Next float?
Steven D'Aprano
given one? By "next float", I mean given a float x, I want the smallest
float larger than x.
Bonus points if I can go in either direction (i.e. the "previous float"
as well as the next).
Note to maths pedants: I am aware that there is no "next real number",
but floats are not reals.
--
Steven
Alex Holkner
<ste...@remove.this.cybersource.com.au> wrote:
> Is there a simple, elegant way in Python to get the next float from a
> given one? By "next float", I mean given a float x, I want the smallest
> float larger than x.
>
> Bonus points if I can go in either direction (i.e. the "previous float"
> as well as the next).
Not so elegant, but you could use ctypes to manipulate the bits
(assumes machine uses IEEE 754 doubles for Python floats, I'm not sure
if that's guaranteed on esoteric platforms):
import ctypes
def inc_float(f):
# Get an int64 pointer to the float data
fv = ctypes.c_double(f)
pfv = ctypes.pointer(fv)
piv = ctypes.cast(pfv, ctypes.POINTER(ctypes.c_uint64))
# Check for NaN or infinity, return unchanged
v = piv.contents.value
if not ~(v & (11 << 52)): # exponent is all 1's
return f
if v == 1 << 63: # -0, treat as +0
v = 1
elif v & (1 << 63): # negative
v -= 1
else: # positive or +0
v += 1
# Set int pointer and return changed float
piv.contents.value = v
return fv.value
def dec_float(f):
# Get an int64 pointer to the float data
fv = ctypes.c_double(f)
pfv = ctypes.pointer(fv)
piv = ctypes.cast(pfv, ctypes.POINTER(ctypes.c_uint64))
# Check for NaN or infinity, return unchanged
v = piv.contents.value
if not ~(v & (11 << 52)): # exponent is all 1's
return f
if v == 0: # +0, treat as -0
v = (1 << 63) | 1
elif v & (1 << 63): # negative
v += 1
else: # positive
v -= 1
# Set int pointer and return changed float
piv.contents.value = v
return fv.value
Mark T
"Steven D'Aprano" <ste...@REMOVE.THIS.cybersource.com.au> wrote in message
news:pan.2007.11...@REMOVE.THIS.cybersource.com.au...
Here's some functions to get the binary representation of a float. Then
just manipulate the bits (an exercise for the reader):
import struct
def f2b(f):
return struct.unpack('I',struct.pack('f',f))[0]
def b2f(b):
return struct.unpack('f',struct.pack('I',b))[0]
>>> f2b(1.0)
1065353216
>>> hex(f2b(1.0))
'0x3f800000'
>>> b2f(0x3f800000)
1.0
>>> b2f(0x3f800001)
1.0000001192092896
>>> b2f(0x3f7fffff)
0.99999994039535522
-Mark
Paul Rubin
> Is there a simple, elegant way in Python to get the next float from a
> given one? By "next float", I mean given a float x, I want the smallest
> float larger than x.
I think you have to do it by bit twiddling. But something like bisection
search could come pretty close, for well-behaved values of x:
def nextfloat(x):
dx = (x, x/2.0)
while x+dx[1] != x:
dx = (dx[1], dx[1]/2.0)
return dx[0]+x
Robert Kern
> Is there a simple, elegant way in Python to get the next float from a
> given one? By "next float", I mean given a float x, I want the smallest
> float larger than x.
Courtesy of Tim Peters:
http://mail.python.org/pipermail/python-list/2001-August/099152.html
> Bonus points if I can go in either direction (i.e. the "previous float"
> as well as the next).
Left as an exercise for the reader. :-)
--
Robert Kern
"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco
Robert Kern
> Is there a simple, elegant way in Python to get the next float from a
> given one? By "next float", I mean given a float x, I want the smallest
> float larger than x.
Heh:
http://mail.python.org/pipermail/python-list/2005-December/357771.html
Steven D'Aprano
> Steven D'Aprano wrote:
>> Is there a simple, elegant way in Python to get the next float from a
>> given one? By "next float", I mean given a float x, I want the smallest
>> float larger than x.
>
> Heh:
>
> http://mail.python.org/pipermail/python-list/2005-December/357771.html
Damn, I don't remember writing that!
:-/
--
Steven.
Fredrik Johansson
You could use the library functions for floating-point math in mpmath
(http://code.google.com/p/mpmath/), which support directed rounding.
Load a floating-point number, add a tiny number and round in the
wanted direction, then convert back to a Python float:
>>> from mpmath.lib import *
>>> eps = (1, -2000, 1) # 2**-2000, smaller than any finite
positive IEEE 754 double
>>> a = from_float(1.0, 53, ROUND_HALF_EVEN) # note: exact
>>> to_float(fadd(a, eps, 53, ROUND_UP))
1.0000000000000002
>>> to_float(fsub(a, eps, 53, ROUND_DOWN))
0.99999999999999989
This currently probably doesn't work if the numbers are subnormal, however.
Fredrik
Mark Dickinson
> Here's some functions to get the binary representation of a float. Then
> just manipulate the bits (an exercise for the reader):
>
> import struct
>
> def f2b(f):
> return struct.unpack('I',struct.pack('f',f))[0]
>
> def b2f(b):
> return struct.unpack('f',struct.pack('I',b))[0]
>
> >>> f2b(1.0)
> 1065353216
> >>> hex(f2b(1.0))
> '0x3f800000'
> >>> b2f(0x3f800000)
> 1.0
> >>> b2f(0x3f800001)
> 1.0000001192092896
> >>> b2f(0x3f7fffff)
>
> 0.99999994039535522
And it's worth noting that thanks to the way the floating-point format
is designed, the 'bit-twiddling' is actually just a matter of adding
or subtracting one from the integer representation above. This works
all the way from zero, through subnormals, up to and including
infinities.
Mark (but not the same Mark)
Mark Dickinson
And also worth noting that the 'f's above should probably be 'd's.
For example, the following works on my machine for positive floats.
Fixing this for negative floats is left as a not-very-hard exercise...
>>> from struct import pack, unpack
>>> def next_float(x): return unpack('d', pack('q', unpack('q', pack('d', x))[0]+1))[0]
...
>>> next_float(1.0)
1.0000000000000002
Mark
Hendrik van Rooyen
> Damn, I don't remember writing that!
It is caused by drinking too much Alzheimer's Light.
: - )
- Hendrik