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
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'
其他回答
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/
我最近提出了一个避免循环的版本,使用log2来确定大小顺序,作为后缀列表的移位和索引:
from math import log2
_suffixes = ['bytes', 'KiB', 'MiB', 'GiB', 'TiB', 'PiB', 'EiB', 'ZiB', 'YiB']
def file_size(size):
# determine binary order in steps of size 10
# (coerce to int, // still returns a float)
order = int(log2(size) / 10) if size else 0
# format file size
# (.4g results in rounded numbers for exact matches and max 3 decimals,
# should never resort to exponent values)
return '{:.4g} {}'.format(size / (1 << (order * 10)), _suffixes[order])
不过,它的可读性很可能被认为是非python化的。
以下工作在Python 3.6+中,在我看来,是这里最容易理解的答案,并允许您自定义使用的小数位数。
def human_readable_size(size, decimal_places=2):
for unit in ['B', 'KiB', 'MiB', 'GiB', 'TiB', 'PiB']:
if size < 1024.0 or unit == 'PiB':
break
size /= 1024.0
return f"{size:.{decimal_places}f} {unit}"
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'
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')