipy.integrate.odeint() for some initial conditions (here I(0)) over some range of times from 0 to t. This range should contain all limits of interest. How finely this is sampled depends on the function and how accurately it needs to be evaluated.
The result will be the value of the integral from 0 to t for the set of t we input. We can turn this into a "continuous" function using interpolation. For example, using a spline we can define i = scipy.interpolate.InterpolatedUnivariateSpline(t,I).
Given a set of upper and lower limits in arrays b and a, respectively, then we can evaluate them all at once as res=i(b)-i(a).
Whether this approach will work in your case will require you to carefully study it over your range of parameters. Also note that the Marcum Q function involves a semi-infinite integral. In principle this is not a problem, just transform the integral to one over a finite range. For example, consider the transformation x->1/x. There is no guarantee this approach will be numerically stable for your problem.