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import re

# See also: Python-Docs-2.4/lib/decimal-recipes.html

def fp_format(num, thousands_separator=',', fraction_separator='.',
              format='%.2f'):
    """Formats a floating point number.
    
    Replaces the floating point with the given `fraction_separator` and groups
    thousands with the given `thousands_separator`.  By default thousands are
    grouped with ``','`` and the fraction is separated by ``'.'``.  The
    fractional part is two digits long.
    
    :bug: Wrong results if `num` is formatted in engeneering notation, i.e.
          with an exponent, and no floating point is present.
    """
    assert thousands_separator != fraction_separator
    
    numstr = format % num
    
    try:
        integer_part, fractional_part = numstr.split('.')
    except ValueError:
        integer_part, fractional_part = numstr, ''
    # 
    # Extract leading non-digit characters like spaces or ``-``/``+``.
    # 
    match = re.match(r'[\D]*', integer_part)
    if match:
        leading_non_digits = match.group()
        integer_part = integer_part[len(leading_non_digits):]
    else:
        leading_non_digits = ''
    # 
    # Split thousands groups and insert separator into `integer_part`.
    # 
    remainder = len(integer_part) % 3
    thousand_groups = list()
    if remainder:
        thousand_groups.append(integer_part[0:remainder])
    for i in xrange(remainder, len(integer_part), 3):
        thousand_groups.append(integer_part[i:i+3])
    integer_part = thousands_separator.join(thousand_groups)
    
    result = [leading_non_digits, integer_part]
    if fractional_part:
        result.append(fraction_separator)
        result.append(fractional_part)
    return ''.join(result)

Code: Alles auswählen

import string

def format_number_0(number,format="%0.2f",comma=",",thousand=".",grouplength=3):
    if number<0:
         vorzeichen="-"
         number=-number
    else:
        vorzeichen=""
    if format.endswith("f"):
        hinter_komma=string.split(format % number,".")[1]
        full_string=comma+hinter_komma
    elif format.endswith("i"):
        full_string=""
    number=int(number)
    x=1
    while (10**grouplength)**x<=number:
        full_string=thousand+str(number%((10**grouplength)**x))[0:grouplength]+full_string
        x+=1
    full_string=vorzeichen+str(number/((10**grouplength)**(x-1)))+full_string
    return full_string

#----------------------------------------------------------------------

def format_number_0b(number,format="%0.2f",comma=",",thousand=".",grouplength=3):
    if number<0:
         vorzeichen="-"
         number=-number
    else:
        vorzeichen=""

    vor_komma,hinter_komma=string.split(format % number,".")
    add=0
    for i in range(grouplength,len(vor_komma),grouplength):
        vor_komma=vor_komma[0:-(i+add)]+thousand+vor_komma[-(i+add):]
    return vor_komma+comma+hinter_komma

Code: Alles auswählen

# -*- coding: cp1252 -*-
#format_number.py
import string

def format_number_0(number,format="%0.2f",comma=",",thousand=".",grouplength=3):
    if number<0:
         vorzeichen="-"
         number=-number
    else:
        vorzeichen=""
    if format.endswith("f"):
        hinter_komma=string.split(format % number,".")[1]
        full_string=comma+hinter_komma
    elif format.endswith("i"):
        full_string=""
    number=int(number)
    x=1
    while (10**grouplength)**x<=number:
        full_string=thousand+str(number%((10**grouplength)**x))[0:grouplength]+full_string
        x+=1
    full_string=vorzeichen+str(number/((10**grouplength)**(x-1)))+full_string
    return full_string

#----------------------------------------------------------------------

def format_number_0b(number,format="%0.2f",comma=",",thousand=".",grouplength=3):
    if number<0:
         vorzeichen="-"
         number=-number
    else:
        vorzeichen=""

    vor_komma,hinter_komma=string.split(format % number,".")
    add=0
    for i in range(grouplength,len(vor_komma),grouplength):
        vor_komma=vor_komma[0:-(i+add)]+thousand+vor_komma[-(i+add):]
    return vor_komma+comma+hinter_komma

#----------------------------------------------------------------------

import re 

#---------------------------------------------------------------------- 
def format_number_1( 
   number, 
   format = "%0.2f", 
   decimal_char = ",", 
   group_char = ".", 
   group_range = 3 
): 
   """
   Nach einer Idee von Voges, HarryH und Jens. 
   http://www.python-forum.de/viewtopic.php?t=371 
   """
    
   def reverse(string): 
      # ersatz für string[::-1] welches erst ab v2.3 gibt :( 
      list = re.findall(".",string) 
      list.reverse() 
      return "".join(list) 
    
   regex = re.compile( 
      "^ *\d*\D\d{1,%s}|\d{1,%s} *" % (group_range, group_range) 
   ) 
    
   return reverse( 
      group_char.join( 
         regex.findall( 
            reverse( (format % number).replace( 
               ".", decimal_char) 
            ) 
         ) 
      ) 
   ) 


#---------------------------------------------------------------------- 
def format_number_2( 
   number, 
   format = "%0.2f", 
   decimal_char = ",", 
   group_char = ".", 
   group_range = 3 
): 
   """
   Nach einer Idee von Milan. 
   http://www.python-forum.de/viewtopic.php?t=371 
    """

   def split(s, size = group_range): 
      return map(lambda i: s[i:i + size], xrange(0, len(s), size)) 

   def reverse(s): 
      l = map(None, s) 
      l.reverse() 
      return ('').join(l) 

   num = (format % number).replace('.', decimal_char) 
   int_part = num[:num.find(decimal_char)] 
   return ( 
      reverse( 
         group_char.join( 
            split(reverse(int_part)) 
         ) 
      ) + num[num.find(decimal_char):] 
   ).replace(" %s" % group_char, "  ")

#---------------------------------------------------------------------- 
def format_number_3(num, thousands_separator=',', fraction_separator='.', 
              format='%.2f'): 
    """Formats a floating point number. 
    
    Replaces the floating point with the given `fraction_separator` and groups 
    thousands with the given `thousands_separator`.  By default thousands are 
    grouped with ``','`` and the fraction is separated by ``'.'``.  The 
    fractional part is two digits long. 
    
    :bug: Wrong results if `num` is formatted in engeneering notation, i.e. 
          with an exponent, and no floating point is present. 
    """ 
    assert thousands_separator != fraction_separator 
    
    numstr = format % num 
    
    try: 
        integer_part, fractional_part = numstr.split('.') 
    except ValueError: 
        integer_part, fractional_part = numstr, '' 
    # 
    # Extract leading non-digit characters like spaces or ``-``/``+``. 
    # 
    match = re.match(r'[\D]*', integer_part) 
    if match: 
        leading_non_digits = match.group() 
        integer_part = integer_part[len(leading_non_digits):] 
    else: 
        leading_non_digits = '' 
    # 
    # Split thousands groups and insert separator into `integer_part`. 
    # 
    remainder = len(integer_part) % 3 
    thousand_groups = list() 
    if remainder: 
        thousand_groups.append(integer_part[0:remainder]) 
    for i in xrange(remainder, len(integer_part), 3): 
        thousand_groups.append(integer_part[i:i+3]) 
    integer_part = thousands_separator.join(thousand_groups) 
    
    result = [leading_non_digits, integer_part] 
    if fractional_part: 
        result.append(fraction_separator) 
        result.append(fractional_part) 
    return ''.join(result) 

#---------------------------------------------------------------------- 

def format_number_4(number, comma=",", thousand="."): 
    n = str(number).split(comma) 
    n[0] = list(str(n[0]))[::-1] 
    n[0] = thousand.join( 
            ["".join(n[0][i:i+3][::-1]) for i in      
             range(0,len(n[0]),3)][::-1]) 
    return comma.join(n) 

#---------------------------------------------------------------------- 

if __name__ == "__main__": 
    # Testen welche Funktion schneller ist 

    import time 

    zahlen = ( 
      66, 
      1, 
      2, 
      3.4, 
      -5.678, 
      612345, 
      612345.555, 
      987654321.067 
    )
    for x in range(2):
        begin = time.time()

        for i in range(10000): 
          for zahl in zahlen: 
             x = format_number_0(zahl)


        print "0:", time.time() - begin 
        print 

        begin = time.time()

        for i in range(10000): 
          for zahl in zahlen: 
             x = format_number_0(zahl)

        print "0b:", time.time() - begin 
        print 

        begin = time.time() 

        for i in range(10000): 
          for zahl in zahlen: 
             x = format_number_1(zahl) 
              
        print "1:", time.time() - begin 
        print

        begin = time.time() 

        for i in range(10000): 
          for zahl in zahlen: 
             x = format_number_2(zahl) 
              
        print "2:", time.time() - begin 
        print 

        begin = time.time() 

        for i in range(10000): 
          for zahl in zahlen: 
             x = format_number_3(zahl) 
              
        print "3:", time.time() - begin 
        print 

        begin = time.time() 

        for i in range(10000): 
          for zahl in zahlen: 
             x = format_number_4(zahl) 
              
        print "4:", time.time() - begin 
        print 

Code: Alles auswählen

0: 1.6099998951

0b: 1.57800006866

1: 3.85900020599

2: 2.93700003624

3: 2.93700003624

4: 2.04699993134

0: 1.7650001049

0b: 1.875

1: 4.45399999619

2: 2.89099979401

3: 3.42199993134

4: 1.89100003242

Code: Alles auswählen

#!/usr/bin/python
# -*- coding: UTF-8 -*-

#format_number.py
import string

def format_number_0(number,format="%0.2f",comma=",",thousand=".",grouplength=3):
    """
    by cime
    http://www.python-forum.de/viewtopic.php?p=20935#20935
    """
    if number<0:
         vorzeichen="-"
         number=-number
    else:
        vorzeichen=""
    if format.endswith("f"):
        hinter_komma=string.split(format % number,".")[1]
        full_string=comma+hinter_komma
    elif format.endswith("i"):
        full_string=""
    number=int(number)
    x=1
    while (10**grouplength)**x<=number:
        full_string=thousand+str(number%((10**grouplength)**x))[0:grouplength]+full_string
        x+=1
    full_string=vorzeichen+str(number/((10**grouplength)**(x-1)))+full_string
    return full_string

#----------------------------------------------------------------------

def format_number_0b(number,format="%0.2f",comma=",",thousand=".",grouplength=3):
    """
    by cime
    http://www.python-forum.de/viewtopic.php?p=20935#20935
    """
    if number<0:
         vorzeichen="-"
         number=-number
    else:
        vorzeichen=""

    vor_komma,hinter_komma=string.split(format % number,".")
    add=0
    for i in range(grouplength,len(vor_komma),grouplength):
        vor_komma=vor_komma[0:-(i+add)]+thousand+vor_komma[-(i+add):]
    return vor_komma+comma+hinter_komma

#----------------------------------------------------------------------

import re

#----------------------------------------------------------------------
format_number_1_regex = re.compile("^ *\d*\D\d{1,3}[ -]*|\d{1,3}[ -]*" )
def format_number_1(
   number,
   format = "%0.2f",
   decimal_char = ",",
   group_char = ".",
   group_range = 3
):
    """
    Nach einer Idee von Voges, HarryH, Milan und Jens.
    http://www.python-forum.de/viewtopic.php?t=371
    """
    def reverse(s):
        # ersatz für string[::-1] welches erst ab v2.3 gibt :(
        l = map(None, s)
        l.reverse()
        return ('').join(l)


    return reverse(
        group_char.join(
            format_number_1_regex.findall(
                reverse( (format % number).replace( ".", decimal_char)
                )
            )
        )
    )


#----------------------------------------------------------------------
def format_number_2(
   number,
   format = "%0.2f",
   decimal_char = ",",
   group_char = ".",
   group_range = 3
):
   """
   Nach einer Idee von Milan.
   http://www.python-forum.de/viewtopic.php?t=371
    """

   def split(s, size = group_range):
      return map(lambda i: s[i:i + size], xrange(0, len(s), size))

   def reverse(s):
      l = map(None, s)
      l.reverse()
      return ('').join(l)

   num = (format % number).replace('.', decimal_char)
   int_part = num[:num.find(decimal_char)]
   return (
      reverse(
         group_char.join(
            split(reverse(int_part))
         )
      ) + num[num.find(decimal_char):]
   ).replace(" %s" % group_char, "  ")

#----------------------------------------------------------------------
def format_number_3(num, thousands_separator=',', fraction_separator='.',
              format='%.2f'):
    """
    by BlackJack
    http://www.python-forum.de/viewtopic.php?p=20917#20917

    Formats a floating point number.

    Replaces the floating point with the given `fraction_separator` and groups
    thousands with the given `thousands_separator`.  By default thousands are
    grouped with ``','`` and the fraction is separated by ``'.'``.  The
    fractional part is two digits long.

    :bug: Wrong results if `num` is formatted in engeneering notation, i.e.
          with an exponent, and no floating point is present.
    """
    assert thousands_separator != fraction_separator

    numstr = format % num

    try:
        integer_part, fractional_part = numstr.split('.')
    except ValueError:
        integer_part, fractional_part = numstr, ''
    #
    # Extract leading non-digit characters like spaces or ``-``/``+``.
    #
    match = re.match(r'[\D]*', integer_part)
    if match:
        leading_non_digits = match.group()
        integer_part = integer_part[len(leading_non_digits):]
    else:
        leading_non_digits = ''
    #
    # Split thousands groups and insert separator into `integer_part`.
    #
    remainder = len(integer_part) % 3
    thousand_groups = list()
    if remainder:
        thousand_groups.append(integer_part[0:remainder])
    for i in xrange(remainder, len(integer_part), 3):
        thousand_groups.append(integer_part[i:i+3])
    integer_part = thousands_separator.join(thousand_groups)

    result = [leading_non_digits, integer_part]
    if fractional_part:
        result.append(fraction_separator)
        result.append(fractional_part)
    return ''.join(result)

#----------------------------------------------------------------------

def format_number_4(number, comma=",", thousand="."):
    n = str(number).split(comma)
    n[0] = list(str(n[0]))[::-1]
    n[0] = thousand.join(
            ["".join(n[0][i:i+3][::-1]) for i in
             range(0,len(n[0]),3)][::-1])
    return comma.join(n)

#----------------------------------------------------------------------

def format_number_5(number, format = "%.2f", decChar = ",", groupChar = ".", groupRange = 3):
    """
    by HarryH
    http://www.python-forum.de/viewtopic.php?p=20888#20888

    Convert to required format
    number =      number to convert
    format =      python string formatting (%),but only for the decimals
    decChar =     decimal char for the converted format
    groupChar =   group char for the converted format
    groupRange =  group range for the converted format

    For example:
    __formatter(1234567890.987, "%.2f", ",", ".", 3)
    ==> 1.234.567.890,99

    """
    if abs(number) < 1:
        return (format % (number)).replace(".", decChar)
    else:
        base = int(abs(number))
        s = (format % (abs(number) % 1))[2:]
        s = s and decChar + s or s
        while base:
            s = groupChar + str(base)[-groupRange:] + s
            base = int(base / 10**groupRange)
        return number < 0 and "-" + s[1:] or s[1:]

#----------------------------------------------------------------------

def format_number_6(number, comma=",", thousand="."):
    """
    by mawe
    http://www.python-forum.de/viewtopic.php?p=20850#20850
    """
    n = str(number).split(comma)
    x = len(n[0]) % 3
    n[0] = "%s%s" % ("0" * (3-x), n[0])
    n[0] = thousand.join([n[0][i:i+3]
              for i in range(0,len(n[0]),3)]).lstrip("0%s" % thousand)
    return comma.join(n)

#----------------------------------------------------------------------

import locale
def format_numer_locale( number ):
    #~ locale.setlocale(locale.LC_ALL, "de_DE.UTF-8")
    return locale.format("%0.1f", number, True)

#----------------------------------------------------------------------

if __name__ == "__main__":
    # Testen welche Funktion schneller ist

    import time

    zahlen = (
        (66, "66,00"),
        (1, "1,00"),
        (2, "2,00"),
        (3.4, "3,40"),
        (-5.678, "-5,68"),
        (612345, "612.345,00"),
        (612345.555, "612.345,56"),
        (987654321.067, "987.654.321,07"),
    )
    defs = [
        ("0", format_number_0),
        ("0b", format_number_0b),
        ("1", format_number_1),
        ("2", format_number_2),
        ("3", format_number_3),
        ("4", format_number_4),
        ("5", format_number_5),
        ("6", format_number_6),
        ("format_numer_locale", format_numer_locale )

    ]
    for i in defs:
        print "_"*80
        print ">>>",i[0]
        for z in zahlen:
            result = i[1](z[0])
            print "%s -> %s " % (z[0],result),
            if str(result) == z[1]:
                print "OK"
            else:
                print "error!"

    print "="*80

    loops = 10000
    for i in defs:
        print i[0],
        begin = time.time()
        for x in range( loops ):
            for z in zahlen:
                i[1](z[0])
        print "%0.3fs" % (time.time() - begin)

Code: Alles auswählen

________________________________________________________________________________
>>> 0
66 -> 66,00  OK
1 -> 1,00  OK
2 -> 2,00  OK
3.4 -> 3,40  OK
-5.678 -> -5,68  OK
612345 -> 612.345,00  OK
612345.555 -> 612.345,56  OK
987654321.067 -> 987.654.321,07  OK
________________________________________________________________________________
>>> 0b
66 -> 66,00  OK
1 -> 1,00  OK
2 -> 2,00  OK
3.4 -> 3,40  OK
-5.678 -> 5,68  error!
612345 -> 612.345,00  OK
612345.555 -> 612.345,56  OK
987654321.067 -> 9876.54.321,07  error!
________________________________________________________________________________
>>> 1
66 -> 66,00  OK
1 -> 1,00  OK
2 -> 2,00  OK
3.4 -> 3,40  OK
-5.678 -> -5,68  OK
612345 -> 612.345,00  OK
612345.555 -> 612.345,56  OK
987654321.067 -> 987.654.321,07  OK
________________________________________________________________________________
>>> 2
66 -> 66,00  OK
1 -> 1,00  OK
2 -> 2,00  OK
3.4 -> 3,40  OK
-5.678 -> -5,68  OK
612345 -> 612.345,00  OK
612345.555 -> 612.345,56  OK
987654321.067 -> 987.654.321,07  OK
________________________________________________________________________________
>>> 3
66 -> 66.00  error!
1 -> 1.00  error!
2 -> 2.00  error!
3.4 -> 3.40  error!
-5.678 -> -5.68  error!
612345 -> 612,345.00  error!
612345.555 -> 612,345.56  error!
987654321.067 -> 987,654,321.07  error!
________________________________________________________________________________
>>> 4
66 -> 66  error!
1 -> 1  error!
2 -> 2  error!
3.4 -> 3.4  error!
-5.678 -> -5..678  error!
612345 -> 612.345  error!
612345.555 -> 6.123.45..555  error!
987654321.067 -> 9.876.543.21..067  error!
________________________________________________________________________________
>>> 5
66 -> 66,00  OK
1 -> 1,00  OK
2 -> 2,00  OK
3.4 -> 3,40  OK
-5.678 -> -5,68  OK
612345 -> 612.345,00  OK
612345.555 -> 612.345,56  OK
987654321.067 -> 987.654.321,07  OK
________________________________________________________________________________
>>> 6
66 -> 66  error!
1 -> 1  error!
2 -> 2  error!
3.4 -> 3.4  error!
-5.678 -> -5..678  error!
612345 -> 612.345  error!
612345.555 -> 6.123.45..555  error!
987654321.067 -> 9.876.543.21..067  error!
________________________________________________________________________________
>>> format_numer_locale
66 -> 66.0  error!
1 -> 1.0  error!
2 -> 2.0  error!
3.4 -> 3.4  error!
-5.678 -> -5.7  error!
612345 -> 612345.0  error!
612345.555 -> 612345.6  error!
987654321.067 -> 987654321.1  error!
================================================================================
0 0.953s
0b 0.703s
1 1.234s
2 1.500s
3 1.407s
4 1.156s
5 0.765s
6 0.875s
format_numer_locale 1.797s

Code: Alles auswählen

def format_number_4(number, comma=".", thousand=",", format="%.2f"):
    number = format % number
    minus = 0
    if number < 0:
        minus = 1
        number = abs(number)
    n = str(number).split(comma)
    n[0] = list(str(n[0]))[::-1]
    n[0] = comma.join(
            ["".join(n[0][i:i+3][::-1]) for i in
             range(0,len(n[0]),3)][::-1])
    if minus:
        return "%s%s" % ('-', thousand.join(n))
    else: return thousand.join(n)

def format_number_6(number, comma=".", thousand=",", format="%.2f"):
    """
    by mawe
    http://www.python-forum.de/viewtopic.php?p=20850#20850
    """
    number = format % number
    minus = 0
    if number < 0:
        minus = 1
        number = abs(number)
    n = str(number).split(comma)
    x = len(n[0]) % 3
    n[0] = "%s%s" % ("0" * (3-x), n[0])
    n[0] = comma.join([n[0][i:i+3]
              for i in range(0,len(n[0]),3)]).lstrip("0%s" % comma)
    if minus: return "%s%s" % ('-', thousand.join(n))
    else: return thousand.join(n)

Code: Alles auswählen

s = s and decChar + s or s 

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s = decChar + (format % (abs(number) % 1))[2:]

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def format_number_5(number, format = "%.2f", decChar = ",", groupChar = ".", groupRange = 3):
    """
    by HarryH
    http://www.python-forum.de/viewtopic.php?p=20888#20888

    Convert to required format
    number =      number to convert
    format =      python string formatting (%),but only for the decimals
    decChar =     decimal char for the converted format
    groupChar =   group char for the converted format
    groupRange =  group range for the converted format

    For example:
    __formatter(1234567890.987, "%.2f", ",", ".", 3)
    ==> 1.234.567.890,99

    """
    if abs(number) < 1000:
        return (format % (number)).replace(".", decChar)
    else:
        base = int(abs(number))
        s = decChar + (format % (abs(number) % 1))[2:]
        while base:
            s = groupChar + str(base)[-groupRange:] + s
            base = int(base / 10**groupRange)
        return number < 0 and "-" + s[1:] or s[1:]

Code: Alles auswählen

if abs(number) < 10**groupRange:

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s = s and decChar + s or s

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def format_number_5(number, format = "%.2f", decChar = ",", groupChar = ".", groupRange = 3):
    """
    by HarryH & mawe
    http://www.python-forum.de/viewtopic.php?p=20943#20943

    Convert to required format
    number =      number to convert
    format =      python string formatting (%),but only for the decimals
    decChar =     decimal char for the converted format
    groupChar =   group char for the converted format
    groupRange =  group range for the converted format

    For example:
    __formatter(1234567890.987, "%.2f", ",", ".", 3)
    ==> 1.234.567.890,99

    """
    if abs(number) < 10**groupRange:
        return (format % (number)).replace(".", decChar)
    else:
        base = int(abs(number))
        s = (format % (abs(number) % 1))[2:]
        s = s and decChar + s or s
        while base:
            s = groupChar + str(base)[-groupRange:] + s
            base = int(base / 10**groupRange)
        return number < 0 and "-" + s[1:] or s[1:]

Code: Alles auswählen

0 0.953s
0b 0.703s
1 1.266s
2 1.500s
3 1.422s
4 1.297s
5 0.687s
6 1.032s
format_numer_locale 1.843s

Code: Alles auswählen

import string

def format_number_0(number,format="%0.2f",comma=",",thousand=".",grouplength=3):
    if abs(number) < 10**grouplength: 
        return (format % (number)).replace(".", comma) 
    if number<0:
         vorzeichen="-"
         number=-number
    else:
        vorzeichen=""
    if format.endswith("f"):
        hinter_komma=string.split(format % number,".")[1]
        full_string=comma+hinter_komma
    elif format.endswith("i"):
        full_string=""
    number=int(number)
    x=1
    while (10**grouplength)**x<=number:
        full_string=thousand+str(number%((10**grouplength)**x))[0:grouplength]+full_string
        x+=1
    full_string=vorzeichen+str(number/((10**grouplength)**(x-1)))+full_string
    return full_string

#----------------------------------------------------------------------

def format_number_0b(number,format="%0.2f",comma=",",thousand=".",grouplength=3):
    if abs(number) < 10**grouplength: 
        return (format % (number)).replace(".", comma) 
    if number<0:
         vorzeichen="-"
         number=-number
    else:
        vorzeichen=""

    vor_komma,hinter_komma=string.split(format % number,".")
    add=0
    for i in range(grouplength,len(vor_komma),grouplength):
        vor_komma=vor_komma[0:-(i+add)]+thousand+vor_komma[-(i+add):]
    return vor_komma+comma+hinter_komma

Code: Alles auswählen

import string


#----------------------------------------------------------------------

def format_number_0b(number,format="%0.2f",comma=",",thousand=".",grouplength=3):
    if abs(number) < 10**grouplength: 
        return (format % (number)).replace(".", comma) 
    if number<0:
         vorzeichen="-"
         number=-number
    else:
        vorzeichen=""

    vor_komma,hinter_komma=string.split(format % number,".")
    add=0
    for i in range(grouplength,len(vor_komma),grouplength):
        vor_komma=vor_komma[0:-(i+add)]+thousand+vor_komma[-(i+add):]
    return vor_komma+comma+hinter_komma


#----------------------------------------------------------------------


def format_number_5(number, format = "%.2f", decChar = ",", groupChar = ".", groupRange = 3): 
    """ 
    by HarryH & mawe 
    http://www.python-forum.de/viewtopic.php?p=20943#20943 

    Convert to required format 
    number =      number to convert 
    format =      python string formatting (%),but only for the decimals 
    decChar =     decimal char for the converted format 
    groupChar =   group char for the converted format 
    groupRange =  group range for the converted format 

    For example: 
    __formatter(1234567890.987, "%.2f", ",", ".", 3) 
    ==> 1.234.567.890,99 

    """ 
    if abs(number) < 10**groupRange: 
        return (format % (number)).replace(".", decChar) 
    else: 
        base = int(abs(number)) 
        s = (format % (abs(number) % 1))[2:] 
        s = s and decChar + s or s 
        while base: 
            s = groupChar + str(base)[-groupRange:] + s 
            base = int(base / 10**groupRange) 
        return number < 0 and "-" + s[1:] or s[1:] 

if __name__ == "__main__": 
    # Testen welche Funktion schneller ist 

    import time 

    zahlen = ( 
      66, 
      1, 
      2, 
      3.4, 
      -5.678, 
      612345, 
      612345.555, 
      987654321.067 
    )

    anz=100000
    for x in range(2):

        begin = time.time()

        for i in range(anz): 
          for zahl in zahlen: 
             x = format_number_0b(zahl)

        print "0b:", time.time() - begin 
        print 

        begin = time.time() 

        for i in range(anz): 
          for zahl in zahlen: 
             x = format_number_5(zahl) 
              
        print "5:", time.time() - begin 
        print 

Code: Alles auswählen

0b: 9.78099989891

5: 10.390999794

0b: 9.81299996376

5: 10.375

Code: Alles auswählen

# -*- coding: cp1252 -*-
#format_number.py
import string


#----------------------------------------------------------------------

def format_number_0b(number,format="%0.2f",comma=",",thousand=".",grouplength=3):
    if abs(number) < 10**grouplength: 
        return (format % (number)).replace(".", comma)
    if format[-1]=="f":
        vor_komma,hinter_komma=string.split(format % number,".")
    else:
        vor_komma=format % number
        comma=""
        hinter_komma=""
    #Hier
    anz_leer=0
    for i in vor_komma:
        if i==" ":
            anz_leer+=1
        else:
            break
    vor_komma=vor_komma[anz_leer:]
    #bis hier
        
    len_vor_komma=len(vor_komma)
    for i in range(grouplength,len_vor_komma+(len_vor_komma-1)/(grouplength+1)-(number<0),grouplength+1):
        vor_komma=vor_komma[0:-(i)]+thousand+vor_komma[-(i):]
    return anz_leer*" "+vor_komma+comma+hinter_komma


#----------------------------------------------------------------------


def format_number_5(number, format = "%.2f", decChar = ",", groupChar = ".", groupRange = 3): 
    """ 
    by HarryH & mawe 
    http://www.python-forum.de/viewtopic.php?p=20943#20943 

    Convert to required format 
    number =      number to convert 
    format =      python string formatting (%),but only for the decimals 
    decChar =     decimal char for the converted format 
    groupChar =   group char for the converted format 
    groupRange =  group range for the converted format 

    For example: 
    __formatter(1234567890.987, "%.2f", ",", ".", 3) 
    ==> 1.234.567.890,99 

    """ 
    if abs(number) < 10**groupRange: 
        return (format % (number)).replace(".", decChar) 
    else: 
        base = int(abs(number)) 
        s = (format % (abs(number) % 1))[2:]
        s = s and decChar + s or s
        while base:
            s = groupChar + str(base)[-groupRange:] + s 
            base = int(base / 10**groupRange)
        return number < 0 and "-" + s[1:] or s[1:] 

if __name__ == "__main__": 
    # Testen welche Funktion schneller ist 

    import time 

    zahlen = ( 
      66, 
      1, 
      2, 
      3.4, 
      -5.678,
      -123456789.348,
      612345, 
      612345.555, 
      987654321.067,
      8756832640,
      354,
      816523
    )

    formate = (
        "%.2f",
        "%30.3f",
        "%10i"
        )

    groups = (
        2,
        3,
        4,
        5
        )
    
    anz=100000
    for x in range(2):

        begin = time.time()

        for i in range(1000):
            for format in formate: 
              for zahl in zahlen:
                  for group in groups:
                     x = format_number_0b(zahl,format=format,grouplength=group)

        print "0b:", time.time() - begin 
        print 

        begin = time.time() 

        for i in range(1000):
            for format in formate: 
              for zahl in zahlen:
                 for group in groups:
                     x = format_number_5(zahl,format=format,groupRange=group)
              
        print "5:", time.time() - begin 
        print 

Code: Alles auswählen

0b: 2.5150001049

5: 2.03099989891

0b: 2.48399996758

5: 2.03100013733

Code: Alles auswählen

format =      python string formatting (%),but only for the decimals 

Code: Alles auswählen

def format_number_5(number, format = "%.2f", decChar = ",", groupChar = ".", groupRange = 3): 
    """ 
    by HarryH & mawe 
    http://www.python-forum.de/viewtopic.php?p=20943#20943 

    Convert to required format 
    number =      number to convert 
    format =      python string formatting (%),but only for the decimals 
    decChar =     decimal char for the converted format 
    groupChar =   group char for the converted format 
    groupRange =  group range for the converted format 

    For example: 
    __formatter(1234567890.987, "%.2f", ",", ".", 3) 
    ==> 1.234.567.890,99 

    """ 
    if abs(number) < 10**groupRange: 
        return (format % (number)).replace(".", decChar) 
    else: 
        base = int(abs(number)) 
        s = (format % (abs(number) % 1))[2:]
        s = s and decChar + s or s
        while base:
            s = groupChar + str(base)[-groupRange:] + s 
            base = int(base / 10**groupRange)
        if format[1]!="." or format[1]!="0":
            if format.endswith("f"):
                new_format=string.split(format,".")[0]+"s"
            else:
                new_format=format[0:-1]+"s"
            return new_format%(number < 0 and "-" + s[1:] or s[1:])
        else:
            return number < 0 and "-" + s[1:] or s[1:] 

Code: Alles auswählen

def __formatter(number, format = "%.2f", decChar = ",", groupChar = ".", groupRange = 3, format_divider = ["^%(0?)([-+# ]*\d*)(.*)"]):
    """
    Convert to required format
    number =         number to convert
    format =         python string formatting (%)
    decChar =        decimal char for the converted format
    groupChar =      group char for the converted format
    groupRange =     group range for the converted format
    format_divider = split the format for exact formatting ( => do not change!)

    For example:
    __formatter(1234567890.987, "%020.2f", ",", ".", 3)
    ==> 00001.234.567.890,99

    """
    if abs(number) < 10**groupRange:
        return (format % (number)).replace(".", decChar)
    else:
        # set formatting strings by first calling of this function
        if type(format_divider[0]) == str:
            result = re.findall(format_divider[0], format)[0]
            format_divider[0] = result[0], "%0" + result[1] + "s", "%" + result[2]
        zeros, base_format, tail_format = format_divider[0]
        # formatting the number
        base = int(abs(number))
        tail = (tail_format % (abs(number) % 1))[2:]
        tail = tail and decChar + tail or tail
        s = ""
        while base:
            s = groupChar + str(base)[-groupRange:] + s
            base = int(base / 10**groupRange)
        # return with left filled zeros
        if zeros == "0":
            return number < 0 and base_format % ("-" + (s[1:] + tail).zfill(int(base_format[1:-1])-1)) \
                    or base_format % ((s[1:] + tail).zfill(int(base_format[1:-1])))
        # return without left filled zeros
        else:
            return number < 0 and base_format % ("-" + s[1:] + tail) or base_format % (s[1:] + tail)

Code: Alles auswählen

def format_number_0b(number,format="%0.2f",comma=",",thousand=".",grouplength=3):
    if abs(number) < 10**grouplength: 
        return (format % (number)).replace(".", comma)
    if format[-1]=="f":
        vor_komma,hinter_komma=(format % number).split(".",-1)
    else:
        vor_komma=format % number
        comma=""
        hinter_komma=""
    #Hier
    anz_leer=0
    for i in vor_komma:
        if i==" ":
            anz_leer+=1
        else:
            break
    vor_komma=vor_komma[anz_leer:]
    #bis hier

    len_vor_komma=len(vor_komma)
    for i in range(grouplength,len_vor_komma+(len_vor_komma-1)/(grouplength+1)-(number<0),grouplength+1):
        vor_komma=vor_komma[0:-(i)]+thousand+vor_komma[-(i):]
    return anz_leer*" "+vor_komma+comma+hinter_komma

Code: Alles auswählen

class test:
    def formatter( number, format="%0.2f", comma=",", thousand=".", grouplength=3):
        """
        Formatierung für Zahlen
        s. http://www.python-forum.de/viewtopic.php?t=371
        """
        if abs(number) < 10**grouplength:
            return (format % (number)).replace(".", comma)
        if format[-1]=="f":
            vor_komma,hinter_komma=(format % number).split(".",-1)
        else:
            vor_komma=format % number
            comma=""
            hinter_komma=""
        #Hier
        anz_leer=0
        for i in vor_komma:
            if i==" ":
                anz_leer+=1
            else:
                break
        vor_komma=vor_komma[anz_leer:]
        #bis hier

        len_vor_komma=len(vor_komma)
        for i in range(grouplength,len_vor_komma+(len_vor_komma-1)/(grouplength+1)-(number<0),grouplength+1):
            vor_komma=vor_komma[0:-(i)]+thousand+vor_komma[-(i):]
        return anz_leer*" "+vor_komma+comma+hinter_komma

print test().formatter( 1028.328 )