Python 有与 java.lang.Math.nextUp 等效的东西吗？ [复制]

我有一个Pythonfloat，我想要floats 为 1ULP http://en.wikipedia.org/wiki/Unit_in_the_last_place更大和更小。

在Java中，我会这样做Math.nextUp(x) http://download.oracle.com/javase/6/docs/api/java/lang/Math.html#nextUp%28double%29 and Math.nextAfter(x, Double.NEGATIVE_INFINITY) http://download.oracle.com/javase/6/docs/api/java/lang/Math.html#nextAfter%28double,%20double%29.

有没有办法在 Python 中做到这一点？我考虑过自己实现它math.frexp and math.ldexp http://docs.python.org/library/math.html但据我所知Python没有指定浮点类型的大小。

Update:在Python 3.9+中有math.nextafter() https://docs.python.org/3/library/math.html#math.nextafter:

>>> import math
>>> x = 4
>>> math.nextafter(x, math.inf)
4.000000000000001

^{Old answer:}

你可以看看如何Decimal.next_plus()/Decimal.next_minus() http://hg.python.org/cpython/file/fecf9e6d7630/Lib/decimal.py#l3448已实施：

>>> from decimal import Decimal as D
>>> d = D.from_float(123456.78901234567890)
>>> d
Decimal('123456.789012345674564130604267120361328125')
>>> d.next_plus()
Decimal('123456.7890123456745641306043')
>>> d.next_minus()
Decimal('123456.7890123456745641306042')
>>> d.next_toward(D('-inf'))
Decimal('123456.7890123456745641306042')

确保十进制上下文 http://docs.python.org/library/decimal.html#decimal.Context有您需要的值：

>>> from decimal import getcontext
>>> getcontext()
Context(prec=28, rounding=ROUND_HALF_EVEN, Emin=-999999999, Emax=999999999,
capitals=1, flags=[], traps=[InvalidOperation, DivisionByZero, Overflow])

备择方案：

• Call C99 nextafter() https://stackoverflow.com/a/6163157/4279 using ctypes:

    >>> import ctypes
    >>> nextafter = ctypes.CDLL(None).nextafter
    >>> nextafter.argtypes = ctypes.c_double, ctypes.c_double
    >>> nextafter.restype = ctypes.c_double
    >>> nextafter(4, float('+inf'))
    4.000000000000001
    >>> _.as_integer_ratio()
    (4503599627370497, 1125899906842624)
  

  Using numpy:

    >>> import numpy
    >>> numpy.nextafter(4, float('+inf'))
    4.0000000000000009
    >>> _.as_integer_ratio()
    (4503599627370497, 1125899906842624)
  

  尽管不同repr()， 结果是一样的。

• 如果我们忽略边缘情况，那么一个简单的 frexp/ldexp 解决方案来自@S.Lott 回答 https://stackoverflow.com/a/6064066/4279 works:

    >>> import math, sys
    >>> m, e = math.frexp(4.0)
    >>> math.ldexp(2 * m + sys.float_info.epsilon, e - 1)
    4.000000000000001
    >>> _.as_integer_ratio()
    (4503599627370497, 1125899906842624)
  

• 纯Pythonnext_after(x, y)由 @Mark Dickinson 实施 https://stackoverflow.com/a/10426033/4279考虑到边缘情况。在本例中结果是相同的。