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How to use test_converge method in molecule

Best Python code snippet using molecule_python

integral.py

Source:integral.py

1# -------------------------------------------------------------------2# Library Program:3# The following program contains various functions to solve ordinary4# integrals using various methods.5# -------------------------------------------------------------------6# PRIMITIVE RULES (SINGLE VARIABLE FUNCTIONS)7def rectangular(a,b,N,f): #removed ELs8    '''9    a ; lower-bound for integration10    b ; upper-bound for integration11    N ; number of iterations/subintervals (must be integer)12    f ; function of x (can be user-defined)13    14    The (left/right) rectangular approximation method computes the15    integral of a function by assuming that the area between the16    function and the axis over which the function is being integrated17    can be defined as the area of a series of rectangles, for a18    series of subintervals bound by successive values of x with19    constant step size ;20        xn - xn-121    For this approximation method, the left/right uppermost corner of22    a rectangle has coordinates computed using the given function,23    then calculates the rectangular area ;24        f(xn) * (xn - xn-1) [for LRAM], or25        f(xn) * (xn+1 - xn) [for RRAM]26    27    These areas are iterated for all subintervals n in N, with the28    sum of the recatangular areas approximating the integral for the29    given function between its lower/upper bounds.30    '''31    dx = (b-a)/N32    Rl = 033    Rr = 034    for n in range(0,N):35        Rl += dx * f(a+n*dx)36    for n in range(1,N+1):37        Rr += dx * f(a+n*dx)38    return Rl,Rr39def midpoint(a,b,N,f): #removed ELs40    '''41    a ; lower-bound for integration42    b ; upper-bound for integration43    N ; number of iterations/subintervals (must be integer)44    f ; function of x (can be user-defined)45    The midpoint approximation method operates similarly to the46    rectangular approximation method, in that it approximates the47    area coinciding to the integral of a given function using48    rectangles over a series of subintervals.49    Where the midpoint approximation method differs is that it50    computes the midpoint of two subinterval bounds ;51        Mn = (xn - xn-1) / 252    53    Once again, like the rectangular method a rectangular area is54    computed using two consecutive subintervals as the x bounds for55    the area and the function of the midpoint as the height ;56        A = f(Mn) * (xn - xn-1)57    58    As for all methods using shapes to approximate the integral of a59    given function within bounds, the areas of the rectangles are60    summed to give a value.61    '''62    dx = (b-a)/N63    M = 064    for n in range(1,N+1):65        M += f((a+n*dx-a+(n-1)*dx)/2) * dx66    return M67def trapezoid(a,b,N,f): #removed ELs68    '''69    a ; lower-bound for integration70    b ; upper-bound for integration71    N ; number of iterations/subintervals (must be integer)72    f ; function of x (can be user-defined)73    '''74    dx = (b-a)/N75    T = dx/2 * (f(a) + f(b))76    for n in range(1,N):77        T += dx * f(a+n*dx)78    return T79def simpson(a,b,N,f): #removed ELs80    '''81    a ; lower-bound for integration82    b ; upper-bound for integration83    N ; number of iterations/subintervals (must be integer)84    f ; function of x (can be user-defined)85    '''86    dx = (b-a)/N87    S = dx/3 * (f(a) + f(b))88    if N%2 == 0:89        for n in range(1,int(N/2)):90            S += dx/3 * (4*f(a+(2*n-1)*dx) + 2*f(a+2*n*dx))91    if N%2 != 0:92        for n in range(1,int((N+1)/2)):93            S += dx/3 * (4*f(a+(2*n-1)*dx) + 2*f(a+2*n*dx))94    return S95# LARGE UNCERTAINTY AT LOW ITERATIONS (REQUIRES HIGH ITERATIONS TO PRODUCE SIMILAR INTEGRALS TO OTHER METHODS/VERY INCONSISTENT)96def boole(a,b,N,f): #removed ELs97    '''98    a ; lower-bound for integration99    b ; upper-bound for integration100    N ; number of iterations/subintervals (must be integer)101    f ; function of x (can be user-defined)102    '''103    # SECOND TERM (32) IS 2,6,10,...,-10,-6,-2104    # THIRD TERM (12) IS 4,8,12,...,-12,-8,-4105    dx = (b-a)/N106    B = 2*dx/45 * 7*(f(a) + f(b))107    108    Ns_1 = []109    Ns_2 = []110    Ns_3 = []111    112    if N%2 == 0:113        Nrange = int(N/2)114    else:115        Nrange = int((N-1)/2 + 1)116    for n in range(1,Nrange):117        if n%2 == 1:118            Ns_1.append(n)119        if n%4 == 2:120            Ns_2.append(n)121        if n%4 == 0:122            Ns_3.append(n)123    124    for n in Ns_1:125        B += 2*dx/45 * (32*f(a+n*dx) + 32*f(b-n*dx))126    for n in Ns_2:127        B += 2*dx/45 * (12*f(a+n*dx) + 12*f(b-n*dx))128    for n in Ns_3:129        B += 2*dx/45 * (14*f(a+n*dx) + 14*f(b-n*dx))130    return B131def romberg(a,b,f): #removed ELs132    '''133    a ; lower-bound for integration134    b ; upper-bound for integration135    f ; function of x (can be user-defined)136    '''137    test_converge = 1E-9138    dx = (b-a)/2139    CRT = dx/2 * (f(a) + f(b))140    for n in range(1,3):141        CRT += dx * f(a+n*dx)142    R_matrix = [CRT]143    144    i = 2145    while i >= 2:146        dx = (b-a)/(2**(i-1))147        R_j = 0148        for n in range(0,int(2**(i-1)+1)):149            R_j += dx * f(a+n*dx)150        151        Rs = [R_j]152        for j in range(2,i+1):153            R = lambda j: (4**(j-1)*R_j - R_matrix[(j-2)]) / (4**(j-1) - 1)154            Rs.append(R(j))155            R_j = R(j)156            if abs(R(j) - R_j) <= test_converge:157                return Rs[-1]158        R_matrix = Rs159        160        i += 1161# MULTI-VARIABLE FUNCTIONS162def double_rectangular(a,b,c,d,N,M,f): #removed ELs163    '''164    a ; lower-bound for integration over x165    b ; upper-bound for integration over x166    c ; lower-bound for integration over y167    d ; upper-bound for integration over y168    N ; number of x iterations/subintervals (must be integer)169    M ; number of y iterations/subintervals (must be integer)170    f ; function of x and y (can be user-defined)171    '''172    dx = (b-a)/N173    dy = (d-c)/M174    dA = dx*dy175    DR = 0176    for n in range(1,N+1):177        for m in range(1,M+1):178            DR += dA * f(a+n*dx,c+m*dy)...

test_jacobian.py

Source:test_jacobian.py

...13                input_vector = (0, 0, 0)))14        npt.assert_almost_equal(matrix, f_solver._jacobian_matrix(15                input_vector = (1, 2, 3)))16class JacobianInverseSolverTestCase(unittest.TestCase):17    def test_converge(self):18        matrix = ((1, 0, 3), (0, 2, 2), (1, 2, 1))19        f = lambda x: np.dot(matrix, x)20        f_solver = JacobianInverseSolver(function = f)21        # For a linear map the unconstrained solver reaches the target22        # output vector after a single iteration.23        npt.assert_almost_equal((1, 1, 0), f_solver.converge(24                input_vector = (0, 0, 0),25                target_output_vector = (1, 2, 3)))26    def test_converge_max_input_fix(self):27        matrix = ((1, 0, 3), (0, 2, 2), (1, 2, 1))28        f = lambda x: np.dot(matrix, x)29        f_solver = JacobianInverseSolver(30                function = f,31                max_input_fix = 0.5)32        # The input vector has to go from (0, 0, 0) to (1, 1, 0). The solver33        # input fix is limited to 0.5 per component. The solver gets halfway34        # there after a single iteration.35        input_vector = (0, 0, 0)36        input_vector = input_vector = f_solver.converge(37               input_vector = input_vector,38               target_output_vector = (1, 2, 3))39        npt.assert_almost_equal((0.5, 0.5, 0), input_vector)40        # And reaches the goal after a second iteration.41        input_vector = f_solver.converge(42                input_vector = input_vector,43                target_output_vector = (1, 2, 3))44        npt.assert_almost_equal((1, 1, 0), input_vector)45    def test_converge_max_output_error(self):46        matrix = ((1, 0, 3), (0, 2, 2), (1, 2, 1))47        f = lambda x: np.dot(matrix, x)48        f_solver = JacobianInverseSolver(49                function = f,50                max_output_error = np.sqrt(14) / 2)51        # The output vector has to go from (0, 0, 0) to (1, 2, 3). The52        # initial output vector error norm is hence sqrt(14). The solver53        # output error norm is limited to sqrt(14)/2. The solver gets54        # halfway there after a single iteration.55        input_vector = (0, 0, 0)56        input_vector = f_solver.converge(57                input_vector = input_vector,58                target_output_vector = (1, 2, 3))59        npt.assert_almost_equal((0.5, 0.5, 0), input_vector)60        # And reaches the goal after a second iteration.61        input_vector = f_solver.converge(62                input_vector = input_vector,63                target_output_vector = (1, 2, 3))64        npt.assert_almost_equal((1, 1, 0), input_vector)65class DampedLeastSquaresSolverTest(unittest.TestCase):66    def test_converge(self):67        matrix = ((1, 0, 3), (0, 2, 2), (1, 2, 1))68        f = lambda x: np.dot(matrix, x)69        # For a linear map the solver reaches the target output vector70        # after enough iterations.71        f_solver = DampedLeastSquaresSolver(function = f, constant = 1)72        input_vector = (0, 0, 0)73        for n in xrange(50):74            input_vector = f_solver.converge(75                    input_vector = input_vector,76                    target_output_vector = (1, 2, 3))...

test1.py

Source:test1.py

...4from .newsvendor import *5class MyTestCase(unittest.TestCase):6    def setUp(self):7        pass8    def test_converge(self):9        self.news = newsvendormodel()10        status = solve_default(self.news,11                               iteration_limit=20,12                               cut_selection_frequency=10,13                               simulation=MonteCarloSimulation(14                                   frequency=10,15                                   steps=list(range(10, 501, 10))16                               ),17                               bound_stalling=BoundStalling(18                                   iterations=5,19                                   atol=1e-320                               )21                               )22        self.assertEqual(status, Staus.stalling_convergence)...

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