Number Format function

[Edward Hartfield]

I'm am relatively new to Python but use it daily. Today, I went looking
for a function, like PHP's number_function , that will take a number and
return a string with number formatted with grouped thousands and the
decimal portion rounded to a given number of places. This is certainly
needed when you want to take a floating-point value from a database and
display it as currency, for instance. I could not find what I was
looking for so I wrote the following function. I hope either (a) I've
provided something useful or (b) someone can tell me what I *should*
have done! Thanks.

import math

def number_format(n um, places=0):
"""Format a number with grouped thousands and given decimal places"""

#is_negative = (num < 0)
#if is_negative:
# num = -num

places = max(0,places)
tmp = "%.*f" % (places, num)
point = tmp.find(".")
integer = (point == -1) and tmp or tmp[:point]
decimal = (point != -1) and tmp[point:] or ""

count = commas = 0
formatted = []
for i in range(len(integ er) - 1, 0, -1):
count += 1
formatted.appen d(integer[i])
if count % 3 == 0:
formatted.appen d(",")

integer = "".join(formatt ed[::-1])
return integer+decimal

[wittempj]

Your code has a little bug, I highly recommend to add a test to your
code, for an idea see below - I fixed your code as well.

#!/usr/bin/env python
import math

def number_format(n um, places=0):
"""Format a number with grouped thousands and given decimal
places"""
#is_negative = (num < 0)
#if is_negative:
# num = -num

places = max(0,places)
tmp = "%.*f" % (places, num)
point = tmp.find(".")
integer = (point == -1) and tmp or tmp[:point]
decimal = (point != -1) and tmp[point:] or ""

count = commas = 0
formatted = []
for i in range(len(integ er) - 1, 0, -1):
count += 1
formatted.appen d(integer[i])
if count % 3 == 0:
formatted.appen d(",")
formatted.appen d(integer[0]) # this misses in your part
integer = "".join(formatt ed[::-1])
return integer+decimal
#
# add something like this: it helps to prevent you break your code
#
import unittest

class test_number_for mat(unittest.Te stCase):
def test(self):
self.assertEqua l(number_format (1000000, 2), '1,000,000.00')
self.assertEqua l(number_format (100000, 2), '100,000.00')
self.assertEqua l(number_format (100, 2), '100.00')
self.assertEqua l(number_format (1000000.33, 2), '1,000,000.33')
self.assertEqua l(number_format (1000000.333, 2), '1,000,000.33')
self.assertEqua l(number_format (1000000.3, 2), '1,000,000.30')
self.assertEqua l(number_format (123456, 2), '123,456.00')
self.assertEqua l(number_format (12345, 2), '12,345.00')
self.assertEqua l(number_format (123, 2), '123.00')
self.assertEqua l(number_format (123456.33, 2), '123,456.33')
self.assertEqua l(number_format (12345.333, 2), '12,345.33')
self.assertEqua l(number_format (123.3, 2), '123.30')

suite = unittest.makeSu ite(test_number _format)
unittest.TextTe stRunner(verbos ity=2).run(suit e)

[Edward Hartfield]

Thanks. I noticed the bugs later. But after talking with my boss, he
suggested something more elegant (again *untested*, yet):

import locale

def number_format(n um, places=0)
"""Format a number according to locality and given places"""
locale.setlocal e(locale.LC_ALL , locale.getdefau ltlocale()[0])
return locale.format(" %.*f", (places, num), 1)

wi******@hotmai l.com wrote:

Your code has a little bug, I highly recommend to add a test to your
code, for an idea see below - I fixed your code as well.

#!/usr/bin/env python
import math

def number_format(n um, places=0):
"""Format a number with grouped thousands and given decimal
places"""
#is_negative = (num < 0)
#if is_negative:
# num = -num

places = max(0,places)
tmp = "%.*f" % (places, num)
point = tmp.find(".")
integer = (point == -1) and tmp or tmp[:point]
decimal = (point != -1) and tmp[point:] or ""

count = commas = 0
formatted = []
for i in range(len(integ er) - 1, 0, -1):
count += 1
formatted.appen d(integer[i])
if count % 3 == 0:
formatted.appen d(",")
formatted.appen d(integer[0]) # this misses in your part
integer = "".join(formatt ed[::-1])
return integer+decimal
#
# add something like this: it helps to prevent you break your code
#
import unittest

class test_number_for mat(unittest.Te stCase):
def test(self):
self.assertEqua l(number_format (1000000, 2), '1,000,000.00')
self.assertEqua l(number_format (100000, 2), '100,000.00')
self.assertEqua l(number_format (100, 2), '100.00')
self.assertEqua l(number_format (1000000.33, 2), '1,000,000.33')
self.assertEqua l(number_format (1000000.333, 2), '1,000,000.33')
self.assertEqua l(number_format (1000000.3, 2), '1,000,000.30')
self.assertEqua l(number_format (123456, 2), '123,456.00')
self.assertEqua l(number_format (12345, 2), '12,345.00')
self.assertEqua l(number_format (123, 2), '123.00')
self.assertEqua l(number_format (123456.33, 2), '123,456.33')
self.assertEqua l(number_format (12345.333, 2), '12,345.33')
self.assertEqua l(number_format (123.3, 2), '123.30')

suite = unittest.makeSu ite(test_number _format)
unittest.TextTe stRunner(verbos ity=2).run(suit e)

[wittempj]

This is a little faster:

def number_format(n um, places=0):
"""Format a number according to locality and given places"""
locale.setlocal e(locale.LC_ALL , "")
return locale.format(" %.*f", (places, num), True)

I tested this ok with my test

[Rick Zantow]

"wi******@hotma il.com" <ma**********@g mail.com> wrote in
news:11******** **************@ f14g2000cwb.goo glegroups.com:

def number_format(n um, places=0):
"""Format a number according to locality and given places"""
locale.setlocal e(locale.LC_ALL , "")
return locale.format(" %.*f", (places, num), True)

There are some edge cases in the format conversion that could present
some issues. For example:

print number_format( 2312753.4450000 0, 2 ) 2,312,753.44 print number_format( 2312753.4450000 1, 2 )

2,312,753.45

I would expect the first to produce the same results as the second, but,
I suppose because of one of floating point's features, it doesn't work
that way.

--
rzed

[Sion Arrowsmith]

Rick Zantow <rz*****@gmail. com> wrote:

print number_format( 2312753.4450000 0, 2 )2,312,753.44 print number_format( 2312753.4450000 1, 2 )2,312,753.45

I would expect the first to produce the same results as the second, but,
I suppose because of one of floating point's features, it doesn't work
that way.

2312753.4450000 0 2312753.4449999 998 2312753.4450000 1

2312753.4450000 101

So, yeah, the nature of floating points is going to make that
first one round "unexpected ly".

--
\S -- si***@chiark.gr eenend.org.uk -- http://www.chaos.org.uk/~sion/
___ | "Frankly I have no feelings towards penguins one way or the other"
\X/ | -- Arthur C. Clarke
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