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
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
其他回答
如果有人想知道,要将@Sridhar Ratnakumar的答案转换回字节,您可以执行以下操作:
import math
def format_back_to_bytes(value):
for power, unit in enumerate(["", "Ki", "Mi", "Gi", "Ti", "Pi", "Ei", "Zi"]):
if value[-3:-1] == unit:
return round(float(value[:-3])*math.pow(2, 10*power))
用法:
>>> format_back_to_bytes('212.4GiB')
228062763418
重复作为匆匆.filesize()替代方案提供的代码段,下面的代码段根据所使用的前缀给出不同的精度数字。它不像某些片段那样简洁,但我喜欢这样的结果。
def human_size(size_bytes):
"""
format a size in bytes into a 'human' file size, e.g. bytes, KB, MB, GB, TB, PB
Note that bytes/KB will be reported in whole numbers but MB and above will have greater precision
e.g. 1 byte, 43 bytes, 443 KB, 4.3 MB, 4.43 GB, etc
"""
if size_bytes == 1:
# because I really hate unnecessary plurals
return "1 byte"
suffixes_table = [('bytes',0),('KB',0),('MB',1),('GB',2),('TB',2), ('PB',2)]
num = float(size_bytes)
for suffix, precision in suffixes_table:
if num < 1024.0:
break
num /= 1024.0
if precision == 0:
formatted_size = "%d" % num
else:
formatted_size = str(round(num, ndigits=precision))
return "%s %s" % (formatted_size, suffix)
这将在几乎任何情况下做你需要做的事情,是可选参数自定义的,正如你所看到的,几乎是自文档化的:
from math import log
def pretty_size(n,pow=0,b=1024,u='B',pre=['']+[p+'i'for p in'KMGTPEZY']):
pow,n=min(int(log(max(n*b**pow,1),b)),len(pre)-1),n*b**pow
return "%%.%if %%s%%s"%abs(pow%(-pow-1))%(n/b**float(pow),pre[pow],u)
示例输出:
>>> pretty_size(42)
'42 B'
>>> pretty_size(2015)
'2.0 KiB'
>>> pretty_size(987654321)
'941.9 MiB'
>>> pretty_size(9876543210)
'9.2 GiB'
>>> pretty_size(0.5,pow=1)
'512 B'
>>> pretty_size(0)
'0 B'
高级定制:
>>> pretty_size(987654321,b=1000,u='bytes',pre=['','kilo','mega','giga'])
'987.7 megabytes'
>>> pretty_size(9876543210,b=1000,u='bytes',pre=['','kilo','mega','giga'])
'9.9 gigabytes'
此代码与Python 2和Python 3兼容。对读者来说,遵从PEP8是一个练习。记住,漂亮的是输出。
更新:
如果你需要数千个逗号,只需应用明显的扩展:
def prettier_size(n,pow=0,b=1024,u='B',pre=['']+[p+'i'for p in'KMGTPEZY']):
r,f=min(int(log(max(n*b**pow,1),b)),len(pre)-1),'{:,.%if} %s%s'
return (f%(abs(r%(-r-1)),pre[r],u)).format(n*b**pow/b**float(r))
例如:
>>> pretty_units(987654321098765432109876543210)
'816,968.5 YiB'
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
下面是一个使用while的选项:
def number_format(n):
n2, n3 = n, 0
while n2 >= 1e3:
n2 /= 1e3
n3 += 1
return '%.3f' % n2 + ('', ' k', ' M', ' G')[n3]
s = number_format(9012345678)
print(s == '9.012 G')
https://docs.python.org/reference/compound_stmts.html#while