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How to use constrained_complex method in hypothesis

Best Python code snippet using hypothesis

core.py

Source:core.py

...1666        # magnitude and therefore no relationship between the real and1667        # imaginary parts.1668        return builds(complex, floats(**allow_kw), floats(**allow_kw))1669    @composite1670    def constrained_complex(draw):1671        # Draw the imaginary part, and determine the maximum real part given1672        # this and the max_magnitude1673        if max_magnitude is None:1674            zi = draw(floats(**allow_kw))1675            rmax = None1676        else:1677            zi = draw(floats(-max_magnitude, max_magnitude, **allow_kw))1678            rmax = cathetus(max_magnitude, zi)1679        # Draw the real part from the allowed range given the imaginary part1680        if min_magnitude == 0 or math.fabs(zi) >= min_magnitude:1681            zr = draw(floats(None if rmax is None else -rmax, rmax, **allow_kw))1682        else:1683            zr = draw(floats(cathetus(min_magnitude, zi), rmax, **allow_kw))1684        # Order of conditions carefully tuned so that for a given pair of1685        # magnitude arguments, we always either draw or do not draw the bool1686        # (crucial for good shrinking behaviour) but only invert when needed.1687        if min_magnitude > 0 and draw(booleans()) and math.fabs(zi) <= min_magnitude:1688            zr = -zr1689        return complex(zr, zi)1690    return constrained_complex()1691@deprecated_posargs1692def shared(1693    base: SearchStrategy[Ex],1694    *,1695    key: Optional[Hashable] = None,1696) -> SearchStrategy[Ex]:1697    """Returns a strategy that draws a single shared value per run, drawn from1698    base. Any two shared instances with the same key will share the same value,1699    otherwise the identity of this strategy will be used. That is:1700    >>> s = integers()  # or any other strategy1701    >>> x = shared(s)1702    >>> y = shared(s)1703    In the above x and y may draw different (or potentially the same) values.1704    In the following they will always draw the same:...

strategies.py

Source:strategies.py

...1594        # magnitude and therefore no relationship between the real and1595        # imaginary parts.1596        return builds(complex, floats(**allow_kw), floats(**allow_kw))1597    @composite1598    def constrained_complex(draw):1599        # Draw the imaginary part, and determine the maximum real part given1600        # this and the max_magnitude1601        if max_magnitude is None:1602            zi = draw(floats(**allow_kw))1603            rmax = float('inf')1604        else:1605            zi = draw(floats(-max_magnitude, max_magnitude, **allow_kw))1606            rmax = cathetus(max_magnitude, zi)1607        # Draw the real part from the allowed range given the imaginary part1608        if min_magnitude is None or math.fabs(zi) >= min_magnitude:1609            zr = draw(floats(-rmax, rmax, **allow_kw))1610        else:1611            zr = draw(floats(cathetus(min_magnitude, zi), rmax, **allow_kw))1612        # Order of conditions carefully tuned so that for a given pair of1613        # magnitude arguments, we always either draw or do not draw the bool1614        # (crucial for good shrinking behaviour) but only invert when needed.1615        if min_magnitude is not None and draw(booleans()) and \1616                math.fabs(zi) <= min_magnitude:1617            zr = -zr1618        return complex(zr, zi)1619    return constrained_complex()1620def shared(base, key=None):1621    # type: (SearchStrategy[Ex], Any) -> SearchStrategy[Ex]1622    """Returns a strategy that draws a single shared value per run, drawn from1623    base. Any two shared instances with the same key will share the same value,1624    otherwise the identity of this strategy will be used. That is:1625    >>> s = integers()  # or any other strategy1626    >>> x = shared(s)1627    >>> y = shared(s)1628    In the above x and y may draw different (or potentially the same) values.1629    In the following they will always draw the same:1630    >>> x = shared(s, key="hi")1631    >>> y = shared(s, key="hi")1632    Examples from this strategy shrink as per their base strategy.1633    """...

test_graphsolver.py

Source:test_graphsolver.py

1"""2Unit tests for module.3"""4import networkx as nx5import pandas as pd6import logging7import pytest8import allocate.solvers.graphsolver9import allocate.network.algorithms10import allocate.network.visualize11import tests.utilities12from allocate.solvers.constrained import BucketSolverConstrained13from allocate.solvers.constrained import BucketSolverSimple14from allocate.solvers import BucketSolver15@pytest.mark.parametrize('starting_frame,expected_graph,solver', [16    # simple_no_addition : the amounts should be redistributed17    (18        pd.DataFrame([19            dict(label='T', current_value=3000.0, optimal_ratio=1.00, amount_to_add=0.0000, children=('H', 'I', 'J')),20            dict(label='H', current_value=3000.0, optimal_ratio=0.50, amount_to_add=0.0000, children=()),21            dict(label='I', current_value=0000.0, optimal_ratio=0.35, amount_to_add=0.0000, children=()),22            dict(label='J', current_value=0000.0, optimal_ratio=0.15, amount_to_add=0.0000, children=()),23        ]),24        tests.utilities.make_graph(nodes=[25            ('T', dict(results_value=3000.0 + 3000.0 * 0.00, amount_to_add=+3000.0 * 0.00)),26            ('H', dict(results_value=3000.0 - 3000.0 * 0.50, amount_to_add=-3000.0 * 0.50)),27            ('I', dict(results_value=0.0000 + 3000.0 * 0.35, amount_to_add=+3000.0 * 0.35)),28            ('J', dict(results_value=0.0000 + 3000.0 * 0.15, amount_to_add=+3000.0 * 0.15)),29        ], edges=[30            ('T', 'H'), ('T', 'I'), ('T', 'J')31        ]),32        BucketSolverSimple33    ),34    # simple_value_added : the amounts should be redistributed and value should be added35    (36        pd.DataFrame([37            dict(label='F', current_value=8000.0 + 0.000, optimal_ratio=1.00, amount_to_add=1000.0,38                 children=('U', 'V', 'W')),39            dict(label='U', current_value=4000.0 + 0.000, optimal_ratio=0.50, amount_to_add=0.0000, children=()),40            dict(label='V', current_value=2800.0 + 256.0, optimal_ratio=0.35, amount_to_add=0.0000, children=()),41            dict(label='W', current_value=1200.0 - 256.0, optimal_ratio=0.15, amount_to_add=0.0000, children=()),42        ]),43        tests.utilities.make_graph(nodes=[44            ('F', dict(results_value=8000.0 + 0.000 + 1000.0 * 1.00 + 0.000, amount_to_add=+1000.0 * 0.00 + 0.000)),45            ('U', dict(results_value=4000.0 + 0.000 + 1000.0 * 0.50 + 0.000, amount_to_add=+1000.0 * 0.50 + 0.000)),46            ('V', dict(results_value=2800.0 + 256.0 + 1000.0 * 0.35 - 256.0, amount_to_add=+1000.0 * 0.35 - 256.0)),47            ('W', dict(results_value=1200.0 - 256.0 + 1000.0 * 0.15 + 256.0, amount_to_add=+1000.0 * 0.15 + 256.0)),48        ], edges=[49            ('F', 'U'), ('F', 'V'), ('F', 'W')50        ]),51        BucketSolverSimple52    ),53    # constrained_simple : values are only added to the final result and are in perfect ratios54    (55        pd.DataFrame([56            dict(label='A', current_value=4000.0, optimal_ratio=1.00, amount_to_add=1000.0, children=('0', '1', '2')),57            dict(label='0', current_value=2000.0, optimal_ratio=0.50, amount_to_add=0.0000, children=()),58            dict(label='1', current_value=1000.0, optimal_ratio=0.25, amount_to_add=0.0000, children=()),59            dict(label='2', current_value=1000.0, optimal_ratio=0.25, amount_to_add=0.0000, children=()),60        ]),61        tests.utilities.make_graph(nodes=[62            ('A', dict(results_value=4000.0 + 1000.0 * 1.00, amount_to_add=1000.0 * 0.00)),63            ('0', dict(results_value=2000.0 + 1000.0 * 0.50, amount_to_add=1000.0 * 0.50)),64            ('1', dict(results_value=1000.0 + 1000.0 * 0.25, amount_to_add=1000.0 * 0.25)),65            ('2', dict(results_value=1000.0 + 1000.0 * 0.25, amount_to_add=1000.0 * 0.25)),66        ], edges=[67            ('A', '0'), ('A', '1'), ('A', '2')68        ]),69        BucketSolverConstrained70    ),71    # constrained_complex : values are only added to the final result and are in perfect ratios72    (73        pd.DataFrame([74            dict(label='B', current_value=8000.0, optimal_ratio=1.00, amount_to_add=4000.0, children=('3', '4', '5')),75            dict(label='3', current_value=4000.0, optimal_ratio=0.50, amount_to_add=0.0000, children=()),76            dict(label='4', current_value=2000.0, optimal_ratio=0.25, amount_to_add=0.0000, children=()),77            dict(label='5', current_value=2000.0, optimal_ratio=0.25, amount_to_add=0.0000, children=('C', 'D')),78            dict(label='C', current_value=1000.0, optimal_ratio=0.50, amount_to_add=0.0000, children=()),79            dict(label='D', current_value=1000.0, optimal_ratio=0.50, amount_to_add=0.0000, children=('6', '7')),80            dict(label='6', current_value=2.50e2, optimal_ratio=0.25, amount_to_add=0.0000, children=()),81            dict(label='7', current_value=7.50e2, optimal_ratio=0.75, amount_to_add=0.0000, children=()),82        ]),83        tests.utilities.make_graph(nodes=[84            ('B', dict(results_value=8000.0 + 4000.0 * 1.00 * 1.00 * 1.00, amount_to_add=4000.0 * 0.00 * 1.00 * 1.00)),85            ('3', dict(results_value=4000.0 + 4000.0 * 0.50 * 1.00 * 1.00, amount_to_add=4000.0 * 0.50 * 1.00 * 1.00)),86            ('4', dict(results_value=2000.0 + 4000.0 * 0.25 * 1.00 * 1.00, amount_to_add=4000.0 * 0.25 * 1.00 * 1.00)),87            ('5', dict(results_value=2000.0 + 4000.0 * 0.25 * 1.00 * 1.00, amount_to_add=4000.0 * 0.00 * 1.00 * 1.00)),88            ('C', dict(results_value=1000.0 + 4000.0 * 0.25 * 0.50 * 1.00, amount_to_add=4000.0 * 0.25 * 0.50 * 1.00)),89            ('D', dict(results_value=1000.0 + 4000.0 * 0.25 * 0.50 * 1.00, amount_to_add=4000.0 * 0.00 * 0.50 * 1.00)),90            ('6', dict(results_value=2.50e2 + 4000.0 * 0.25 * 0.50 * 0.25, amount_to_add=4000.0 * 0.25 * 0.50 * 0.25)),91            ('7', dict(results_value=7.50e2 + 4000.0 * 0.25 * 0.50 * 0.75, amount_to_add=4000.0 * 0.25 * 0.50 * 0.75)),92        ], edges=[93            ('B', '3'), ('B', '4'), ('B', '5'), ('5', 'C'), ('5', 'D'), ('D', '6'), ('D', '7')94        ]),95        BucketSolverConstrained96    ),97], ids=[98    'simple_no_addition',99    'simple_value_added',100    'constrained_simple',101    'constrained_complex',102])103def test_solve(starting_frame: pd.DataFrame, expected_graph: nx.DiGraph, solver: BucketSolver):104    logging.debug('starting_frame:\n%s', starting_frame)105    starting_graph: nx.DiGraph = allocate.network.algorithms.create(starting_frame)106    tests.utilities.show_graph('starting_graph', starting_graph, **allocate.network.visualize.formats_inp)107    tests.utilities.show_graph('expected_graph', expected_graph, **allocate.network.visualize.formats_out)108    observed_graph: nx.DiGraph = allocate.solvers.graphsolver.solve(starting_graph, solver=solver, inplace=False)109    tests.utilities.show_graph('observed_graph', observed_graph, **allocate.network.visualize.formats_out)110    node_match = nx.algorithms.isomorphism.numerical_node_match(111        ['results_value', 'amount_to_add'], [-1000, -1000])...

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