Related
Say I have an arbitrary float x = 123.123456 and want to remove the last n decimal digits from the float i.e. n = 1 then x = 123.12345, n = 2 then x = 123.1234 and so on. How can this be achieved in python?
This will do the trick you are asking, but be mindful of the issues with floating point numbers.
# let's cut 2 digits
n = 2
# naively we can do this
f = 123.123456
short_f = float(str(f)[:-n])
# but watch out for floating point error
f = 1.2 - 1.0 # f should be 0.2 but is actually 0.19999999999999996
short_f = float(str(f)[:-n]) # so this gives 0.199999999999999
This sounds like an XY problem, maybe you are looking for round or string formatting.
I have numbers that range from very small to very large, and I'd like to format them using "engineering notation" with a magnitude and a suffix:
n.nnn S
where 1.0 <= n.nnn < 1000., and S is a metric (SI) prefix. So:
1234.5e+13 => 12.35P
12345678 => 12.35M
1234 => 1.234K
1.234 => 1.234
0.1234 => 123.4m
1234.5e-16 => 1.235f
etc. How can I do that, e.g. using Python?
(Posted here in Q&A style because I keep re-inventing this code, and others might find it helpful. Feel free to tweak it if you see improvements...)
Here is one implementation that lets you choose a long suffix (e.g. "peta") or a short suffix (e.g. "P"), and also lets you choose how many total digits are displayed (i.e. the precision):
def si_classifier(val):
suffixes = {
24:{'long_suffix':'yotta', 'short_suffix':'Y', 'scalar':10**24},
21:{'long_suffix':'zetta', 'short_suffix':'Z', 'scalar':10**21},
18:{'long_suffix':'exa', 'short_suffix':'E', 'scalar':10**18},
15:{'long_suffix':'peta', 'short_suffix':'P', 'scalar':10**15},
12:{'long_suffix':'tera', 'short_suffix':'T', 'scalar':10**12},
9:{'long_suffix':'giga', 'short_suffix':'G', 'scalar':10**9},
6:{'long_suffix':'mega', 'short_suffix':'M', 'scalar':10**6},
3:{'long_suffix':'kilo', 'short_suffix':'k', 'scalar':10**3},
0:{'long_suffix':'', 'short_suffix':'', 'scalar':10**0},
-3:{'long_suffix':'milli', 'short_suffix':'m', 'scalar':10**-3},
-6:{'long_suffix':'micro', 'short_suffix':'µ', 'scalar':10**-6},
-9:{'long_suffix':'nano', 'short_suffix':'n', 'scalar':10**-9},
-12:{'long_suffix':'pico', 'short_suffix':'p', 'scalar':10**-12},
-15:{'long_suffix':'femto', 'short_suffix':'f', 'scalar':10**-15},
-18:{'long_suffix':'atto', 'short_suffix':'a', 'scalar':10**-18},
-21:{'long_suffix':'zepto', 'short_suffix':'z', 'scalar':10**-21},
-24:{'long_suffix':'yocto', 'short_suffix':'y', 'scalar':10**-24}
}
exponent = int(math.floor(math.log10(abs(val))/3.0)*3)
return suffixes.get(exponent, None)
def si_formatter(value):
'''
Return a triple of scaled value, short suffix, long suffix, or None if
the value cannot be classified.
'''
classifier = si_classifier(value)
if classifier == None:
# Don't know how to classify this value
return None
scaled = value / classifier['scalar']
return (scaled, classifier['short_suffix'], classifier['long_suffix'])
def si_format(value, precision=4, long_form=False, separator=''):
'''
"SI prefix" formatted string: return a string with the given precision
and an appropriate order-of-3-magnitudes suffix, e.g.:
si_format(1001.0) => '1.00K'
si_format(0.00000000123, long_form=True, separator=' ') => '1.230 nano'
'''
scaled, short_suffix, long_suffix = si_formatter(value)
if scaled == None:
# Don't know how to format this value
return value
suffix = long_suffix if long_form else short_suffix
if abs(scaled) < 10:
precision = precision - 1
elif abs(scaled) < 100:
precision = precision - 2
else:
precision = precision - 3
return '{scaled:.{precision}f}{separator}{suffix}'.format(
scaled=scaled, precision=precision, separator=separator, suffix=suffix)
You can use Prefixed which has a float type with additional formatting options.
>>> from prefixed import Float
>>> f'{Float(1234.5e+13):.2h}'
'12.35P'
>>> f'{Float(12345678):.2h}'
'12.35M'
>>> f'{Float(1234):.2h}'
'1.23k'
>>> f'{Float(1.234):.2h}'
'1.23'
>>> f'{Float(0.1234):.2h}'
'123.40m'
>>> f'{Float(1234.5e-16):.2h}'
'123.45f'
I am having problem rounding floating value to a decimal number greater than 1.
def float_generator(size=6, chars = 6):
return random.uniform(size,chars)
and i tried
target.write(str(float_generator(7,3)))
the above returns random float to 1 decimal num like 1.2222 , 1.333333
I want to generate 7 decimal and 3 points as illustrated in the 7,3
i tried rounding off below. It round up the points instead .
target.write(str(format(float_generator(7,3),'.3f')))
Please how do i achieve 12345432.123 instead of 1.222343223 ? Any help would be appreciated.
You could use decimal and random to return a Decimal object if you want fine-tuned control of decimal places:
import random, decimal
digits = list('0123456789')
def float_generator(size = 6, chars = 6):
base_str = ''.join(random.choice(digits) for i in range(size))
dec_str = ''.join(random.choice(digits) for i in range(chars))
return decimal.Decimal(base_str + '.' + dec_str)
>>> float_generator(7,3)
Decimal('4768087.977')
Decimal objects can be fed to float:
>>> float(decimal.Decimal('4768087.977'))
4768087.977
If you always want a float, you can replace the last line of the function by:
return float(decimal.Decimal(base_str + '.' + dec_str))
here is a solution for you:
import random
def float_generator(size = 6, chars = 6):
return round(random.uniform(10**(size-1),10**(size)), chars)
call it:
float_generator(7,3)
output example:
2296871.988
or better:
float_generator = lambda size, chars: round(random.uniform(10**(size-1),10**(size)), chars)
call it the same way.
Here is the example which is bothering me:
>>> x = decimal.Decimal('0.0001')
>>> print x.normalize()
>>> print x.normalize().to_eng_string()
0.0001
0.0001
Is there a way to have engineering notation for representing mili (10e-3) and micro (10e-6)?
Here's a function that does things explicitly, and also has support for using SI suffixes for the exponent:
def eng_string( x, format='%s', si=False):
'''
Returns float/int value <x> formatted in a simplified engineering format -
using an exponent that is a multiple of 3.
format: printf-style string used to format the value before the exponent.
si: if true, use SI suffix for exponent, e.g. k instead of e3, n instead of
e-9 etc.
E.g. with format='%.2f':
1.23e-08 => 12.30e-9
123 => 123.00
1230.0 => 1.23e3
-1230000.0 => -1.23e6
and with si=True:
1230.0 => 1.23k
-1230000.0 => -1.23M
'''
sign = ''
if x < 0:
x = -x
sign = '-'
exp = int( math.floor( math.log10( x)))
exp3 = exp - ( exp % 3)
x3 = x / ( 10 ** exp3)
if si and exp3 >= -24 and exp3 <= 24 and exp3 != 0:
exp3_text = 'yzafpnum kMGTPEZY'[ ( exp3 - (-24)) / 3]
elif exp3 == 0:
exp3_text = ''
else:
exp3_text = 'e%s' % exp3
return ( '%s'+format+'%s') % ( sign, x3, exp3_text)
EDIT:
Matplotlib implemented the engineering formatter, so one option is to directly use Matplotlibs formatter, e.g.:
import matplotlib as mpl
formatter = mpl.ticker.EngFormatter()
formatter(10000)
result: '10 k'
Original answer:
Based on Julian Smith's excellent answer (and this answer), I changed the function to improve on the following points:
Python3 compatible (integer division)
Compatible for 0 input
Rounding to significant number of digits, by default 3, no trailing zeros printed
so here's the updated function:
import math
def eng_string( x, sig_figs=3, si=True):
"""
Returns float/int value <x> formatted in a simplified engineering format -
using an exponent that is a multiple of 3.
sig_figs: number of significant figures
si: if true, use SI suffix for exponent, e.g. k instead of e3, n instead of
e-9 etc.
"""
x = float(x)
sign = ''
if x < 0:
x = -x
sign = '-'
if x == 0:
exp = 0
exp3 = 0
x3 = 0
else:
exp = int(math.floor(math.log10( x )))
exp3 = exp - ( exp % 3)
x3 = x / ( 10 ** exp3)
x3 = round( x3, -int( math.floor(math.log10( x3 )) - (sig_figs-1)) )
if x3 == int(x3): # prevent from displaying .0
x3 = int(x3)
if si and exp3 >= -24 and exp3 <= 24 and exp3 != 0:
exp3_text = 'yzafpnum kMGTPEZY'[ exp3 // 3 + 8]
elif exp3 == 0:
exp3_text = ''
else:
exp3_text = 'e%s' % exp3
return ( '%s%s%s') % ( sign, x3, exp3_text)
The decimal module is following the Decimal Arithmetic Specification, which states:
This is outdated - see below
to-scientific-string – conversion to numeric string
[...]
The coefficient is first converted to a string in base ten using the characters 0 through 9 with no leading zeros (except if its value is zero, in which case a single 0 character is used).
Next, the adjusted exponent is calculated; this is the exponent, plus the number of characters in the converted coefficient, less one. That is, exponent+(clength-1), where clength is the length of the coefficient in decimal digits.
If the exponent is less than or equal to zero and the adjusted exponent is greater than or equal to -6, the number will be converted
to a character form without using exponential notation.
[...]
to-engineering-string – conversion to numeric string
This operation converts a number to a string, using engineering
notation if an exponent is needed.
The conversion exactly follows the rules for conversion to scientific
numeric string except in the case of finite numbers where exponential
notation is used. In this case, the converted exponent is adjusted to be a multiple of three (engineering notation) by positioning the decimal point with one, two, or three characters preceding it (that is, the part before the decimal point will range from 1 through 999).
This may require the addition of either one or two trailing zeros.
If after the adjustment the decimal point would not be followed by a digit then it is not added. If the final exponent is zero then no indicator letter and exponent is suffixed.
Examples:
For each abstract representation [sign, coefficient, exponent] on the left, the resulting string is shown on the right.
Representation
String
[0,123,1]
"1.23E+3"
[0,123,3]
"123E+3"
[0,123,-10]
"12.3E-9"
[1,123,-12]
"-123E-12"
[0,7,-7]
"700E-9"
[0,7,1]
"70"
Or, in other words:
>>> for n in (10 ** e for e in range(-1, -8, -1)):
... d = Decimal(str(n))
... print d.to_eng_string()
...
0.1
0.01
0.001
0.0001
0.00001
0.000001
100E-9
I realize that this is an old thread, but it does come near the top of a search for python engineering notation and it seems prudent to have this information located here.
I am an engineer who likes the "engineering 101" engineering units. I don't even like designations such as 0.1uF, I want that to read 100nF. I played with the Decimal class and didn't really like its behavior over the range of possible values, so I rolled a package called engineering_notation that is pip-installable.
pip install engineering_notation
From within Python:
>>> from engineering_notation import EngNumber
>>> EngNumber('1000000')
1M
>>> EngNumber(1000000)
1M
>>> EngNumber(1000000.0)
1M
>>> EngNumber('0.1u')
100n
>>> EngNumber('1000m')
1
This package also supports comparisons and other simple numerical operations.
https://github.com/slightlynybbled/engineering_notation
The «full» quote shows what is wrong!
The decimal module is indeed following the proprietary (IBM) Decimal Arithmetic Specification.
Quoting this IBM specification in its entirety clearly shows what is wrong with decimal.to_eng_string() (emphasis added):
to-engineering-string – conversion to numeric string
This operation converts a number to a string, using engineering
notation if an exponent is needed.
The conversion exactly follows the rules for conversion to scientific
numeric string except in the case of finite numbers where exponential
notation is used. In this case, the converted exponent is adjusted to be a multiple of three (engineering notation) by positioning the decimal point with one, two, or three characters preceding it (that is, the part before the decimal point will range from 1 through 999). This may require the addition of either one or two trailing zeros.
If after the adjustment the decimal point would not be followed by a digit then it is not added. If the final exponent is zero then no indicator letter and exponent is suffixed.
This proprietary IBM specification actually admits to not applying the engineering notation for numbers with an infinite decimal representation, for which ordinary scientific notation is used instead! This is obviously incorrect behaviour for which a Python bug report was opened.
Solution
from math import floor, log10
def powerise10(x):
""" Returns x as a*10**b with 0 <= a < 10
"""
if x == 0: return 0,0
Neg = x < 0
if Neg: x = -x
a = 1.0 * x / 10**(floor(log10(x)))
b = int(floor(log10(x)))
if Neg: a = -a
return a,b
def eng(x):
"""Return a string representing x in an engineer friendly notation"""
a,b = powerise10(x)
if -3 < b < 3: return "%.4g" % x
a = a * 10**(b % 3)
b = b - b % 3
return "%.4gE%s" % (a,b)
Source: https://code.activestate.com/recipes/578238-engineering-notation/
Test result
>>> eng(0.0001)
100E-6
Like the answers above, but a bit more compact:
from math import log10, floor
def eng_format(x,precision=3):
"""Returns string in engineering format, i.e. 100.1e-3"""
x = float(x) # inplace copy
if x == 0:
a,b = 0,0
else:
sgn = 1.0 if x > 0 else -1.0
x = abs(x)
a = sgn * x / 10**(floor(log10(x)))
b = int(floor(log10(x)))
if -3 < b < 3:
return ("%." + str(precision) + "g") % x
else:
a = a * 10**(b % 3)
b = b - b % 3
return ("%." + str(precision) + "gE%s") % (a,b)
Trial:
In [10]: eng_format(-1.2345e-4,precision=5)
Out[10]: '-123.45E-6'
I am wondering how to convert the result returned by float.hex() to binary, for example, from 0x1.a000000000000p+2 to 110.1.
Can anyone please help? Thanks.
def float_to_binary(num):
exponent=0
shifted_num=num
while shifted_num != int(shifted_num):
shifted_num*=2
exponent+=1
if exponent==0:
return '{0:0b}'.format(int(shifted_num))
binary='{0:0{1}b}'.format(int(shifted_num),exponent+1)
integer_part=binary[:-exponent]
fractional_part=binary[-exponent:].rstrip('0')
return '{0}.{1}'.format(integer_part,fractional_part)
def floathex_to_binary(floathex):
num = float.fromhex(floathex)
return float_to_binary(num)
print(floathex_to_binary('0x1.a000000000000p+2'))
# 110.1
print(floathex_to_binary('0x1.b5c2000000000p+1'))
# 11.01101011100001
Explanation:
float.fromhex returns a float num. We'd like its binary representation.
{0:b}.format(...) returns binary representations of integers, but not floats.
But if we multiply the float by enough powers of 2, that is, shift the binary representation to the left enough places, we end up with an integer, shifted_num.
Once we have that integer, we are home free, because now we can use {0:b}.format(...).
We can re-insert the decimal point (err, binary point?) by using a bit of string slicing based on the number of places we had shifted to the left (exponent).
Technical point: The number of digits in the binary representation of shifted_num may be smaller than exponent. In that case, we need to pad the binary representation with more 0's on the left, so binary slicing with binary[:-exponent] won't be empty. We manage that with '{0:0{1}b}'.format(...). The 0{1} in the format string sets the width of the formated string to {1}, padded on the left with zeros. (The {1} gets replaced by the number exponent.)
Note that the binary form of 0x1.a000000000000p+2 isn't 101.1 (or more exactly 0b101.1 )
but 0b110.1 (in my Python 2.7, binary numbers are displayed like that)
.
First, a useful method of float instances float.hex() and its inverse function, a float class method float.fromhex()
fh = 12.34.hex()
print fh
print float.fromhex(fh)
result
0x1.8ae147ae147aep+3 # hexadecimal representation of a float
12.34
"Note that float.hex() is an instance method, while float.fromhex() is a class method."
http://docs.python.org/library/stdtypes.html#float.fromhex
.
Secondly, I didn't find a Python's function to transform an hexadecimal representation of a float into a binary representation of this float, that is to say with a dot ( nor than one to transform directly a decimal representation of a float into a binary one).
So I created a function for that purpose.
Before hand, this function transforms the hexadecimal representation into a decimal representation (string) of the input float.
Then there are two problems:
how to transform the part before the dot ?
This part being an integer, it's easy to use bin()
how to transform the part after the dot ??
The problem of this transformation has been asked several times on SO, but I didn't understand the solutions, so I wrote my own.
Then, here's the function you wish, Qiang Li:
def hexf2binf(x):
'''Transforms an hexadecimal float with a dot into a binary float with a dot'''
a,_,p = str(float.fromhex(x)).partition('.')
# the following part transforms the part after the dot into a binary after the dot
tinies = [ Decimal(1) / Decimal(2**i) for i in xrange(1,400)]
bits = []
pdec = Decimal('.'+p)
for tin in tinies:
if pdec-tin==0:
bits.append('1')
break
elif pdec-tin>0:
bits.append('1')
pdec -= tin
else:
bits.append('0')
pbin = ''.join(bits) # it's the binary after the dot
# the integer before the dot is easily transformed into a binary
return '.'.join((bin(int(a)),pbin))
.
In order to perform verification, I wrote a function to transform the part of a binary float after a dot into its decimal representation:
from decimal import Decimal, getcontext()
getcontext().prec = 500
# precision == 500 , to be large !
tinies = [ Decimal(1) / Decimal(2**i) for i in xrange(1,400)]
com = dict((i,tin) for i,tin in enumerate(tinies,1))
def afterdotbinary2float(sbin, com = com):
'''Transforms a binary lying after a dot into a float after a dot'''
if sbin.startswith('0b.') or sbin.startswith('.'):
sbin = sbin.split('.')[1]
if all(c in '01' for c in sbin):
return sum(int(c)*com[i] for i,c in enumerate(sbin,1))
else:
return None
.
.
Finally, applying these functions:
from decimal import Decimal
getcontext().prec = 500
# precision == 500 , to be large !
tinies = [ Decimal(1) / Decimal(2**i) for i in xrange(1,400)]
com = dict((i,tin) for i,tin in enumerate(tinies,1))
def afterdotbinary2float(sbin, com = com):
'''Transforms a binary lying after a dot into a float after a dot'''
if sbin.startswith('0b.') or sbin.startswith('.'):
sbin = sbin.split('.')[1]
if all(c in '01' for c in sbin):
return sum(int(c)*com[i] for i,c in enumerate(sbin,1))
else:
return None
def hexf2binf(x):
'''Transforms an hexadecimal float with a dot into a binary float with a dot'''
a,_,p = str(float.fromhex(x)).partition('.')
# the following part transforms the float after the dot into a binary after the dot
tinies = [ Decimal(1) / Decimal(2**i) for i in xrange(1,400)]
bits = []
pdec = Decimal('.'+p)
for tin in tinies:
if pdec-tin==0:
bits.append('1')
break
elif pdec-tin>0:
bits.append('1')
pdec -= tin
else:
bits.append('0')
pbin = ''.join(bits) # it's the binary after the dot
# the float before the dot is easily transformed into a binary
return '.'.join((bin(int(a)),pbin))
for n in (45.625 , 780.2265625 , 1022.796875):
print 'n ==',n,' transformed with its method hex() to:'
nhexed = n.hex()
print 'nhexed = n.hex() ==',nhexed
print '\nhexf2binf(nhexed) ==',hexf2binf(nhexed)
print "\nVerification:\nbefore,_,after = hexf2binf(nhexed).partition('.')"
before,_,after = hexf2binf(nhexed).partition('.')
print 'before ==',before,' after ==',after
print 'int(before,2) ==',int(before,2)
print 'afterdotbinary2float(after) ==',afterdotbinary2float(after)
print '\n---------------------------------------------------------------\n'
result
n == 45.625 transformed with its method hex() to:
nhexed = n.hex() == 0x1.6d00000000000p+5
hexf2binf(nhexed) == 0b101101.101
Verification:
before,_,after = hexf2binf(nhexed).partition('.')
before == 0b101101 after == 101
int(before,2) == 45
afterdotbinary2float(after) == 0.625
---------------------------------------------------------------
n == 780.2265625 transformed with its method hex() to:
nhexed = n.hex() == 0x1.861d000000000p+9
hexf2binf(nhexed) == 0b1100001100.0011101
Verification:
before,_,after = hexf2binf(nhexed).partition('.')
before == 0b1100001100 after == 0011101
int(before,2) == 780
afterdotbinary2float(after) == 0.2265625
---------------------------------------------------------------
n == 1022.796875 transformed with its method hex() to:
nhexed = n.hex() == 0x1.ff66000000000p+9
hexf2binf(nhexed) == 0b1111111110.110011
Verification:
before,_,after = hexf2binf(nhexed).partition('.')
before == 0b1111111110 after == 110011
int(before,2) == 1022
afterdotbinary2float(after) == 0.796875
---------------------------------------------------------------
.
For the two numbers:
from decimal import Decimal
tinies = [ Decimal(1) / Decimal(2**i) for i in xrange(1,400)]
com = dict((i,tin) for i,tin in enumerate(tinies,1))
def hexf2binf(x, tinies = tinies):
'''Transforms an hexadecimal float with a dot into a binary float with a dot'''
a,_,p = str(float.fromhex(x)).partition('.')
# the following part transforms the float after the dot into a binary after the dot
bits = []
pdec = Decimal('.'+p)
for tin in tinies:
if pdec-tin==0:
bits.append('1')
break
elif pdec-tin>0:
bits.append('1')
pdec -= tin
else:
bits.append('0')
pbin = ''.join(bits) # it's the binary after the dot
# the float before the dot is easily transformed into a binary
return '.'.join((bin(int(a)),pbin))
print hexf2binf('0x1.a000000000000p+2')
print
print hexf2binf('0x1.b5c2000000000p+1')
as a result, this displays:
0b110.1
0b11.011010111000010000000000000000000000010000011111100001111111101001000110110000101101110000000001111110011100110101011100011110011101111010001011111001011101011000000011001111111010111011000101000111100100110000010110001000111010111101110111011111000100100110110101011100101001110011000100000000010001101111111010001110100100101110111000111111001011101101010011011111011001010011111111010101011010110
Given the a hexadecimal string h, you can find the corresponding float with
x = float.fromhex(h)
So really you're interesting in being able to produce a "fixed point" binary representation of any float. Its possible there is no finite representation so you probably want to restrict the length it can be. (eg the binary representation of math.pi wouldn't end ...)
So something like the following might work
def binaryRepresentation(x, n=8):
# the base and remainder ... handle negatives as well ...
base = int(x)
fraction = abs(int(round( (x - base) * (2**n) )))
# format and remove redundant zeros
return "{0:b}.{1:b}".format(base, fraction).rstrip("0")
Big EDIT about big numbers
.
The following code shows a problem with my solution in my other answer.
Note that I changed the parameter of my function hexf2binf(floathex) from h to floathex, to make it the same as the parameter used by unutbu in his function floathex_to_binary(floathex)
from decimal import Decimal,getcontext
getcontext.prec = 500
tinies = [ Decimal(1) / Decimal(2**i) for i in xrange(1,400)]
com = dict((i,tin) for i,tin in enumerate(tinies,1))
def hexf2binf(floathex, tinies = tinies):
fromh = float.fromhex(floathex)
print 'fromh = float.fromhex(h) DONE'
print 'fromh ==',fromh
print "str(float.fromhex(floathex)) ==",str(float.fromhex(floathex))
a,_,p = str(float.fromhex(floathex)).partition('.')
print 'before the dot ==',a
print 'after the dot ==',p
# the following part transforms the float after the dot into a binary after the dot
bits = []
pdec = Decimal('.'+p)
for tin in tinies:
if pdec-tin==0:
bits.append('1')
break
elif pdec-tin>0:
bits.append('1')
pdec -= tin
else:
bits.append('0')
pbin = ''.join(bits) # it's the binary after the dot
# the float before the dot is easily transformed into a binary
return '.'.join((bin(int(a)),pbin))
x = x = 123456789012345685803008.0
print ' x = {:f}'.format(x)
h = x.hex()
print ' h = x.hex() ==',h
print '\nENTERING hexf2binf(floathex) with h as argument'
v = hexf2binf(h)
print '\nhexf2binf(x)==',v
result
x = 123456789012345685803008.000000
h = x.hex() == 0x1.a249b1f10a06dp+76
ENTERING hexf2binf(floathex) with h as argument
fromh = float.fromhex(h) DONE
fromh == 1.23456789012e+23
str(float.fromhex(floathex)) == 1.23456789012e+23
before the dot == 1
after the dot == 23456789012e+23
hexf2binf(x)== 0b1.111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
The problem is due to the instruction str(float.fromhex(x)) in the instruction a,_,p = str(float.fromhex(x)).partition('.') that produces , for a big number, a representation of float.fromhex(x) with an exponent.
Then THE PARTS BEFORE THE DOT (a as ante) AND AFTER THE DOT (p as post) ARE FALSE.
Correcting this is easy: replacing the inaccurate instruction with this one:
a,_,p = '{:f}'.format(float.fromhex(x)).partition('.')
.
Nota bene:
On a typical machine running Python, there are 53 bits of precision
available for a Python float, so the value stored internally when you
enter the decimal number 0.1 is the binary fraction
0.00011001100110011001100110011001100110011001100110011010
http://docs.python.org/tutorial/floatingpoint.html
That means that when a big value for a float is written in a code, its internal representation is in fact an approximation of the written value.
That is shown by the following code:
x1 = 123456789012345685803008.0
print 'x1 == 123456789012345685803008.0'
h1 = x1.hex()
print 'h1 = x1.hex() ==',h1
y1 = float.fromhex(h1)
print 'y1 = float.fromhex(h1) == {:f}'.format(y1)
print
x2 = 123456789012345678901234.64655
print 'x2 == 123456789012345678901234.64655'
h2 = x2.hex()
print 'h2 = x2.hex() ==',h2
y2 = float.fromhex(h2)
print 'y2 = float.fromhex(h2) == {:f}'.format(y2)
print
result
x1 == 123456789012345685803008.0
h1 = x1.hex() == 0x1.a249b1f10a06dp+76
y1 = float.fromhex(h1) == 123456789012345685803008.000000
x2 == 123456789012345678901234.64655
h2 = x2.hex() == 0x1.a249b1f10a06dp+76
y2 = float.fromhex(h2) == 123456789012345685803008.000000
Values of h1 and h2 are the same because, though different values are assigned to identifiers x1 and x2 in the script, the OBJECTS x1 and x2 are represented with the same approximation in the machine.
The internal representation of 123456789012345685803008.0 is the exact value of 123456789012345685803008.0 and is the internal representation of 123456789012345678901234.64655 but its approximation, hence deduction of h1 and h2 from x1 and x2 gives the same value to h1 and h2.
This problem exists when we write a number in decimal representation in a script. It doesn't exist when we write a number directly in hexadecimal or binary representation.
What I wanted to underline
is that I wrote a function afterdotbinary2float(sbin, com = com) to perform verification on the results yielded by hexf2binf( ). This verification works well when the number passed to hexf2binf( ) isn't big, but because of the internal approximation of big numbers (= having a lot of digits), I wonder if this verification isn't distorted. Indeed, when a big number arrives in the function, it has already been approximated : the digits after the dot have been transformed into a series of zeros;
as it is shown here after:
from decimal import Decimal, getcontext
getcontext().prec = 500
tinies = [ Decimal(1) / Decimal(2**i) for i in xrange(1,400)]
com = dict((i,tin) for i,tin in enumerate(tinies,1))
def afterdotbinary2float(sbin, com = com):
'''Transforms a binary lying after a dot into a float after a dot'''
if sbin.startswith('0b.') or sbin.startswith('.'):
sbin = sbin.split('.')[1]
if all(c in '01' for c in sbin):
return sum(int(c)*com[i] for i,c in enumerate(sbin,1))
else:
return None
def hexf2binf(floathex, tinies = tinies):
'''Transforms an hexadecimal float with a dot into a binary float with a dot'''
a,_,p = '{:.400f}'.format(float.fromhex(floathex)).partition('.')
# the following part transforms the float after the dot into a binary after the dot
bits = []
pdec = Decimal('.'+p)
for tin in tinies:
if pdec-tin==0:
bits.append('1')
break
elif pdec-tin>0:
bits.append('1')
pdec -= tin
else:
bits.append('0')
pbin = ''.join(bits) # it's the binary after the dot
# the float before the dot is easily transformed into a binary
return '.'.join((bin(int(a)),pbin))
for n in (123456789012345685803008.0, 123456789012345678901234.64655, Decimal('123456789012345.2546') ):
print 'n == {:f} transformed with its method hex() to:'.format(n)
nhexed = n.hex()
print 'nhexed = n.hex() ==',nhexed
print '\nhexf2binf(nhexed) ==',hexf2binf(nhexed)
print "\nVerification:\nbefore,_,after = hexf2binf(nhexed).partition('.')"
before,_,after = hexf2binf(nhexed).partition('.')
print 'before ==',before,' after ==',after
print 'int(before,2) ==',int(before,2)
print 'afterdotbinary2float(after) ==',afterdotbinary2float(after)
print '\n---------------------------------------------------------------\n'
result
n == 123456789012345685803008.000000 transformed with its method hex() to:
nhexed = n.hex() == 0x1.a249b1f10a06dp+76
hexf2binf(nhexed) == 0b11010001001001001101100011111000100001010000001101101000000000000000000000000.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
Verification:
before,_,after = hexf2binf(nhexed).partition('.')
before == 0b11010001001001001101100011111000100001010000001101101000000000000000000000000 after == 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
int(before,2) == 123456789012345685803008
afterdotbinary2float(after) == 0E-399
---------------------------------------------------------------
n == 123456789012345685803008.000000 transformed with its method hex() to:
nhexed = n.hex() == 0x1.a249b1f10a06dp+76
hexf2binf(nhexed) == 0b11010001001001001101100011111000100001010000001101101000000000000000000000000.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
Verification:
before,_,after = hexf2binf(nhexed).partition('.')
before == 0b11010001001001001101100011111000100001010000001101101000000000000000000000000 after == 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
int(before,2) == 123456789012345685803008
afterdotbinary2float(after) == 0E-399
---------------------------------------------------------------
n == 123456789012345.2546 transformed with its method hex() to:
Traceback (most recent call last):
File "I:\verfitruc.py", line 41, in <module>
nhexed = n.hex()
AttributeError: 'Decimal' object has no attribute 'hex'
Conclusion: testing with numbers 123456789012345685803008.0 and 123456789012345678901234.64655 makes no difference and no interest.
So, I wanted to test un-approximated numbers, and I passed a Decimal float number. As you see, the problem is that such an instance hasn't the hex() method.
.
Finally, I'm not entirely sure of my function for the big numbers, but it works correctly for common numbers after I corrected the inaccurate instruction.
.
EDIT
I added '.400' in the instruction:
a,_,p = '{:.400f}'.format(fromh).partition('.')
otherwise the value of p could be truncated, thus giving a binary representation of a slightly different number than the one passed to the function.
I put 400 because it is the length I defined for the list tinies that contains the Decimal instances corresponding to 1/2 , 1/4 , 1/8 , 1/16 etc
However, though it is rare that a number with more than 400 digits after the comma has any sense, this adding remains unsatisfactory for me: the code isn't absolutely general, which is the case of unutbu 's code.