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/

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'

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

如果有人想知道，要将@Sridhar Ratnakumar的答案转换回字节，您可以执行以下操作:

import math

def format_back_to_bytes(value):
    for power, unit in enumerate(["", "Ki", "Mi", "Gi", "Ti", "Pi", "Ei", "Zi"]):
        if value[-3:-1] == unit:
            return round(float(value[:-3])*math.pow(2, 10*power))

用法:

>>> format_back_to_bytes('212.4GiB')
228062763418

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)