问题
Scipy offers several seemingly equivalent functions for finding the root of a function in a given interval:
brentq(f, a, b[, args, xtol, rtol, maxiter, ...]) Find a root of a function in given interval.
brenth(f, a, b[, args, xtol, rtol, maxiter, ...]) Find root of f in [a,b].
ridder(f, a, b[, args, xtol, rtol, maxiter, ...]) Find a root of a function in an interval.
bisect(f, a, b[, args, xtol, rtol, maxiter, ...]) Find root of a function within an interval.
(See this webpage.)
Can anyone provide some guidelines for choosing one of these over the others? Is the best strategy for finding the one that works for my case simple trial and error?
回答1:
brentq
brentq purports to be the best of the four functions in the question. Its docstring reads
Generally considered the best of the rootfinding routines here.
However, it has (at least) two annoying features:
1) It requires that f(a) have a different sign than f(b).
2) If a is a very small positive number (as large as 1e-3), it occasionally returns 0.0 as a solution -- i.e., it returns a solution outside the submitted bounds.
brenth
brenth shares brentq's feature 1, above.
ridder
ridder shares brentq's feature 1, above.
bisect
bisect shares brentq's feature 1, above, and is slower than the other functions.
Minimization methods
I realized I could turn my root-finding problem into a minimization problem by taking the absolute value of the output of my function f. (Another option is to take the square of the output of f.) Scipy offers several functions for bounded minimization of a scalar function:
fminbound(func, x1, x2[, args, xtol, ...]) Bounded minimization for scalar functions.
brent(func[, args, brack, tol, full_output, ...]) Given a function of one-variable and a possible bracketing interval, return the minimum of the function isolated to a fractional precision of tol.
brute(func, ranges[, args, Ns, full_output, ...]) Minimize a function over a given range by brute force.
fminbound
My only complaint is that it's slow. It does not have the limitation of requiring that f(a) have a different sign than f(b).
brent
For its bracketing interval [a, b], brent requires that f(a) be less than f(b). Its solution is not guaranteed to be within [a, b].
brute
brute is of course very slow (depending on the value of the Ns argument), and, oddly, may return a solution outside the submitted bounds.
Conclusion
All that said, I get the best results with the method in this answer -- i.e., using the function least_squares in a yet-to-be-released version of scipy. This function has none of the limitations of those above.
来源：https://stackoverflow.com/questions/32814173/bounded-root-finding-in-scipy