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
如果你安装了Django,你也可以试试filesizeformat:
from django.template.defaultfilters import filesizeformat
filesizeformat(1073741824)
=>
"1.0 GB"
其他回答
根据之前所有的答案,以下是我的看法。它是一个以字节为单位以整数形式存储文件大小的对象。但是当你尝试打印对象时,你会自动得到一个人类可读的版本。
class Filesize(object):
"""
Container for a size in bytes with a human readable representation
Use it like this::
>>> size = Filesize(123123123)
>>> print size
'117.4 MB'
"""
chunk = 1024
units = ['bytes', 'KB', 'MB', 'GB', 'TB', 'PB']
precisions = [0, 0, 1, 2, 2, 2]
def __init__(self, size):
self.size = size
def __int__(self):
return self.size
def __str__(self):
if self.size == 0: return '0 bytes'
from math import log
unit = self.units[min(int(log(self.size, self.chunk)), len(self.units) - 1)]
return self.format(unit)
def format(self, unit):
if unit not in self.units: raise Exception("Not a valid file size unit: %s" % unit)
if self.size == 1 and unit == 'bytes': return '1 byte'
exponent = self.units.index(unit)
quotient = float(self.size) / self.chunk**exponent
precision = self.precisions[exponent]
format_string = '{:.%sf} {}' % (precision)
return format_string.format(quotient, unit)
如果你安装了Django,你也可以试试filesizeformat:
from django.template.defaultfilters import filesizeformat
filesizeformat(1073741824)
=>
"1.0 GB"
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
使用1000或kibibytes的幂将更符合标准:
def sizeof_fmt(num, use_kibibyte=True):
base, suffix = [(1000.,'B'),(1024.,'iB')][use_kibibyte]
for x in ['B'] + map(lambda x: x+suffix, list('kMGTP')):
if -base < num < base:
return "%3.1f %s" % (num, x)
num /= base
return "%3.1f %s" % (num, x)
附注:永远不要相信一个以K(大写)后缀打印数千的库。
一个拥有你所寻找的所有功能的库似乎是人性化的。Humanize.naturalsize()似乎可以做您所寻找的所有事情。
示例代码(python 3.10)
import humanize
disk_sizes_list = [1, 100, 999, 1000,1024, 2000,2048, 3000, 9999, 10000, 2048000000, 9990000000, 9000000000000000000000]
for size in disk_sizes_list:
natural_size = humanize.naturalsize(size)
binary_size = humanize.naturalsize(size, binary=True)
print(f" {natural_size} \t| {binary_size}\t|{size}")
输出
1 Byte | 1 Byte |1
100 Bytes | 100 Bytes |100
999 Bytes | 999 Bytes |999
1.0 kB | 1000 Bytes |1000
1.0 kB | 1.0 KiB |1024
2.0 kB | 2.0 KiB |2000
2.0 kB | 2.0 KiB |2048
3.0 kB | 2.9 KiB |3000
10.0 kB | 9.8 KiB |9999
10.0 kB | 9.8 KiB |10000
2.0 GB | 1.9 GiB |2048000000
10.0 GB | 9.3 GiB |9990000000
9.0 ZB | 7.6 ZiB |9000000000000000000000