Утверждают, что два словаря почти равны

Комментарий @dano ответил на мой вопрос:

Я скопировал функцию по ссылке, предоставленной dano

import unittest
import numpy

def assertDeepAlmostEqual(test_case, expected, actual, *args, **kwargs):
    """
    Assert that two complex structures have almost equal contents.

    Compares lists, dicts and tuples recursively. Checks numeric values
    using test_case's :py:meth:`unittest.TestCase.assertAlmostEqual` and
    checks all other values with :py:meth:`unittest.TestCase.assertEqual`.
    Accepts additional positional and keyword arguments and pass those
    intact to assertAlmostEqual() (that's how you specify comparison
    precision).

    :param test_case: TestCase object on which we can call all of the basic
    'assert' methods.
    :type test_case: :py:class:`unittest.TestCase` object
    """
    is_root = not '__trace' in kwargs
    trace = kwargs.pop('__trace', 'ROOT')
    try:
        if isinstance(expected, (int, float, long, complex)):
            test_case.assertAlmostEqual(expected, actual, *args, **kwargs)
        elif isinstance(expected, (list, tuple, numpy.ndarray)):
            test_case.assertEqual(len(expected), len(actual))
            for index in xrange(len(expected)):
                v1, v2 = expected[index], actual[index]
                assertDeepAlmostEqual(test_case, v1, v2,
                                      __trace=repr(index), *args, **kwargs)
        elif isinstance(expected, dict):
            test_case.assertEqual(set(expected), set(actual))
            for key in expected:
                assertDeepAlmostEqual(test_case, expected[key], actual[key],
                                      __trace=repr(key), *args, **kwargs)
        else:
            test_case.assertEqual(expected, actual)
    except AssertionError as exc:
        exc.__dict__.setdefault('traces', []).append(trace)
        if is_root:
            trace = ' -> '.join(reversed(exc.traces))
            exc = AssertionError("%s\nTRACE: %s" % (exc.message, trace))
        raise exc

# My part, using the function

class TestMyClass(unittest.TestCase):
    def test_dicts(self):
        assertDeepAlmostEqual(self, {'a' : 12.4}, {'a' : 5.6 + 6.8})
    def test_dicts_2(self):
        dict_1 = {'a' : {'b' : [12.4, 0.3]}}
        dict_2 = {'a' : {'b' : [5.6 + 6.8, 0.1 + 0.2]}}

        assertDeepAlmostEqual(self, dict_1, dict_2)

def main():
    unittest.main()

if __name__ == "__main__":
    main()

Результат:

Ran 2 tests in 0.000s

OK

Я понимаю, что вы не стали бы импортировать pandas только для этого, но если вы используете pandas, вы можете преобразовать dicts в серию и использовать assert_series_equal из pandas.testing, который по умолчанию имеет check_exact=False.

>>> import pandas as pd
>>> from pandas.testing import assert_series_equal
>>> a = pd.Series({'a' : 12.4})
>>> b = pd.Series({'a': 12.399999999999999})
>>> assert_series_equal(a, b)

Я не мог запустить функцию Akavall, поэтому сделал свою собственную. Это слишком просто, но работает для моих целей. Код для проверки работы этой функции, написанный с использованием pytest

from numbers import Number
from math import isclose

def dictsAlmostEqual(dict1, dict2, rel_tol=1e-8):
    """
    If dictionary value is a number, then check that the numbers are almost equal, otherwise check if values are exactly equal
    Note: does not currently try converting strings to digits and comparing them. Does not care about ordering of keys in dictionaries
    Just returns true or false
    """
    if len(dict1) != len(dict2):
        return False
    # Loop through each item in the first dict and compare it to the second dict
    for key, item in dict1.items():
        # If it is a nested dictionary, need to call the function again
        if isinstance(item, dict):
            # If the nested dictionaries are not almost equal, return False
            if not dictsAlmostEqual(dict1[key], dict2[key], rel_tol=rel_tol):
                return False
        # If it's not a dictionary, then continue comparing
        # Put in else statement or else the nested dictionary will get compared twice and
        # On the second time will check for exactly equal and will fail
        else:
            # If the value is a number, check if they are approximately equal
            if isinstance(item, Number):
                # if not abs(dict1[key] - dict2[key]) <= rel_tol:
                # https://stackoverflow.com/questions/5595425/what-is-the-best-way-to-compare-floats-for-almost-equality-in-python
                if not isclose(dict1[key], dict2[key], rel_tol=rel_tol):
                    return False
            else:
                if not (dict1[key] == dict2[key]):
                    return False
    return True

Проверить вывод функции с помощью pytest

import pytest
import dictsAlmostEqual
def test_dictsAlmostEqual():
    a = {}
    b = {}
    assert dictsAlmostEqual(a, b)
    a = {"1": "a"}
    b = {}
    assert not dictsAlmostEqual(a, b)
    a = {"1": "a"}
    b = {"1": "a"}
    assert dictsAlmostEqual(a, b)
    a = {"1": "a"}
    b = {"1": "b"}
    assert not dictsAlmostEqual(a, b)
    a = {"1": "1.23"}
    b = {"1": "1.23"}
    assert dictsAlmostEqual(a, b)
    a = {"1": "1.234"}
    b = {"1": "1.23"}
    assert not dictsAlmostEqual(a, b)
    a = {"1": 1.000000000000001, "2": "a"}
    b = {"1": 1.000000000000002, "2": "a"}
    assert not dictsAlmostEqual(a, b, rel_tol=1e-20)
    assert dictsAlmostEqual(a, b, rel_tol=1e-8)
    assert dictsAlmostEqual(a, b)
    # Nested dicts
    a = {"1": {"2": 1.000000000000001}}
    b = {"1": {"2": 1.000000000000002}}
    assert not dictsAlmostEqual(a, b, rel_tol=1e-20)
    assert dictsAlmostEqual(a, b, rel_tol=1e-8)
    assert dictsAlmostEqual(a, b)
    a = {"1": {"2": 1.000000000000001, "3": "a"}, "2": "1.23"}
    b = {"1": {"2": 1.000000000000002, "3": "a"}, "2": "1.23"}
    assert not dictsAlmostEqual(a, b, rel_tol=1e-20)
    assert dictsAlmostEqual(a, b, rel_tol=1e-8)
    assert dictsAlmostEqual(a, b)

Pytest "приблизительно" делает свою работу

In [10]: {'a': 2.000001} == pytest.approx({'a': 2}) Out[10]: True