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
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)
其他回答
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'
为了以人类可读的形式获取文件大小,我创建了这个函数:
import os
def get_size(path):
size = os.path.getsize(path)
if size < 1024:
return f"{size} bytes"
elif size < pow(1024,2):
return f"{round(size/1024, 2)} KB"
elif size < pow(1024,3):
return f"{round(size/(pow(1024,2)), 2)} MB"
elif size < pow(1024,4):
return f"{round(size/(pow(1024,3)), 2)} GB"
>>> get_size("a.txt")
1.4KB
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'
这个解决方案可能也会吸引你,这取决于你的思维方式:
from pathlib import Path
def get_size(path = Path('.')):
""" Gets file size, or total directory size """
if path.is_file():
size = path.stat().st_size
elif path.is_dir():
size = sum(file.stat().st_size for file in path.glob('*.*'))
return size
def format_size(path, unit="MB"):
""" Converts integers to common size units used in computing """
bit_shift = {"B": 0,
"kb": 7,
"KB": 10,
"mb": 17,
"MB": 20,
"gb": 27,
"GB": 30,
"TB": 40,}
return "{:,.0f}".format(get_size(path) / float(1 << bit_shift[unit])) + " " + unit
# Tests and test results
>>> get_size("d:\\media\\bags of fun.avi")
'38 MB'
>>> get_size("d:\\media\\bags of fun.avi","KB")
'38,763 KB'
>>> get_size("d:\\media\\bags of fun.avi","kb")
'310,104 kb'
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