# Next float?

Discussion in 'Python' started by Steven D'Aprano, Nov 22, 2007.

1. ### Steven D'ApranoGuest

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).

Note to maths pedants: I am aware that there is no "next real number",
but floats are not reals.

--
Steven

Steven D'Aprano, Nov 22, 2007

2. ### Alex HolknerGuest

On Nov 22, 2007 2:04 PM, 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.
>
> 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

Alex Holkner, Nov 22, 2007

3. ### Mark TGuest

"Steven D'Aprano" <> wrote in message
news...
> 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).
>
> Note to maths pedants: I am aware that there is no "next real number",
> but floats are not reals.
>
>
> --
> Steven

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

Mark T, Nov 22, 2007
4. ### Paul RubinGuest

Steven D'Aprano <> writes:
> 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

Paul Rubin, Nov 22, 2007
5. ### Robert KernGuest

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.

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
an underlying truth."
-- Umberto Eco

Robert Kern, Nov 22, 2007
6. ### Robert KernGuest

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

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
an underlying truth."
-- Umberto Eco

Robert Kern, Nov 22, 2007
7. ### Steven D'ApranoGuest

On Thu, 22 Nov 2007 00:44:34 -0600, Robert Kern wrote:

> 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.

Steven D'Aprano, Nov 22, 2007
8. ### Hendrik van RooyenGuest

"Steven D'Aprano" <steve..IS.cybersource.com.au> wrote:

> Damn, I don't remember writing that!

It is caused by drinking too much Alzheimer's Light.

: - )

- Hendrik

Hendrik van Rooyen, Nov 22, 2007
9. ### Fredrik JohanssonGuest

On Nov 22, 2007 4:04 AM, 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.
>
> 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

You could use the library functions for floating-point math in mpmath
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

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

Fredrik Johansson, Nov 22, 2007
10. ### Mark DickinsonGuest

On Nov 21, 11:48 pm, "Mark T" <> wrote:
> 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, Nov 22, 2007
11. ### Mark DickinsonGuest

On Nov 22, 5:41 pm, Mark Dickinson <> wrote:
> On Nov 21, 11:48 pm, "Mark T" <> wrote:
>
>
>
> > 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)

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

Mark Dickinson, Nov 22, 2007