从字节大小返回人类可读大小的函数:

>>> human_readable(2048)
'2 kilobytes'
>>>

如何做到这一点?


当前回答

其中一个库是hurry.filesize。

>>> from hurry.filesize import alternative
>>> size(1, system=alternative)
'1 byte'
>>> size(10, system=alternative)
'10 bytes'
>>> size(1024, system=alternative)
'1 KB'

其他回答

该功能在Boltons中可用,这对于大多数项目来说都是一个非常方便的库。

>>> bytes2human(128991)
'126K'
>>> bytes2human(100001221)
'95M'
>>> bytes2human(0, 2)
'0.00B'

其中一个库是hurry.filesize。

>>> from hurry.filesize import alternative
>>> size(1, system=alternative)
'1 byte'
>>> size(10, system=alternative)
'10 bytes'
>>> size(1024, system=alternative)
'1 KB'

参考Sridhar Ratnakumar的回答,更新为:

def formatSize(sizeInBytes, decimalNum=1, isUnitWithI=False, sizeUnitSeperator=""):
  """format size to human readable string"""
  # https://en.wikipedia.org/wiki/Binary_prefix#Specific_units_of_IEC_60027-2_A.2_and_ISO.2FIEC_80000
  # K=kilo, M=mega, G=giga, T=tera, P=peta, E=exa, Z=zetta, Y=yotta
  sizeUnitList = ['','K','M','G','T','P','E','Z']
  largestUnit = 'Y'

  if isUnitWithI:
    sizeUnitListWithI = []
    for curIdx, eachUnit in enumerate(sizeUnitList):
      unitWithI = eachUnit
      if curIdx >= 1:
        unitWithI += 'i'
      sizeUnitListWithI.append(unitWithI)

    # sizeUnitListWithI = ['','Ki','Mi','Gi','Ti','Pi','Ei','Zi']
    sizeUnitList = sizeUnitListWithI

    largestUnit += 'i'

  suffix = "B"
  decimalFormat = "." + str(decimalNum) + "f" # ".1f"
  finalFormat = "%" + decimalFormat + sizeUnitSeperator + "%s%s" # "%.1f%s%s"
  sizeNum = sizeInBytes
  for sizeUnit in sizeUnitList:
      if abs(sizeNum) < 1024.0:
        return finalFormat % (sizeNum, sizeUnit, suffix)
      sizeNum /= 1024.0
  return finalFormat % (sizeNum, largestUnit, suffix)

示例输出如下:

def testKb():
  kbSize = 3746
  kbStr = formatSize(kbSize)
  print("%s -> %s" % (kbSize, kbStr))

def testI():
  iSize = 87533
  iStr = formatSize(iSize, isUnitWithI=True)
  print("%s -> %s" % (iSize, iStr))

def testSeparator():
  seperatorSize = 98654
  seperatorStr = formatSize(seperatorSize, sizeUnitSeperator=" ")
  print("%s -> %s" % (seperatorSize, seperatorStr))

def testBytes():
  bytesSize = 352
  bytesStr = formatSize(bytesSize)
  print("%s -> %s" % (bytesSize, bytesStr))

def testMb():
  mbSize = 76383285
  mbStr = formatSize(mbSize, decimalNum=2)
  print("%s -> %s" % (mbSize, mbStr))

def testTb():
  tbSize = 763832854988542
  tbStr = formatSize(tbSize, decimalNum=2)
  print("%s -> %s" % (tbSize, tbStr))

def testPb():
  pbSize = 763832854988542665
  pbStr = formatSize(pbSize, decimalNum=4)
  print("%s -> %s" % (pbSize, pbStr))


def demoFormatSize():
  testKb()
  testI()
  testSeparator()
  testBytes()
  testMb()
  testTb()
  testPb()

  # 3746 -> 3.7KB
  # 87533 -> 85.5KiB
  # 98654 -> 96.3 KB
  # 352 -> 352.0B
  # 76383285 -> 72.84MB
  # 763832854988542 -> 694.70TB
  # 763832854988542665 -> 678.4199PB

通过简单的实现(使用f-strings,所以Python 3.6+)解决上述“任务太小,不需要库”的问题:

def sizeof_fmt(num, suffix="B"):
    for unit in ["", "Ki", "Mi", "Gi", "Ti", "Pi", "Ei", "Zi"]:
        if abs(num) < 1024.0:
            return f"{num:3.1f}{unit}{suffix}"
        num /= 1024.0
    return f"{num:.1f}Yi{suffix}"

支持:

所有当前已知的二进制前缀 负数和正数 大于1000约字节的数字 任意单位(也许你喜欢用吉比特来计数!)

例子:

>>> sizeof_fmt(168963795964)
'157.4GiB'

作者:Fred Cirera

“人类友好”项目有助于解决这一问题。

import humanfriendly
humanfriendly.format_size(1024)

上面的代码将给出1KB的答案。 例子可以在这里找到。