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)
这是我的版本。它不使用for循环。它具有常数复杂度O(1),理论上比这里使用for循环的答案更有效。
from math import log
unit_list = zip(['bytes', 'kB', 'MB', 'GB', 'TB', 'PB'], [0, 0, 1, 2, 2, 2])
def sizeof_fmt(num):
"""Human friendly file size"""
if num > 1:
exponent = min(int(log(num, 1024)), len(unit_list) - 1)
quotient = float(num) / 1024**exponent
unit, num_decimals = unit_list[exponent]
format_string = '{:.%sf} {}' % (num_decimals)
return format_string.format(quotient, unit)
if num == 0:
return '0 bytes'
if num == 1:
return '1 byte'
为了更清楚地说明发生了什么,我们可以省略字符串格式化的代码。以下是真正起作用的台词:
exponent = int(log(num, 1024))
quotient = num / 1024**exponent
unit_list[exponent]
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
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')
通过简单的实现(使用f-strings,所以Python 3.6+)解决上述“任务太小,不需要库”的问题:
def sizeof_fmt(num, suffix="B"):
for unit in ["", "Ki", "Mi", "Gi", "Ti", "Pi", "Ei", "Zi"]:
if abs(num) < 1024.0:
return f"{num:3.1f}{unit}{suffix}"
num /= 1024.0
return f"{num:.1f}Yi{suffix}"
支持:
所有当前已知的二进制前缀 负数和正数 大于1000约字节的数字 任意单位(也许你喜欢用吉比特来计数!)
例子:
>>> sizeof_fmt(168963795964)
'157.4GiB'
作者:Fred Cirera