断言两个字典几乎相等

Question

我想要检查两个字典（就是一种数据结构，像个装东西的箱子）是否几乎相等，但我似乎找不到办法。

这里有个例子：

>>> import nose.tools as nt
>>> nt.assert_dict_equal({'a' : 12.4}, {'a' : 5.6 + 6.8})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.7/unittest/case.py", line 838, in assertDictEqual
    self.fail(self._formatMessage(msg, standardMsg))
  File "/usr/lib/python2.7/unittest/case.py", line 413, in fail
    raise self.failureException(msg)
AssertionError: {'a': 12.4} != {'a': 12.399999999999999}
- {'a': 12.4}
+ {'a': 12.399999999999999}

我希望这个检查能通过，像这样：

>>> nt.assert_almost_equal(12.4, 5.6 + 6.8)

我在想，可能我遗漏了一些简单的东西，比如 nt.assert_almost_dict_equal，或者也许有个参数可以传给 nt.assert_dict_equal，用来指定浮点数（就是带小数的数字）应该有多接近，但我找不到相关的信息。

当然，我可以遍历字典，逐个比较值，用 nt.assert_almost_equal 来检查；不过在我的情况下，字典比较复杂，所以我希望能避免这样做。

那么，检查两个字典是否几乎相等的最佳方法是什么呢？

Answer 1

我需要一个递归版本的 round，这样可以让输出看起来更好：

def recursive_round(data, ndigits=5):
    if isinstance(data, dict):
        return {k: recursive_round(v, ndigits) for k, v in data.items()}
    if isinstance(data, list):
        return [recursive_round(v, ndigits) for v in data]
    if isinstance(data, tuple):
        return tuple(recursive_round(v, ndigits) for v in data)
    if isinstance(data, set):
        return {recursive_round(v, ndigits) for v in data}
    if isinstance(data, float):
        if data.is_integer():
            return int(data)
        return round(data, ndigits)
    return data

举个例子：

from pprint import pprint
import math

DATA = {
    'test': [1.23456, 'whatever', (2.34567, math.pi)],
    'another_key': [1.0, 2.0, 0.0, {math.e, math.inf}],
    'last_key': [0.123456789, 9.87654321]
}

pprint(recursive_round(DATA, 3))
# {'another_key': [1, 2, 0, {2.718, inf}],
#  'last_key': [0.123, 9.877],
#  'test': [1.235, 'whatever', (2.346, 3.142)]}

这个也可以用来做单元测试，配合 assertEqual 使用：

import unittest
import math


def recursive_round(data, ndigits=5):
    if isinstance(data, dict):
        return {k: recursive_round(v, ndigits) for k, v in data.items()}
    if isinstance(data, list):
        return [recursive_round(v, ndigits) for v in data]
    if isinstance(data, tuple):
        return tuple(recursive_round(v, ndigits) for v in data)
    if isinstance(data, set):
        return {recursive_round(v, ndigits) for v in data}
    if isinstance(data, float):
        if data.is_integer():
            return int(data)
        return round(data, ndigits)
    return data


DATA = {
    'test': [1.23456, 'whatever', (2.34567, math.pi)],
    'another_key': [1.0, 2.0, 0.0, {math.e, math.inf}],
    'last_key': [0.123456789, 9.87654321]
}


class TestClass(unittest.TestCase):

    def test_method(self):
        expected = {'another_key': [1, 2, 0, {2.718, math.inf}],
                    'last_key': [0.123, 9.877],
                    'test': [1.235, 'whatever', (2.346, 3.142)]}
        self.assertEqual(expected, recursive_round(DATA, 3))


if __name__ == '__main__':
    unittest.main()

Answer 2

我没能让Akavall的函数运行，所以我自己写了一个。虽然有点简单，但对我来说足够用了。下面是用pytest测试这个函数是否正常工作的代码。

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

使用pytest来验证函数的输出。

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)

Answer 3

>>> 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)

我知道你不会单纯为了这个去导入pandas库，但如果你正好在使用pandas的话，你可以把字典转换成序列，然后使用

pandas.testing里的assert_series_equal函数。这个函数默认情况下是check_exact=False。

Answer 4

Pytest的"approx"功能可以解决这个问题

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

输出结果是: True

Answer 5

@dano的评论回答了我的问题：

我从dano提供的链接复制了一个函数。

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()

结果：

Ran 2 tests in 0.000s

OK