我发现下面的比较很有帮助:

str(f1) == str(f2)

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

我发现下面的比较很有帮助:

str(f1) == str(f2)

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

不带atol/rtol与给定小数进行比较:

def almost_equal(a, b, decimal=6):
    return '{0:.{1}f}'.format(a, decimal) == '{0:.{1}f}'.format(b, decimal)

print(almost_equal(0.0, 0.0001, decimal=5)) # False
print(almost_equal(0.0, 0.0001, decimal=4)) # True 

我同意Gareth的答案可能是最合适的轻量级函数/解决方案。

但我认为，如果您正在使用NumPy或正在考虑使用NumPy，那么有一个打包的函数用于此，这将是有帮助的。

numpy.isclose(a, b, rtol=1e-05, atol=1e-08, equal_nan=False)

不过有一点免责声明:根据您的平台，安装NumPy可能是一种非常重要的体验。