German number separators using format language on

问题:

Update: The answers show so far that it seems to be a platform-related bug on OSX that has to do with the specific locale settings as they don't fully support grouping numbers.

Update 2: I have just opened an issue on Python's bug tracker. Let's see if there is a solution to this problem.

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I want to format integer and float numbers according to the German numbering convention. This is possible using the format language and the presentation type n but fails on my platform.

• Platform: OS X 10.8.2 (Mountain Lion)
• Python: 2.7.3 64-bit (v2.7.3:70274d53c1dd, Apr 9 2012, 20:52:43) [GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin

Examples:

• 1234 => 1.234
• 1234.56 => 1.234,56
• 1000000 => 1.000.000

What I have tried so far:

1. Setting the German locale

   import locale
   locale.setlocale(locale.LC_ALL, 'de_DE')
   

2. The format specification option , only recognizes the English format.

   '{:,}'.format(1234)
   '1,234'
   
   '{:,}'.format(1234.56)
   '1,234.56'
   
   '{:,}'.format(1000000)
   '1,000,000'
   

3. According to the Python docs, the integer and float presentation type n is supposed to do what I want but it doesn't.

    '{:n}'.format(1234)
    '1234'
   
    '{:n}'.format(1234.56)
    '1234,56'  # at least the comma was set correctly here
   
    '{:n}'.format(1000000)
    '1000000'
   
    '{:n}'.format(12345769.56)
    '1,23458e+07'  # it's doing weird things for large floats
   

4. Some more examples and comparisons inspired by @J.F.Sebastian:

   for n in [1234, 1234.56, 1000000, 12345769.56]:
       print('{0:,} {0:n}'.format(n))
       fmt, val = "%d %f", (n, n)
       print(fmt % val)
       print(locale.format_string(fmt, val))
       print(locale.format_string(fmt, val, grouping=True))
       print('-'*60)
   

   This yields the following incorrect results on my platform:

       1,234 1234
       1234 1234.000000
       1234 1234,000000
       1234 1234,000000
       ------------------------------------------------------------
       1,234.56 1234,56
       1234 1234.560000
       1234 1234,560000
       1234 1234,560000
       ------------------------------------------------------------
       1,000,000 1000000
       1000000 1000000.000000
       1000000 1000000,000000
       1000000 1000000,000000
       ------------------------------------------------------------
       12,345,769.56 1,23458e+07
       12345769 12345769.560000
       12345769 12345769,560000
       12345769 12345769,560000
       ------------------------------------------------------------
   

   The correct results which I'm not getting would look like that:

       1,234 1.234
       1234 1234.000000
       1234 1234,000000
       1.234 1.234,000000
       ------------------------------------------------------------
       1,234.56 1.234,56
       1234 1234.560000
       1234 1234,560000
       1.234 1.234,560000
       ------------------------------------------------------------
       1,000,000 1.000.000
       1000000 1000000.000000
       1000000 1000000,000000
       1.000.000 1.000.000,000000
       ------------------------------------------------------------
       12,345,769.56 1,23458e+07 
       12345769 12345769.560000
       12345769 12345769,560000
       12.345.769 12.345.769,560000
       ------------------------------------------------------------
   

Do you have a solution for me using the format language only? Is there any way to trick the locale settings on my platform to accept grouping?

回答1:

Super ugly, but technically answers the question:

From PEP 378:

'{:,}'.format(1234.56).replace(",", "X").replace(".", ",").replace("X", ".")
'1.234,56'

回答2:

Python's locale module's implementation unfortunately varies quite a bit across platforms. It's really just a light wrapper around the C library vendor's notion of locales.

So, on Windows 7, with Python 2.7.3 64-bit, this happens to work (note: locales have different names in Windows):

>>> import locale
>>> locale.setlocale(locale.LC_ALL, 'deu_deu')
'German_Germany.1252'
>>> '{0:n}'.format(1234.56)
'1.234,56'

Whether the thousands separator will be used can be determined by examining the "local conventions":

>>> locale.localeconv()['grouping'] # On Windows, 'deu_deu'.
[3, 0] # Insert separator every three digits.

>>> locale.localeconv()['grouping'] # On OS X, 'de_DE'.
[127] # No separator (locale.CHAR_MAX == 127).

>>> locale.localeconv()['grouping'] # Default C locale.
[] # Also no separator.

回答3:

This worked for me when used with the German locale:

>>> import locale
>>> locale.setlocale(locale.LC_ALL, 'de_DE')
'de_DE'
>>> '{0:n}'.format(1234.56)
'1.234,56'

This is in Cygwin under Windows 7:

>>> import sys
>>> print sys.version
2.6.5 (r265:79063, Jun 12 2010, 17:07:01)
[GCC 4.3.4 20090804 (release) 1]

回答4:

I was asked by @Lattyware to provide my own solution for including separators according to the German numbering convention without using the format language. Here is the best solution that I can come up with:

import re

def group_num(num):
    if isinstance(num, (int, float)):
        if isinstance(num, float):
            head, tail = str(num).split('.')
        elif isinstance(num, int):
            head, tail = str(num), ''
        digit_parts = re.findall(r'\d{1,3}\-?', ''.join(head[::-1]))
        num = '.'.join(part[::-1] for part in digit_parts[::-1])
        if tail:
            num = ','.join((num, tail))
        return num
    else:
        raise TypeError(num, 'is not of type int or float')

>>> group_num(1234)
'1.234'
>>> group_num(123456.7890)
'123.456,789'
>>> group_num(-1000000000.12)
'-1.000.000.000,12'

The performance is also quite okay, compared to the solution given by @Jon-Eric.

%timeit group_num(1000000000.12)
10000 loops, best of 3: 20.6 us per loop

# For integers, it's faster since several steps are not necessary
%timeit group_num(100000000012)
100000 loops, best of 3: 18.2 us per loop

%timeit '{:,}'.format(1000000000.12).replace(",", "X").replace(".", ",").replace("X", ".")
100000 loops, best of 3: 2.63 us per loop

%timeit '{:,}'.format(100000000012).replace(",", "X").replace(".", ",").replace("X", ".")
100000 loops, best of 3: 2.01 us per loop

If you know how my solution could be optimized, please let me know.

回答5:

Even more ugly with split, join and replace:

>>> amount = '{0:,}'.format(12345.67)
>>> amount
'12,345.67'
>>> ','.join([s.replace(',','.') for s in amount.split('.')])
'12.345,67'