Increment a Python floating point value by the smallest possible amount

Question

How can I increment a floating point value in python by the smallest possible amount?

---

Background: I'm using floating point values as dictionary keys.

Occasionally, very occasionally (and perhaps never, but not certainly never), there will be collisions. I would like to resolve these by incrementing the floating point value by as small an amount as possible. How can I do this?

In C, I would twiddle the bits of the mantissa to achieve this, but I assume that isn't possible in Python.

Answer 1

Since Python 3.9 there is math.nextafter in the stdlib. Read on for alternatives in older Python versions.

Increment a python floating point value by the smallest possible amount

The nextafter(x,y) functions return the next discretely different representable floating-point value following x in the direction of y. The nextafter() functions are guaranteed to work on the platform or to return a sensible value to indicate that the next value is not possible.

The nextafter() functions are part of POSIX and ISO C99 standards and is _nextafter() in Visual C. C99 compliant standard math libraries, Visual C, C++, Boost and Java all implement the IEEE recommended nextafter() functions or methods. (I do not honestly know if .NET has nextafter(). Microsoft does not care much about C99 or POSIX.)

None of the bit twiddling functions here fully or correctly deal with the edge cases, such as values going though 0.0, negative 0.0, subnormals, infinities, negative values, over or underflows, etc. Here is a reference implementation of nextafter() in C to give an idea of how to do the correct bit twiddling if that is your direction.

There are two solid work arounds to get nextafter() or other excluded POSIX math functions in Python < 3.9:

Use Numpy:

>>> import numpy
>>> numpy.nextafter(0,1)
4.9406564584124654e-324
>>> numpy.nextafter(.1, 1)
0.10000000000000002
>>> numpy.nextafter(1e6, -1)
999999.99999999988
>>> numpy.nextafter(-.1, 1)
-0.099999999999999992

Link directly to the system math DLL:

import ctypes
import sys
from sys import platform as _platform

if _platform == "linux" or _platform == "linux2":
    _libm = ctypes.cdll.LoadLibrary('libm.so.6')
    _funcname = 'nextafter'
elif _platform == "darwin":
    _libm = ctypes.cdll.LoadLibrary('libSystem.dylib')
    _funcname = 'nextafter'
elif _platform == "win32":
    _libm = ctypes.cdll.LoadLibrary('msvcrt.dll')
    _funcname = '_nextafter'
else:
    # these are the ones I have access to...
    # fill in library and function name for your system math dll
    print("Platform", repr(_platform), "is not supported")
    sys.exit(0)

_nextafter = getattr(_libm, _funcname)
_nextafter.restype = ctypes.c_double
_nextafter.argtypes = [ctypes.c_double, ctypes.c_double]

def nextafter(x, y):
    "Returns the next floating-point number after x in the direction of y."
    return _nextafter(x, y)

assert nextafter(0, 1) - nextafter(0, 1) == 0
assert 0.0 + nextafter(0, 1) > 0.0

And if you really really want a pure Python solution:

# handles edge cases correctly on MY computer 
# not extensively QA'd...
import math
# 'double' means IEEE 754 double precision -- c 'double'
epsilon  = math.ldexp(1.0, -53) # smallest double that 0.5+epsilon != 0.5
maxDouble = float(2**1024 - 2**971)  # From the IEEE 754 standard
minDouble  = math.ldexp(1.0, -1022) # min positive normalized double
smallEpsilon  = math.ldexp(1.0, -1074) # smallest increment for doubles < minFloat
infinity = math.ldexp(1.0, 1023) * 2

def nextafter(x,y):    
    """returns the next IEEE double after x in the direction of y if possible"""
    if y==x:
       return y         #if x==y, no increment

    # handle NaN
    if x!=x or y!=y:
        return x + y       

    if x >= infinity:
        return infinity

    if x <= -infinity:
        return -infinity

    if -minDouble < x < minDouble:
        if y > x:
            return x + smallEpsilon
        else:
            return x - smallEpsilon  

    m, e = math.frexp(x)        
    if y > x:
        m += epsilon
    else:
        m -= epsilon

    return math.ldexp(m,e)

Or, use Mark Dickinson's excellent solution

Obviously the Numpy solution is the easiest.