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Splitting arrays in unequal chunks using np.array_split: overcomplicating?

I am trying to convert an octal number to decimal.
The inputs are a set of strings as numbers such as "23", or "23 24", or "23 24 25". My code works for inputs like this, but cannot handle say "23 240", or "23 240 1" I.e. when the inputs are of different lengths, the array splits them incorrectly.
I think I've overcomplicated it by using arrays. Is there a way to assess each input individually (i.e. "23" then "240" then "1"), and then put these back into the desired output "19 160 1"?
Code:
import numpy as np
def decode(code):
decimals = []
code = code.split()
for n in code:
p = len(n)
for digit in n:
decimal = int(digit) * (8**(p - 1))
decimals.append(decimal)
p -= 1
split_input = np.array_split(decimals, len(code))
sum_decimals = []
for number in split_input:
sum_decimal = sum(map(int, number))
sum_decimals.append(str(sum_decimal))
separate_outputs = " ".join(sum_decimals)
return str(separate_outputs)

Does this work? Try running it in an instant interpreter such as Google Colab
CODE:
import numpy as np
def decode(code):
if isinstance(code, str):
decimals = []
code = code.split(' ')
for n in code:
p = len(n)
for digit in n:
decimal = int(digit) * (8**(p - 1))
decimals.append(decimal)
p -= 1
split_input = np.array_split(decimals, len(code))
sum_decimals = []
for number in split_input:
sum_decimal = sum(map(int, number))
sum_decimals.append(str(sum_decimal))
separate_outputs = " ".join(sum_decimals)
return str(separate_outputs)
INPUT:
str1 = input('Enter octals:') #23 240 1
decode(str1)
OUTPUT:
19 160 1
EXTRAS:
If you copy code make sure to adjust tabs because that could lead to
multiple errors.
Also, you could put an else statement to raise
error if function does not receive a string array. OR, you could add
multiple if...else statements to handle different data formats
like: str, int, list, tuple, etc.

Find length of a string that includes its own length?

I want to get the length of a string including a part of the string that represents its own length without padding or using structs or anything like that that forces fixed lengths.
So for example I want to be able to take this string as input:
"A string|"
And return this:
"A string|11"

On the basis of the OP tolerating such an approach (and to provide an implementation technique for the eventual python answer), here's a solution in Java.
final String s = "A String|";
int n = s.length(); // `length()` returns the length of the string.
String t; // the result
do {
t = s + n; // append the stringified n to the original string
if (n == t.length()){
return t; // string length no longer changing; we're good.
}
n = t.length(); // n must hold the total length
} while (true); // round again
The problem of, course, is that in appending n, the string length changes. But luckily, the length only ever increases or stays the same. So it will converge very quickly: due to the logarithmic nature of the length of n. In this particular case, the attempted values of n are 9, 10, and 11. And that's a pernicious case.

A simple solution is :
def addlength(string):
n1=len(string)
n2=len(str(n1))+n1
n2 += len(str(n2))-len(str(n1)) # a carry can arise
return string+str(n2)
Since a possible carry will increase the length by at most one unit.
Examples :
In [2]: addlength('a'*8)
Out[2]: 'aaaaaaaa9'
In [3]: addlength('a'*9)
Out[3]: 'aaaaaaaaa11'
In [4]: addlength('a'*99)
Out[4]: 'aaaaa...aaa102'
In [5]: addlength('a'*999)
Out[5]: 'aaaa...aaa1003'

Here is a simple python port of Bathsheba's answer :
def str_len(s):
n = len(s)
t = ''
while True:
t = s + str(n)
if n == len(t):
return t
n = len(t)
This is a much more clever and simple way than anything I was thinking of trying!
Suppose you had s = 'abcdefgh|, On the first pass through, t = 'abcdefgh|9
Since n != len(t) ( which is now 10 ) it goes through again : t = 'abcdefgh|' + str(n) and str(n)='10' so you have abcdefgh|10 which is still not quite right! Now n=len(t) which is finally n=11 you get it right then. Pretty clever solution!

It is a tricky one, but I think I've figured it out.
Done in a hurry in Python 2.7, please fully test - this should handle strings up to 998 characters:
import sys
orig = sys.argv[1]
origLen = len(orig)
if (origLen >= 98):
extra = str(origLen + 3)
elif (origLen >= 8):
extra = str(origLen + 2)
else:
extra = str(origLen + 1)
final = orig + extra
print final
Results of very brief testing
C:\Users\PH\Desktop>python test.py "tiny|"
tiny|6
C:\Users\PH\Desktop>python test.py "myString|"
myString|11
C:\Users\PH\Desktop>python test.py "myStringWith98Characters.........................................................................|"
myStringWith98Characters.........................................................................|101

Just find the length of the string. Then iterate through each value of the number of digits the length of the resulting string can possibly have. While iterating, check if the sum of the number of digits to be appended and the initial string length is equal to the length of the resulting string.
def get_length(s):
s = s + "|"
result = ""
len_s = len(s)
i = 1
while True:
candidate = len_s + i
if len(str(candidate)) == i:
result = s + str(len_s + i)
break
i += 1

This code gives the result.
I used a few var, but at the end it shows the output you want:
def len_s(s):
s = s + '|'
b = len(s)
z = s + str(b)
length = len(z)
new_s = s + str(length)
new_len = len(new_s)
return s + str(new_len)
s = "A string"
print len_s(s)

Here's a direct equation for this (so it's not necessary to construct the string). If s is the string, then the length of the string including the length of the appended length will be:
L1 = len(s) + 1 + int(log10(len(s) + 1 + int(log10(len(s)))))
The idea here is that a direct calculation is only problematic when the appended length will push the length past a power of ten; that is, at 9, 98, 99, 997, 998, 999, 9996, etc. To work this through, 1 + int(log10(len(s))) is the number of digits in the length of s. If we add that to len(s), then 9->10, 98->100, 99->101, etc, but still 8->9, 97->99, etc, so we can push past the power of ten exactly as needed. That is, adding this produces a number with the correct number of digits after the addition. Then do the log again to find the length of that number and that's the answer.
To test this:
from math import log10
def find_length(s):
L1 = len(s) + 1 + int(log10(len(s) + 1 + int(log10(len(s)))))
return L1
# test, just looking at lengths around 10**n
for i in range(9):
for j in range(30):
L = abs(10**i - j + 10) + 1
s = "a"*L
x0 = find_length(s)
new0 = s+`x0`
if len(new0)!=x0:
print "error", len(s), x0, log10(len(s)), log10(x0)

Python Decimal - engineering notation for mili (10e-3) and micro (10e-6)

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'

How to format a float with a maximum number of decimal places and without extra zero padding?

I need to do some decimal place formatting in python. Preferably, the floating point value should always show at least a starting 0 and one decimal place. Example:
Input: 0
Output: 0.0
Values with more decimal places should continue to show them, until it gets 4 out. So:
Input: 65.53
Output: 65.53
Input: 40.355435
Output: 40.3554
I know that I can use {0.4f} to get it to print out to four decimal places, but it will pad with unwanted 0s. Is there a formatting code to tell it to print out up to a certain number of decimals, but to leave them blank if there is no data? I believe C# accomplishes this with something like:
floatValue.ToString("0.0###")
Where the # symbols represent a place that can be left blank.

What you're asking for should be addressed by rounding methods like the built-in round function. Then let the float number be naturally displayed with its string representation.
>>> round(65.53, 4) # num decimal <= precision, do nothing
'65.53'
>>> round(40.355435, 4) # num decimal > precision, round
'40.3554'
>>> round(0, 4) # note: converts int to float
'0.0'

Sorry, the best I can do:
' {:0.4f}'.format(1./2.).rstrip('0')
Corrected:
ff=1./2.
' {:0.4f}'.format(ff).rstrip('0')+'0'[0:(ff%1==0)]

From trial and error I think :.15g is what you want:
In: f"{3/4:.15g}"
Out: '0.75'
In f"{355/113:.15g}"
Out: '3.14159292035398'
(while f"{3/4:.15f}" == '0.750000000000000')

>>> def pad(float, front = 0, end = 4):
s = '%%%s.%sf' % (front, end) % float
i = len(s)
while i > 0 and s[i - 1] == '0':
i-= 1
if s[i - 1] == '.' and len(s) > i:
i+= 1 # for 0.0
return s[:i] + ' ' * (len(s) - i)
>>> pad(0, 3, 4)
'0.0 '
>>> pad(65.53, 3, 4)
'65.53 '
>>> pad(40.355435, 3, 4)
'40.3554'

Converting float.hex() value to binary in Python

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.