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
这是我为另一个问题写的东西……
与xApple的答案非常相似,该对象总是以人类可读的格式打印。不同的是,它也是一个适当的int,所以你可以用它做数学! 它将格式说明符直接传递给数字格式,并附加后缀,因此几乎可以保证请求的长度将超出两到三个字符。我从来没有使用过这个代码,所以我没有费心去修复它!
class ByteSize(int):
_KB = 1024
_suffixes = 'B', 'KB', 'MB', 'GB', 'PB'
def __new__(cls, *args, **kwargs):
return super().__new__(cls, *args, **kwargs)
def __init__(self, *args, **kwargs):
self.bytes = self.B = int(self)
self.kilobytes = self.KB = self / self._KB**1
self.megabytes = self.MB = self / self._KB**2
self.gigabytes = self.GB = self / self._KB**3
self.petabytes = self.PB = self / self._KB**4
*suffixes, last = self._suffixes
suffix = next((
suffix
for suffix in suffixes
if 1 < getattr(self, suffix) < self._KB
), last)
self.readable = suffix, getattr(self, suffix)
super().__init__()
def __str__(self):
return self.__format__('.2f')
def __repr__(self):
return '{}({})'.format(self.__class__.__name__, super().__repr__())
def __format__(self, format_spec):
suffix, val = self.readable
return '{val:{fmt}} {suf}'.format(val=val, fmt=format_spec, suf=suffix)
def __sub__(self, other):
return self.__class__(super().__sub__(other))
def __add__(self, other):
return self.__class__(super().__add__(other))
def __mul__(self, other):
return self.__class__(super().__mul__(other))
def __rsub__(self, other):
return self.__class__(super().__sub__(other))
def __radd__(self, other):
return self.__class__(super().__add__(other))
def __rmul__(self, other):
return self.__class__(super().__rmul__(other))
用法:
>>> size = 6239397620
>>> print(size)
5.81 GB
>>> size.GB
5.810891855508089
>>> size.gigabytes
5.810891855508089
>>> size.PB
0.005674699077644618
>>> size.MB
5950.353260040283
>>> size
ByteSize(6239397620)
其他回答
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。
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。
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'
def human_readable_data_quantity(quantity, multiple=1024):
if quantity == 0:
quantity = +0
SUFFIXES = ["B"] + [i + {1000: "B", 1024: "iB"}[multiple] for i in "KMGTPEZY"]
for suffix in SUFFIXES:
if quantity < multiple or suffix == SUFFIXES[-1]:
if suffix == SUFFIXES[0]:
return "%d%s" % (quantity, suffix)
else:
return "%.1f%s" % (quantity, suffix)
else:
quantity /= multiple
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}"