157

I am using a library that reads a file and returns its size in bytes.

This file size is then displayed to the end user; to make it easier for them to understand it, I am explicitly converting the file size to MB by dividing it by 1024.0 * 1024.0. Of course this works, but I am wondering is there a better way to do this in Python?

By better, I mean perhaps a stdlib function that can manipulate sizes according to the type I want. Like if I specify MB, it automatically divides it by 1024.0 * 1024.0. Somethign on these lines.

8
  • 6
    So write one. Also note that many systems now use MB to mean 10^6 instead of 2^20. Commented Mar 4, 2011 at 13:08
  • 8
    @A A, @tc: Please keep in mind that the SI and IEC Norm is kB (Kilo) for 1.000 Byte and KiB (Kibi) for 1.024 Byte. See en.wikipedia.org/wiki/Kibibyte . Commented Mar 4, 2011 at 13:12
  • 2
    @Bobby: kB actually means "kilobel", equal to 10000 dB. There is no SI unit for byte. IIRC, the IEC recommends KiB but does not define kB or KB. Commented Mar 12, 2011 at 4:41
  • 2
    @tc. The prefix kilo is defined by SI to mean 1000. The IEC defined kB, etc. to use the SI prefix instead of 2^10. Commented Feb 11, 2013 at 23:03
  • 1
    I mean the prefixes are defined generally by SI, but the abbreviations for data size are not: physics.nist.gov/cuu/Units/prefixes.html. Those are defined by IEC: physics.nist.gov/cuu/Units/binary.html Commented Feb 19, 2013 at 16:40

12 Answers 12

281

Here is what I use:

import math

def convert_size(size_bytes):
   if size_bytes == 0:
       return "0B"
   size_name = ("B", "KB", "MB", "GB", "TB", "PB", "EB", "ZB", "YB")
   i = int(math.floor(math.log(size_bytes, 1024)))
   p = math.pow(1024, i)
   s = round(size_bytes / p, 2)
   return "%s %s" % (s, size_name[i])

NB : size should be sent in Bytes.

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6 Comments

If you're sending size in bytes then just add "B" as the first element of size_name.
When you have 0 sized byte of file, it fails. log(0, 1024) is not defined! You should check 0 byte case before this statement i = int(math.floor(math.log(size,1024))).
genclik - you're right. I've just submitted a minor edit which will fix this, and enable conversion from bytes. Thanks, Sapam, for the original
HI @WHK as tuxGurl mentioned its an easy fix.
Actually the size names would need to be ("B", "KiB", "MiB", "GiB", "TiB", "PiB", "EiB", "ZiB", "YiB"). See en.wikipedia.org/wiki/Mebibyte for more information.
|
152

There is hurry.filesize that will take the size in bytes and make a nice string out if it.

>>> from hurry.filesize import size
>>> size(11000)
'10K'
>>> size(198283722)
'189M'

Or if you want 1K == 1000 (which is what most users assume):

>>> from hurry.filesize import size, si
>>> size(11000, system=si)
'11K'
>>> size(198283722, system=si)
'198M'

It has IEC support as well (but that wasn't documented):

>>> from hurry.filesize import size, iec
>>> size(11000, system=iec)
'10Ki'
>>> size(198283722, system=iec)
'189Mi'

Because it's written by the Awesome Martijn Faassen, the code is small, clear and extensible. Writing your own systems is dead easy.

Here is one:

mysystem = [
    (1024 ** 5, ' Megamanys'),
    (1024 ** 4, ' Lotses'),
    (1024 ** 3, ' Tons'), 
    (1024 ** 2, ' Heaps'), 
    (1024 ** 1, ' Bunches'),
    (1024 ** 0, ' Thingies'),
    ]

Used like so:

>>> from hurry.filesize import size
>>> size(11000, system=mysystem)
'10 Bunches'
>>> size(198283722, system=mysystem)
'189 Heaps'

10 Comments

Hm, now I need one to go the other way. From "1 kb" to 1024 (an int).
Works only in python 2
This package might be cool but the odd license and the fact that there is no online available source code make it something I'd be very happy to avoid. And also it seems to support python2 only.
@AlmogCohen the source is online, available straight from PyPI (some packages do not have a Github repository, just a PyPI page) and the license is not that obscure, ZPL is the Zope Public License, which is, to the best of my knowledge, BSD-like. I do agree that the licensing itself is odd: there is no standard 'LICENSE.txt' file, nor is there a preamble at the top of each source file.
In order to get megabyte I did following equation using bitwise shifting operator: MBFACTOR = float(1 << 20); mb= int(size_in_bytes) / MBFACTOR @LennartRegebro
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62

Instead of a size divisor of 1024 * 1024 you could use the << bitwise shifting operator, i.e. 1<<20 to get megabytes, 1<<30 to get gigabytes, etc.

In the simplest scenario you can have e.g. a constant MBFACTOR = float(1<<20) which can then be used with bytes, i.e.: megas = size_in_bytes/MBFACTOR.

Megabytes are usually all that you need, or otherwise something like this can be used:

# bytes pretty-printing
UNITS_MAPPING = [
    (1<<50, ' PB'),
    (1<<40, ' TB'),
    (1<<30, ' GB'),
    (1<<20, ' MB'),
    (1<<10, ' KB'),
    (1, (' byte', ' bytes')),
]


def pretty_size(bytes, units=UNITS_MAPPING):
    """Get human-readable file sizes.
    simplified version of https://pypi.python.org/pypi/hurry.filesize/
    """
    for factor, suffix in units:
        if bytes >= factor:
            break
    amount = int(bytes / factor)

    if isinstance(suffix, tuple):
        singular, multiple = suffix
        if amount == 1:
            suffix = singular
        else:
            suffix = multiple
    return str(amount) + suffix

print(pretty_size(1))
print(pretty_size(42))
print(pretty_size(4096))
print(pretty_size(238048577))
print(pretty_size(334073741824))
print(pretty_size(96995116277763))
print(pretty_size(3125899904842624))

## [Out] ###########################
1 byte
42 bytes
4 KB
227 MB
311 GB
88 TB
2 PB

4 Comments

Is it not >>?
@Tjorriemorrie: it must be a left shift, right shifting will drop the only bit off and will result in 0.
i know this is old, but would this be correct usage? def convert_to_mb(data_b): print(data_b/(1 << 20))
@roastbeef yes that's correct. And I like this answer for this purpose, I also had to change bytes to megabytes only.
49

Here are some easy-to-copy one liners to use if you already know what unit size you want. If you're looking for in a more generic function with a few nice options, see my FEB 2021 update further on...

Bytes

print(f"{os.path.getsize(filepath):,} B") 

Kilobits

print(f"{os.path.getsize(filepath)/(1<<7):,.0f} kb")

Kilobytes

print(f"{os.path.getsize(filepath)/(1<<10):,.0f} KB")

Megabits

print(f"{os.path.getsize(filepath)/(1<<17):,.0f} mb")

Megabytes

print(f"{os.path.getsize(filepath)/(1<<20):,.0f} MB")

Gigabits

print(f"{os.path.getsize(filepath)/(1<<27):,.0f} gb")

Gigabytes

print(f"{os.path.getsize(filepath)/(1<<30):,.0f} GB")

Terabytes

print(f"{os.path.getsize(filepath)/(1<<40):,.0f} TB")

UPDATE FEB 2021 Here are my updated and fleshed-out functions to a) get file/folder size, b) convert into desired units:

from pathlib import Path

def get_path_size(path = Path('.'), recursive=False):
    """
    Gets file size, or total directory size

    Parameters
    ----------
    path: str | pathlib.Path
        File path or directory/folder path

    recursive: bool
        True -> use .rglob i.e. include nested files and directories
        False -> use .glob i.e. only process current directory/folder

    Returns
    -------
    int:
        File size or recursive directory size in bytes
        Use cleverutils.format_bytes to convert to other units e.g. MB
    """
    path = Path(path)
    if path.is_file():
        size = path.stat().st_size
    elif path.is_dir():
        path_glob = path.rglob('*.*') if recursive else path.glob('*.*')
        size = sum(file.stat().st_size for file in path_glob)
    return size


def format_bytes(bytes, unit, SI=False):
    """
    Converts bytes to common units such as kb, kib, KB, mb, mib, MB

    Parameters
    ---------
    bytes: int
        Number of bytes to be converted

    unit: str
        Desired unit of measure for output


    SI: bool
        True -> Use SI standard e.g. KB = 1000 bytes
        False -> Use JEDEC standard e.g. KB = 1024 bytes

    Returns
    -------
    str:
        E.g. "7 MiB" where MiB is the original unit abbreviation supplied
    """
    if unit.lower() in "b bit bits".split():
        return f"{bytes*8} {unit}"
    unitN = unit[0].upper()+unit[1:].replace("s","")  # Normalised
    reference = {"Kb Kib Kibibit Kilobit": (7, 1),
                 "KB KiB Kibibyte Kilobyte": (10, 1),
                 "Mb Mib Mebibit Megabit": (17, 2),
                 "MB MiB Mebibyte Megabyte": (20, 2),
                 "Gb Gib Gibibit Gigabit": (27, 3),
                 "GB GiB Gibibyte Gigabyte": (30, 3),
                 "Tb Tib Tebibit Terabit": (37, 4),
                 "TB TiB Tebibyte Terabyte": (40, 4),
                 "Pb Pib Pebibit Petabit": (47, 5),
                 "PB PiB Pebibyte Petabyte": (50, 5),
                 "Eb Eib Exbibit Exabit": (57, 6),
                 "EB EiB Exbibyte Exabyte": (60, 6),
                 "Zb Zib Zebibit Zettabit": (67, 7),
                 "ZB ZiB Zebibyte Zettabyte": (70, 7),
                 "Yb Yib Yobibit Yottabit": (77, 8),
                 "YB YiB Yobibyte Yottabyte": (80, 8),
                 }
    key_list = '\n'.join(["     b Bit"] + [x for x in reference.keys()]) +"\n"
    if unitN not in key_list:
        raise IndexError(f"\n\nConversion unit must be one of:\n\n{key_list}")
    units, divisors = [(k,v) for k,v in reference.items() if unitN in k][0]
    if SI:
        divisor = 1000**divisors[1]/8 if "bit" in units else 1000**divisors[1]
    else:
        divisor = float(1 << divisors[0])
    value = bytes / divisor
    return f"{value:,.0f} {unitN}{(value != 1 and len(unitN) > 3)*'s'}"


# Tests 
>>> assert format_bytes(1,"b") == '8 b'
>>> assert format_bytes(1,"bits") == '8 bits'
>>> assert format_bytes(1024, "kilobyte") == "1 Kilobyte"
>>> assert format_bytes(1024, "kB") == "1 KB"
>>> assert format_bytes(7141000, "mb") == '54 Mb'
>>> assert format_bytes(7141000, "mib") == '54 Mib'
>>> assert format_bytes(7141000, "Mb") == '54 Mb'
>>> assert format_bytes(7141000, "MB") == '7 MB'
>>> assert format_bytes(7141000, "mebibytes") == '7 Mebibytes'
>>> assert format_bytes(7141000, "gb") == '0 Gb'
>>> assert format_bytes(1000000, "kB") == '977 KB'
>>> assert format_bytes(1000000, "kB", SI=True) == '1,000 KB'
>>> assert format_bytes(1000000, "kb") == '7,812 Kb'
>>> assert format_bytes(1000000, "kb", SI=True) == '8,000 Kb'
>>> assert format_bytes(125000, "kb") == '977 Kb'
>>> assert format_bytes(125000, "kb", SI=True) == '1,000 Kb'
>>> assert format_bytes(125*1024, "kb") == '1,000 Kb'
>>> assert format_bytes(125*1024, "kb", SI=True) == '1,024 Kb'

UPDATE OCT 2022

My answer to a recent comment was too long, so here's some further explanation of the 1<<20 magic! I also notice that float isn't needed so I've removed that from the examples above.

As stated in another reply (above) "<<" is called a "bitwise operator". It converts the left hand side to binary and moves the binary digits 20 places to the left (in this case). When we count normally in decimal, the total number of digits dictates whether we've reached the tens, hundreds, thousands, millions etc. Similar thing in binary except the number of digits dictates whether we're talking bits, bytes, kilobytes, megabytes etc. So.... 1<<20 is actually the same as (binary) 1 with 20 (binary) zeros after it, or if you remember how to convert from binary to decimal: 2 to the power of 20 (2**20) which equals 1048576. In the snippets above, os.path.getsize returns a value in BYTES and 1048576 bytes are strictly speaking a Mebibyte (MiB) and casually speaking a Megabyte (MB).

7 Comments

That's a pretty clever way to do that. I wonder if you could put these into a function where you pass in whether you want kb's. mb's and so-on. You could even have an input command that asks which one you want, which would be pretty convenient if you do this a lot.
See above, Hildy... You can also customise the dictionary line like @lennart-regebro outlined above... which could be useful for storage management e.g. "Partition", "Cluster", "4TB Disks", "DVD_RW", "Blu-Ray Disk", "1GB memory sticks" or whatever.
I've also just added Kb (Kilobit), Mb (Megabit), and Gb (Gigabit) - users often get those confused in terms of network or file-transfer speeds, so thought it might be handy.
I love the one-liners, consider condensing with f-strings, e.g.: f'{os.path.getsize(filepath)/float(1<<20):.0f} MB'
Many thanks @pan0ramic!
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33

Here is the compact function to calculate size

def GetHumanReadable(size,precision=2):
    suffixes=['B','KB','MB','GB','TB']
    suffixIndex = 0
    while size > 1024 and suffixIndex < 4:
        suffixIndex += 1 #increment the index of the suffix
        size = size/1024.0 #apply the division
    return "%.*f%s"%(precision,size,suffixes[suffixIndex])

For more detailed output and vice versa operation please refer: http://code.activestate.com/recipes/578019-bytes-to-human-human-to-bytes-converter/

1 Comment

The while statement should be changed to while size >= 1024 and index < len(suffixes):, otherwise the function would return 1024.0KB instead of 1.0MB for example.
26

Here it is:

def convert_bytes(size):
    for x in ['bytes', 'KB', 'MB', 'GB', 'TB']:
        if size < 1024.0:
            return "%3.1f %s" % (size, x)
        size /= 1024.0

    return size

Output

>>> convert_bytes(1024)
'1.0 KB'
>>> convert_bytes(102400)
'100.0 KB'

2 Comments

That MiB, not MB and so on ...
it converting bit to byte, if you already have byte change 1024 to 1000
12

Just in case anyone's searching for the reverse of this problem (as I sure did) here's what works for me:

def get_bytes(size, suffix):
    size = int(float(size))
    suffix = suffix.lower()

    if suffix == 'kb' or suffix == 'kib':
        return size << 10
    elif suffix == 'mb' or suffix == 'mib':
        return size << 20
    elif suffix == 'gb' or suffix == 'gib':
        return size << 30

    return False

1 Comment

You are not handling the case of decimal numbers like 1.5GB. To fix it just change the << 10 to * 1024, << 20 to * 1024**2 and << 30 to * 1024**3.
9
UNITS = {1000: ['KB', 'MB', 'GB'],
            1024: ['KiB', 'MiB', 'GiB']}

def approximate_size(size, use_base_1024=True):
    mult = 1024 if use_base_1024 else 1000
    for unit in UNITS[mult]:
        size = size / mult
        if size < mult:
            return '{0:.3f} {1}'.format(size, unit)
            
approximate_size(2123, False)

2 Comments

this is usable in so many settings. glad i came across this comment. thanks a lot.
yah this is pretty sweet and does not require outside libs
3

Here my two cents, which permits casting up and down, and adds customizable precision:

def convertFloatToDecimal(f=0.0, precision=2):
    '''
    Convert a float to string of decimal.
    precision: by default 2.
    If no arg provided, return "0.00".
    '''
    return ("%." + str(precision) + "f") % f

def formatFileSize(size, sizeIn, sizeOut, precision=0):
    '''
    Convert file size to a string representing its value in B, KB, MB and GB.
    The convention is based on sizeIn as original unit and sizeOut
    as final unit. 
    '''
    assert sizeIn.upper() in {"B", "KB", "MB", "GB"}, "sizeIn type error"
    assert sizeOut.upper() in {"B", "KB", "MB", "GB"}, "sizeOut type error"
    if sizeIn == "B":
        if sizeOut == "KB":
            return convertFloatToDecimal((size/1024.0), precision)
        elif sizeOut == "MB":
            return convertFloatToDecimal((size/1024.0**2), precision)
        elif sizeOut == "GB":
            return convertFloatToDecimal((size/1024.0**3), precision)
    elif sizeIn == "KB":
        if sizeOut == "B":
            return convertFloatToDecimal((size*1024.0), precision)
        elif sizeOut == "MB":
            return convertFloatToDecimal((size/1024.0), precision)
        elif sizeOut == "GB":
            return convertFloatToDecimal((size/1024.0**2), precision)
    elif sizeIn == "MB":
        if sizeOut == "B":
            return convertFloatToDecimal((size*1024.0**2), precision)
        elif sizeOut == "KB":
            return convertFloatToDecimal((size*1024.0), precision)
        elif sizeOut == "GB":
            return convertFloatToDecimal((size/1024.0), precision)
    elif sizeIn == "GB":
        if sizeOut == "B":
            return convertFloatToDecimal((size*1024.0**3), precision)
        elif sizeOut == "KB":
            return convertFloatToDecimal((size*1024.0**2), precision)
        elif sizeOut == "MB":
            return convertFloatToDecimal((size*1024.0), precision)

Add TB, etc, as you wish.

1 Comment

I will vote this up because it can be worked out just with the python standard library
2

Here's a version that matches the output of ls -lh.

def human_size(num: int) -> str:
    base = 1
    for unit in ['B', 'K', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']:
        n = num / base
        if n < 9.95 and unit != 'B':
            # Less than 10 then keep 1 decimal place
            value = "{:.1f}{}".format(n, unit)
            return value
        if round(n) < 1000:
            # Less than 4 digits so use this
            value = "{}{}".format(round(n), unit)
            return value
        base *= 1024
    value = "{}{}".format(round(n), unit)
    return value

Comments

2

I wanted 2 way conversion, and I wanted to use Python 3 format() support to be most pythonic. Maybe try datasize library module? https://pypi.org/project/datasize/

$ pip install -qqq datasize
$ python
...
>>> from datasize import DataSize
>>> 'My new {:GB} SSD really only stores {:.2GiB} of data.'.format(DataSize('750GB'),DataSize(DataSize('750GB') * 0.8))
'My new 750GB SSD really only stores 558.79GiB of data.'

Comments

-1

Here is my implementation:

from bisect import bisect

def to_filesize(bytes_num, si=True):
    decade = 1000 if si else 1024
    partitions = tuple(decade ** n for n in range(1, 6))
    suffixes = tuple('BKMGTP')

    i = bisect(partitions, bytes_num)
    s = suffixes[i]

    for n in range(i):
        bytes_num /= decade

    f = '{:.3f}'.format(bytes_num)

    return '{}{}'.format(f.rstrip('0').rstrip('.'), s)

It will print up to three decimals and it strips trailing zeros and periods. The boolean parameter si will toggle usage of 10-based vs. 2-based size magnitude.

This is its counterpart. It allows to write clean configuration files like {'maximum_filesize': from_filesize('10M'). It returns an integer that approximates the intended filesize. I am not using bit shifting because the source value is a floating point number (it will accept from_filesize('2.15M') just fine). Converting it to an integer/decimal would work but makes the code more complicated and it already works as it is.

def from_filesize(spec, si=True):
    decade = 1000 if si else 1024
    suffixes = tuple('BKMGTP')

    num = float(spec[:-1])
    s = spec[-1]
    i = suffixes.index(s)

    for n in range(i):
        num *= decade

    return int(num)

Comments

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