How can I print any float, like 1.234e22, in a long, nicely rounded and localized format in python?

Question

There are a few existing questions about float formatting, but none answer the following question, I think.

I'm looking for a way to print large floats in a long, nicely rounded and localized format:

>>> print magic_format(1.234e22, locale="en_US")
12,340,000,000,000,000,000,000
>>> print magic_format(1.234e22, locale="fr_FR")
12 340 000 000 000 000 000 000

Unfortunately, magic_format does not exist. ;-) How can I implement it?

Details

Here are a few ways to print floats. None of them produces the above output:

>>> x = 1.234e22
>>> print str(x)
1.234e+22
>>> print repr(x)
1.234e+22
>>> print "%f" % x
12339999999999998951424.000000
>>> print "%g" % x
1.234e+22

Fail: either I get the short version, or a non-grouping non-localized non-rounded output.

BTW, I understand that 1.234e22 cannot be stored exactly as a float, there's a necessary rounding error (that explains the odd output above). But since str, repr and "%g" % x are able to properly round that to the proper value, I would like to have the same friendly rounded number, but in a long and localized form.

Let's try localizing now...

>>> import locale
>>> locale.setlocale(locale.LC_ALL, "en_US")
'en_US'
>>> locale.format("%g", x, grouping = True)
'1.234e+22'
>>> locale.format("%f", x, grouping = True)
'12,339,999,999,999,998,951,424.000000'
>>> locale.setlocale(locale.LC_ALL, "fr_FR")
'fr_FR'
>>> locale.format("%g", x, grouping = True)
'1,234e+22'
>>> locale.format("%f", x, grouping = True)
'12339999999999998951424,000000'

Closer, but not ok. I still have the annoying rounding error, and the French localization sucks, it does not allow grouping at all.

So let's use the excellent Babel library, perhaps it can do everything I want:

>>> from babel.numbers import format_number
>>> format_number(x, locale = "en_US")
u'12,339,999,999,999,998,951,424'
>>> format_number(x, locale = "fr_FR")
u'12\xa0339\xa0999\xa0999\xa0999\xa0998\xa0951\xa0424'

Wow, really close. They even use non-breakable spaces for grouping in French, I love it. It's really too bad they still have the rounding issue.

Hey!? What if I used python Decimals?

>>> from decimal import Decimal
>>> Decimal(x)
Decimal('12339999999999998951424')
>>> Decimal("%g" % x)
Decimal('1.234E+22')
>>> "%g" % Decimal("%g" % x)
'1.234e+22'
>>> "%f" % Decimal("%g" % x)
'12339999999999998951424.000000'

Nope. I can get an exact representation of the number I want with Decimal("%g" % x), but whenever I try to display it, it's either short or converted to a bad float before it's printed.

But what if I mixed Babel and Decimals?

>>> Decimal("%g" % 1.234e22)
Decimal('1.234E+22')
>>> dx = _
>>> format_number(dx, locale = "en_US")
Traceback (most recent call last):
...
TypeError: bad operand type for abs(): 'str'

Ouch. But Babel's got a function called format_decimal, let's use that instead:

>>> from babel.numbers import format_decimal
>>> format_decimal(dx, locale = "en_US")
Traceback (most recent call last):
...
TypeError: bad operand type for abs(): 'str'

Oops, format_decimal can't format python Decimals. :-(

Ok, one last idea: I could try converting to a long.

>>> x = 1.234e22
>>> long(x)
12339999999999998951424L
>>> long(Decimal(x))
12339999999999998951424L
>>> long(Decimal("%g" % x))
12340000000000000000000L

Yes! I've got the exact number I want to format. Let's give that to Babel:

>>> format_number(long(Decimal("%g" % x)), locale = "en_US")
u'12,339,999,999,999,998,951,424'

Oh, no... Apparently Babel converts the long to a float before trying to format it. I'm out of luck, and out of ideas. :-(

If you think that this is tough, then try answering the same question for x = 1.234e-22. So far all I can print is either the short form 1.234e-22 or 0.0!

I would prefer this:

>>> print magic_format(1.234e-22, locale="en_US")
0.0000000000000000000001234
>>> print magic_format(1.234e-22, locale="fr_FR")
0,0000000000000000000001234
>>> print magic_format(1.234e-22, locale="en_US", group_frac=True)
0.000,000,000,000,000,000,000,123,400
>>> print magic_format(1.234e-22, locale="fr_FR", group_frac=True)
0,000 000 000 000 000 000 000 123 400

I can imagine writing a little function that would parse "1.234e-22" and format it nicely, but I would have to know all about the rules of number localization, and I'd rather not reinvent the wheel, Babel is supposed to do that. What should I do?

Thanks for your help. :-)

Answer 1

This takes a large chunk of code from selected answer from Nicely representing a floating-point number in python but incorporates Babel to handle L10N.

NOTE : Babel uses a weird unicode version of the space character for a lot of locales. Hence the if loop that mentions 'fr_FR' directly to convert it to a normal space character.

import locale
from babel.numbers import get_decimal_symbol,get_group_symbol
import decimal

# https://stackoverflow.com/questions/2663612/nicely-representing-a-floating-point-number-in-python/2663623#2663623
def float_to_decimal(f):
    # http://docs.python.org/library/decimal.html#decimal-faq
    "Convert a floating point number to a Decimal with no loss of information"
    n, d = f.as_integer_ratio()
    numerator, denominator = decimal.Decimal(n), decimal.Decimal(d)
    ctx = decimal.Context(prec=60)
    result = ctx.divide(numerator, denominator)
    while ctx.flags[decimal.Inexact]:
        ctx.flags[decimal.Inexact] = False
        ctx.prec *= 2
        result = ctx.divide(numerator, denominator)
    return result 

def f(number, sigfig):
    assert(sigfig>0)
    try:
        d=decimal.Decimal(number)
    except TypeError:
        d=float_to_decimal(float(number))
    sign,digits,exponent=d.as_tuple()
    if len(digits) < sigfig:
        digits = list(digits)
        digits.extend([0] * (sigfig - len(digits)))    
    shift=d.adjusted()
    result=int(''.join(map(str,digits[:sigfig])))
    # Round the result
    if len(digits)>sigfig and digits[sigfig]>=5: result+=1
    result=list(str(result))
    # Rounding can change the length of result
    # If so, adjust shift
    shift+=len(result)-sigfig
    # reset len of result to sigfig
    result=result[:sigfig]
    if shift >= sigfig-1:
        # Tack more zeros on the end
        result+=['0']*(shift-sigfig+1)
    elif 0<=shift:
        # Place the decimal point in between digits
        result.insert(shift+1,'.')
    else:
        # Tack zeros on the front
        assert(shift<0)
        result=['0.']+['0']*(-shift-1)+result
    if sign:
        result.insert(0,'-')
    return ''.join(result)

def magic_format(num, locale="en_US", group_frac=True):
    sep = get_group_symbol(locale)
    if sep == get_group_symbol('fr_FR'): 
        sep = ' '
    else:
        sep = str(sep)
    dec = str(get_decimal_symbol(locale))

    n = float(('%E' % num)[:-4:])
    sigfig = len(str(n)) - (1 if '.' in str(n) else 0) 

    s = f(num,sigfig)

    if group_frac:
        ans = ""
        if '.' not in s:
            point = None
            new_d = ""
            new_s = s[::-1]
        else:
            point = s.index('.')
            new_d = s[point+1::]
            new_s = s[:point:][::-1]
        for idx,char in enumerate(new_d):
            ans += char
            if (idx+1)%3 == 0 and (idx+1) != len(new_d): 
                ans += sep
        else: ans = ans[::-1] + (dec if point != None else '')
        for idx,char in enumerate(new_s):
            ans += char
            if (idx+1)%3 == 0 and (idx+1) != len(new_s): 
                ans += sep 
        else:
            ans = ans[::-1]
    else:
        ans = s
    return ans

This chuck of code can be used as follows:

>>> magic_format(num2, locale = 'fr_FR')
'0,000 000 000 000 000 000 000 123 456 0'
>>> magic_format(num2, locale = 'de_DE')
'0,000.000.000.000.000.000.000.123.456.0'
>>> magic_format(num2)
'0.000,000,000,000,000,000,000,123,456'
>>> f(num,6)
'12345600000000000000000'
>>> f(num2,6)
'0.000000000000000000000123456'

with the f function coming from the link.