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

下面是一个使用while的选项:

def number_format(n):
   n2, n3 = n, 0
   while n2 >= 1e3:
      n2 /= 1e3
      n3 += 1
   return '%.3f' % n2 + ('', ' k', ' M', ' G')[n3]

s = number_format(9012345678)
print(s == '9.012 G')

https://docs.python.org/reference/compound_stmts.html#while

其他回答

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

简单的两行字怎么样:

def humanizeFileSize(filesize):
    p = int(math.floor(math.log(filesize, 2)/10))
    return "%.3f%s" % (filesize/math.pow(1024,p), ['B','KiB','MiB','GiB','TiB','PiB','EiB','ZiB','YiB'][p])

下面是它的工作原理:

计算log2(文件大小) 除以10得到最接近的单位。(例如,如果大小是5000字节,最接近的单位是Kb,所以答案应该是X KiB) 返回file_size/value_of_closest_unit和unit。

然而,如果filesize为0或负(因为log对于0和-ve数字是未定义的),它就不起作用。你可以为他们添加额外的检查:

def humanizeFileSize(filesize):
    filesize = abs(filesize)
    if (filesize==0):
        return "0 Bytes"
    p = int(math.floor(math.log(filesize, 2)/10))
    return "%0.2f %s" % (filesize/math.pow(1024,p), ['Bytes','KiB','MiB','GiB','TiB','PiB','EiB','ZiB','YiB'][p])

例子:

>>> humanizeFileSize(538244835492574234)
'478.06 PiB'
>>> humanizeFileSize(-924372537)
'881.55 MiB'
>>> humanizeFileSize(0)
'0 Bytes'

—Kb与KiB有区别。KB表示1000字节,而KiB表示1024字节。KB、MB、GB都是1000的倍数,KiB、MiB、GiB等都是1024的倍数。更多信息请点击这里

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')

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)

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/