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
你应该用humanize。
>>> humanize.naturalsize(1000000)
'1.0 MB'
>>> humanize.naturalsize(1000000, binary=True)
'976.6 KiB'
>>> humanize.naturalsize(1000000, gnu=True)
'976.6K'
参考: https://pypi.org/project/humanize/
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'
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'
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
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
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)