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

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

这里是一个在线lambda,没有任何导入来转换为人类可读的文件大小。以字节为单位传递值。

to_human = lambda v : str(v >> ((max(v.bit_length()-1, 0)//10)*10)) +["", "K", "M", "G", "T", "P", "E"][max(v.bit_length()-1, 0)//10]
>>> to_human(1024)
'1K'
>>> to_human(1024*1024*3)
'3M'

重复作为匆匆.filesize()替代方案提供的代码段,下面的代码段根据所使用的前缀给出不同的精度数字。它不像某些片段那样简洁,但我喜欢这样的结果。

def human_size(size_bytes):
    """
    format a size in bytes into a 'human' file size, e.g. bytes, KB, MB, GB, TB, PB
    Note that bytes/KB will be reported in whole numbers but MB and above will have greater precision
    e.g. 1 byte, 43 bytes, 443 KB, 4.3 MB, 4.43 GB, etc
    """
    if size_bytes == 1:
        # because I really hate unnecessary plurals
        return "1 byte"

    suffixes_table = [('bytes',0),('KB',0),('MB',1),('GB',2),('TB',2), ('PB',2)]

    num = float(size_bytes)
    for suffix, precision in suffixes_table:
        if num < 1024.0:
            break
        num /= 1024.0

    if precision == 0:
        formatted_size = "%d" % num
    else:
        formatted_size = str(round(num, ndigits=precision))

    return "%s %s" % (formatted_size, suffix)

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]

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"