Python optimize.rosen方法代码示例

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def test_l_bfgs_b_maxiter(self):
        # gh7854
        # Ensure that not more than maxiters are ever run.
        class Callback(object):
            def __init__(self):
                self.nit = 0
                self.fun = None
                self.x = None

            def __call__(self, x):
                self.x = x
                self.fun = optimize.rosen(x)
                self.nit += 1

        c = Callback()
        res = optimize.minimize(optimize.rosen, [0., 0.], method='l-bfgs-b',
                                callback=c, options={'maxiter': 5})

        assert_equal(res.nit, 5)
        assert_almost_equal(res.x, c.x)
        assert_almost_equal(res.fun, c.fun)
        assert_equal(res.status, 1)
        assert_(res.success is False)
        assert_equal(res.message.decode(), 'STOP: TOTAL NO. of ITERATIONS REACHED LIMIT') 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def test_minimize_l_bfgs_b_maxfun_interruption(self):
        # gh-6162
        f = optimize.rosen
        g = optimize.rosen_der
        values = []
        x0 = np.ones(7) * 1000

        def objfun(x):
            value = f(x)
            values.append(value)
            return value

        # Look for an interesting test case.
        # Request a maxfun that stops at a particularly bad function
        # evaluation somewhere between 100 and 300 evaluations.
        low, medium, high = 30, 100, 300
        optimize.fmin_l_bfgs_b(objfun, x0, fprime=g, maxfun=high)
        v, k = max((y, i) for i, y in enumerate(values[medium:]))
        maxfun = medium + k
        # If the minimization strategy is reasonable,
        # the minimize() result should not be worse than the best
        # of the first 30 function evaluations.
        target = min(values[:low])
        xmin, fmin, d = optimize.fmin_l_bfgs_b(f, x0, fprime=g, maxfun=maxfun)
        assert_array_less(fmin, target) 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def test_bounds_checking(self):
        # test that the bounds checking works
        func = rosen
        bounds = [(-3, None)]
        assert_raises(ValueError,
                          differential_evolution,
                          func,
                          bounds)
        bounds = [(-3)]
        assert_raises(ValueError,
                          differential_evolution,
                          func,
                          bounds)
        bounds = [(-3, 3), (3, 4, 5)]
        assert_raises(ValueError,
                          differential_evolution,
                          func,
                          bounds) 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def test_iteration(self):
        # test that DifferentialEvolutionSolver is iterable
        # if popsize is 3 then the overall generation has size (6,)
        solver = DifferentialEvolutionSolver(rosen, self.bounds, popsize=3,
                                             maxfun=12)
        x, fun = next(solver)
        assert_equal(np.size(x, 0), 2)

        # 6 nfev are required for initial calculation of energies, 6 nfev are
        # required for the evolution of the 6 population members.
        assert_equal(solver._nfev, 12)

        # the next generation should halt because it exceeds maxfun
        assert_raises(StopIteration, next, solver)

        # check a proper minimisation can be done by an iterable solver
        solver = DifferentialEvolutionSolver(rosen, self.bounds)
        for i, soln in enumerate(solver):
            x_current, fun_current = soln
            # need to have this otherwise the solver would never stop.
            if i == 1000:
                break

        assert_almost_equal(fun_current, 0) 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def rosenbrock_problem():
    """Problem based on rosenbrock objective.

    Features
    --------
    * 3-dim
    * has fixed parameters
    """
    objective = pypesto.Objective(fun=so.rosen)

    dim_full = 2
    lb = -5 * np.ones((dim_full, 1))
    ub = 5 * np.ones((dim_full, 1))

    problem = pypesto.Problem(
            objective=objective, lb=lb, ub=ub,
            x_fixed_indices=[1], x_fixed_vals=[2])
    return problem 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def fun(self, x, *args):
        self.nfev += 1

        return rosen(x) 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def test_minimize_l_bfgs_maxls(self):
        # check that the maxls is passed down to the Fortran routine
        sol = optimize.minimize(optimize.rosen, np.array([-1.2,1.0]),
                                method='L-BFGS-B', jac=optimize.rosen_der,
                                options={'disp': False, 'maxls': 1})
        assert_(not sol.success) 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def test_custom(self):
        # This function comes from the documentation example.
        def custmin(fun, x0, args=(), maxfev=None, stepsize=0.1,
                maxiter=100, callback=None, **options):
            bestx = x0
            besty = fun(x0)
            funcalls = 1
            niter = 0
            improved = True
            stop = False

            while improved and not stop and niter < maxiter:
                improved = False
                niter += 1
                for dim in range(np.size(x0)):
                    for s in [bestx[dim] - stepsize, bestx[dim] + stepsize]:
                        testx = np.copy(bestx)
                        testx[dim] = s
                        testy = fun(testx, *args)
                        funcalls += 1
                        if testy < besty:
                            besty = testy
                            bestx = testx
                            improved = True
                    if callback is not None:
                        callback(bestx)
                    if maxfev is not None and funcalls >= maxfev:
                        stop = True
                        break

            return optimize.OptimizeResult(fun=besty, x=bestx, nit=niter,
                                           nfev=funcalls, success=(niter > 1))

        x0 = [1.35, 0.9, 0.8, 1.1, 1.2]
        res = optimize.minimize(optimize.rosen, x0, method=custmin,
                                options=dict(stepsize=0.05))
        assert_allclose(res.x, 1.0, rtol=1e-4, atol=1e-4) 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def setup_method(self):
        self.x0 = [5, 5]
        self.func = optimize.rosen
        self.jac = optimize.rosen_der
        self.hess = optimize.rosen_hess
        self.hessp = optimize.rosen_hess_prod
        self.bounds = [(0., 10.), (0., 10.)] 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def test_invalid_mutation_values_arent_accepted(self):
        func = rosen
        mutation = (0.5, 3)
        assert_raises(ValueError,
                          DifferentialEvolutionSolver,
                          func,
                          self.bounds,
                          mutation=mutation)

        mutation = (-1, 1)
        assert_raises(ValueError,
                          DifferentialEvolutionSolver,
                          func,
                          self.bounds,
                          mutation=mutation)

        mutation = (0.1, np.nan)
        assert_raises(ValueError,
                          DifferentialEvolutionSolver,
                          func,
                          self.bounds,
                          mutation=mutation)

        mutation = 0.5
        solver = DifferentialEvolutionSolver(func,
                                             self.bounds,
                                             mutation=mutation)
        assert_equal(0.5, solver.scale)
        assert_equal(None, solver.dither) 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def test_callback_terminates(self):
        # test that if the callback returns true, then the minimization halts
        bounds = [(0, 2), (0, 2)]

        def callback(param, convergence=0.):
            return True

        result = differential_evolution(rosen, bounds, callback=callback)

        assert_string_equal(result.message,
                                'callback function requested stop early '
                                'by returning True') 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def test_init_with_invalid_strategy(self):
        # test that passing an invalid strategy raises ValueError
        func = rosen
        bounds = [(-3, 3)]
        assert_raises(ValueError,
                          differential_evolution,
                          func,
                          bounds,
                          strategy='abc') 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def test_maxfun_stops_solve(self):
        # test that if the maximum number of function evaluations is exceeded
        # during initialisation the solver stops
        solver = DifferentialEvolutionSolver(rosen, self.bounds, maxfun=1,
                                             polish=False)
        result = solver.solve()

        assert_equal(result.nfev, 2)
        assert_equal(result.success, False)
        assert_equal(result.message,
                     'Maximum number of function evaluations has '
                     'been exceeded.')

        # test that if the maximum number of function evaluations is exceeded
        # during the actual minimisation, then the solver stops.
        # Have to turn polishing off, as this will still occur even if maxfun
        # is reached. For popsize=5 and len(bounds)=2, then there are only 10
        # function evaluations during initialisation.
        solver = DifferentialEvolutionSolver(rosen,
                                             self.bounds,
                                             popsize=5,
                                             polish=False,
                                             maxfun=40)
        result = solver.solve()

        assert_equal(result.nfev, 41)
        assert_equal(result.success, False)
        assert_equal(result.message,
                         'Maximum number of function evaluations has '
                              'been exceeded.') 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def test_exp_runs(self):
        # test whether exponential mutation loop runs
        solver = DifferentialEvolutionSolver(rosen,
                                             self.bounds,
                                             strategy='best1exp',
                                             maxiter=1)

        solver.solve() 

# 需要导入模块: from scipy import optimize [as 别名]
# 或者: from scipy.optimize import rosen [as 别名]
def test_gh_4511_regression(self):
        # This modification of the differential evolution docstring example
        # uses a custom popsize that had triggered an off-by-one error.
        # Because we do not care about solving the optimization problem in
        # this test, we use maxiter=1 to reduce the testing time.
        bounds = [(-5, 5), (-5, 5)]
        result = differential_evolution(rosen, bounds, popsize=1815, maxiter=1)