使用Python的decimal模块，该模块提供decimal类。

评论如下:

值得注意的是，如果你 做繁重的数学工作，而你没有 绝对需要精准的 小数，这很麻烦 下来。浮点数要快得多 处理，但不精确。小数是 非常精确但很慢。

对于一些可以影响源数表示的情况，可以使用整数分子和整数分母将它们表示为分数而不是浮点数。这样你就可以进行准确的比较。

详见分数模块中的分数。

如果你想在测试/TDD环境中使用它，我认为这是一种标准方法:

from nose.tools import assert_almost_equals

assert_almost_equals(x, y, places=7) # The default is 7

如果你想在测试或TDD上下文中使用pytest包，下面是如何做到的:

import pytest


PRECISION = 1e-3

def assert_almost_equal():
    obtained_value = 99.99
    expected_value = 100.00
    assert obtained_value == pytest.approx(expected_value, PRECISION)

I'm not aware of anything in the Python standard library (or elsewhere) that implements Dawson's AlmostEqual2sComplement function. If that's the sort of behaviour you want, you'll have to implement it yourself. (In which case, rather than using Dawson's clever bitwise hacks you'd probably do better to use more conventional tests of the form if abs(a-b) <= eps1*(abs(a)+abs(b)) + eps2 or similar. To get Dawson-like behaviour you might say something like if abs(a-b) <= eps*max(EPS,abs(a),abs(b)) for some small fixed EPS; this isn't exactly the same as Dawson, but it's similar in spirit.

math.isclose()已为此添加到Python 3.5(源代码)。这里是它到Python 2的一个端口。它与Mark Ransom的单行程序的不同之处在于它可以正确地处理“inf”和“-inf”。

def isclose(a, b, rel_tol=1e-09, abs_tol=0.0):
    '''
    Python 2 implementation of Python 3.5 math.isclose()
    https://github.com/python/cpython/blob/v3.5.10/Modules/mathmodule.c#L1993
    '''
    # sanity check on the inputs
    if rel_tol < 0 or abs_tol < 0:
        raise ValueError("tolerances must be non-negative")

    # short circuit exact equality -- needed to catch two infinities of
    # the same sign. And perhaps speeds things up a bit sometimes.
    if a == b:
        return True

    # This catches the case of two infinities of opposite sign, or
    # one infinity and one finite number. Two infinities of opposite
    # sign would otherwise have an infinite relative tolerance.
    # Two infinities of the same sign are caught by the equality check
    # above.
    if math.isinf(a) or math.isinf(b):
        return False

    # now do the regular computation
    # this is essentially the "weak" test from the Boost library
    diff = math.fabs(b - a)
    result = (((diff <= math.fabs(rel_tol * b)) or
               (diff <= math.fabs(rel_tol * a))) or
              (diff <= abs_tol))
    return result