better formatThousands
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parent
ed10852e4e
commit
03607ad139
1 changed files with 17 additions and 21 deletions
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@ -48,14 +48,28 @@ def floatValue(strValue, default=''):
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def formatNumber(number, longName, shortName):
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"""
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Return the number in a human-readable format (23 KB, 42.7 MB, 68.77 GB)
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Return the number in a human-readable format (23 KB, 23.4 MB, 23.42 GB)
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"""
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if number < 1024:
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return "%d %s%s" % (number, longName, number != 1 and 's' or '')
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return '%s %s%s' % (formatThousands(number), longName, number != 1 and 's' or '')
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prefix = ['K', 'M', 'G', 'T', 'P']
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for i in range(5):
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if number < math.pow(1024, i + 2) or i == 4:
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return '%.*f %s%s' % (i, bytes / math.pow(1024, i + 1), prefix[i], shortName)
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n = number / math.pow(1024, i + 1)
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return '%s %s%s' % (formatThousands('%.*f' % (i, n)), prefix[i], shortName)
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def formatThousands(number, separator = ','):
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"""
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Return the number with separators (1,000,000)
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"""
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string = str(number).split('.')
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l = []
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for i, character in enumerate(reversed(string[0])):
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if i and (not (i % 3)):
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l.insert(0, separator)
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l.insert(0, character)
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string[0] = ''.join(l)
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return '.'.join(string)
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def formatBits(number):
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return formatNumber(number, 'bit', 'b')
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@ -66,24 +80,6 @@ def formatBytes(number):
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def formatPixels(number):
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return formatNumber(number, 'pixel', 'px')
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'''
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seperate number with thousand comma
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'''
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def numberThousands(n, sep=','):
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if n < 1000:
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return "%s" % n
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ln = list(str(n))
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ln.reverse()
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newn = []
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while len(ln) > 3:
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newn.extend(ln[:3])
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newn.append(sep)
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ln = ln[3:]
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newn.extend(ln)
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newn.reverse()
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return "".join(newn)
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def plural(amount, unit, plural='s'):
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if abs(amount) != 1:
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if plural == 's':
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