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?
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
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
其他回答
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'
“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。
简单的两行字怎么样:
def humanizeFileSize(filesize):
p = int(math.floor(math.log(filesize, 2)/10))
return "%.3f%s" % (filesize/math.pow(1024,p), ['B','KiB','MiB','GiB','TiB','PiB','EiB','ZiB','YiB'][p])
下面是它的工作原理:
计算log2(文件大小) 除以10得到最接近的单位。(例如,如果大小是5000字节,最接近的单位是Kb,所以答案应该是X KiB) 返回file_size/value_of_closest_unit和unit。
然而,如果filesize为0或负(因为log对于0和-ve数字是未定义的),它就不起作用。你可以为他们添加额外的检查:
def humanizeFileSize(filesize):
filesize = abs(filesize)
if (filesize==0):
return "0 Bytes"
p = int(math.floor(math.log(filesize, 2)/10))
return "%0.2f %s" % (filesize/math.pow(1024,p), ['Bytes','KiB','MiB','GiB','TiB','PiB','EiB','ZiB','YiB'][p])
例子:
>>> humanizeFileSize(538244835492574234)
'478.06 PiB'
>>> humanizeFileSize(-924372537)
'881.55 MiB'
>>> humanizeFileSize(0)
'0 Bytes'
—Kb与KiB有区别。KB表示1000字节,而KiB表示1024字节。KB、MB、GB都是1000的倍数,KiB、MiB、GiB等都是1024的倍数。更多信息请点击这里
现代Django有自模板标签filesizeformat:
将值格式化为人类可读的文件大小(即:“13 KB”,“4.1 MB”,“102字节”等)。
例如:
{{ value|filesizeformat }}
如果值是123456789,输出将是117.7 MB。
更多信息:https://docs.djangoproject.com/en/1.10/ref/templates/builtins/#filesizeformat
我喜欢senderle的十进制版本的固定精度,所以这里有一种与上面joctee的答案的混合(你知道你可以取非整数底数的对数吗?):
from math import log
def human_readable_bytes(x):
# hybrid of https://stackoverflow.com/a/10171475/2595465
# with https://stackoverflow.com/a/5414105/2595465
if x == 0: return '0'
magnitude = int(log(abs(x),10.24))
if magnitude > 16:
format_str = '%iP'
denominator_mag = 15
else:
float_fmt = '%2.1f' if magnitude % 3 == 1 else '%1.2f'
illion = (magnitude + 1) // 3
format_str = float_fmt + ['', 'K', 'M', 'G', 'T', 'P'][illion]
return (format_str % (x * 1.0 / (1024 ** illion))).lstrip('0')