better formatThousands

This commit is contained in:
Rolux 2008-04-28 11:25:52 +02:00
parent ed10852e4e
commit 03607ad139

View file

@ -48,14 +48,28 @@ def floatValue(strValue, default=''):
def formatNumber(number, longName, shortName):
"""
Return the number in a human-readable format (23 KB, 42.7 MB, 68.77 GB)
Return the number in a human-readable format (23 KB, 23.4 MB, 23.42 GB)
"""
if number < 1024:
return "%d %s%s" % (number, longName, number != 1 and 's' or '')
return '%s %s%s' % (formatThousands(number), longName, number != 1 and 's' or '')
prefix = ['K', 'M', 'G', 'T', 'P']
for i in range(5):
if number < math.pow(1024, i + 2) or i == 4:
return '%.*f %s%s' % (i, bytes / math.pow(1024, i + 1), prefix[i], shortName)
n = number / math.pow(1024, i + 1)
return '%s %s%s' % (formatThousands('%.*f' % (i, n)), prefix[i], shortName)
def formatThousands(number, separator = ','):
"""
Return the number with separators (1,000,000)
"""
string = str(number).split('.')
l = []
for i, character in enumerate(reversed(string[0])):
if i and (not (i % 3)):
l.insert(0, separator)
l.insert(0, character)
string[0] = ''.join(l)
return '.'.join(string)
def formatBits(number):
return formatNumber(number, 'bit', 'b')
@ -66,24 +80,6 @@ def formatBytes(number):
def formatPixels(number):
return formatNumber(number, 'pixel', 'px')
'''
seperate number with thousand comma
'''
def numberThousands(n, sep=','):
if n < 1000:
return "%s" % n
ln = list(str(n))
ln.reverse()
newn = []
while len(ln) > 3:
newn.extend(ln[:3])
newn.append(sep)
ln = ln[3:]
newn.extend(ln)
newn.reverse()
return "".join(newn)
def plural(amount, unit, plural='s'):
if abs(amount) != 1:
if plural == 's':