SCIPY MINIMIZE CODE
The x portion is passed in by the optimizer, and the args tuple is given as the remaining arguments. from scipy.optimize import minimize from math import def f(c): return sqrt((sin(pi/2) + sin(0) + sin(c) - 2)2 + (cos(pi/2) + cos(0) + cos(c) - 1)2) print minimize(f, 3.14/2 + 3.14/7) The above code does try to minimize the function f, but for my task I need to minimize with respect to three variables. The documentation tries to explain how the args tuple is usedĮffectively, will pass whatever is in args as the remainder of the arguments to fun, using the asterisk arguments notation: the function is then called as fun(x, *args) during optimization. Parameters funcallable The objective function to be minimized. it should look something like this: def obj(arguments) '''objective function, to be solved. The optimizer is responsible for creating values of x and passing them to fun for evaluation.īut what if you happen to have a function fun(x, y) that has some additional parameter y that needs to be passed in separately (but is considered a constant for the purposes of the optimization)? This is what the args tuple is for. ¶ (fun, x0, args(), methodNone, jacNone, hessNone, hesspNone, boundsNone, constraints(), tolNone, callbackNone, optionsNone) source ¶ Minimization of scalar function of one or more variables. for, multiple arguments should be packed into a tuple, which will then be unpacked by the objective function during numerical optimization. then finds an argument value xp such that fun(xp) is less than fun(x) for other values of x. For detailed control, use solver-specific.
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It defines a tol argument, for which the docs say: Tolerance for termination. "By default, takes a function fun(x) that accepts one argument x (which might be an array or the like) and returns a scalar. The method offers an interface to several minimizers. Hope it will not cause some IP problem, quoted the essential part of the answer here:įrom at Structure of inputs to scipy minimize function