A function to return human readable size from bytes size:
>>> human_readable(2048)
'2 kilobytes'
>>>
How to do this?
24 Answers
Addressing the above "too small a task to require a library" issue by a straightforward implementation (using f-strings, so Python 3.6+):
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}"
Supports:
- all currently known binary prefixes
- negative and positive numbers
- numbers larger than 1000 Yobibytes
- arbitrary units (maybe you like to count in Gibibits!)
Example:
>>> sizeof_fmt(168963795964)
'157.4GiB'
by Fred Cirera
Here's my version. It does not use a for-loop. It has constant complexity, O(1), and is in theory more efficient than the answers here that use a for-loop.
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'
To make it more clear what is going on, we can omit the code for the string formatting. Here are the lines that actually do the work:
exponent = int(log(num, 1024))
quotient = num / 1024**exponent
unit_list[exponent]
You should use "humanize".
>>> humanize.naturalsize(1000000)
'1.0 MB'
>>> humanize.naturalsize(1000000, binary=True)
'976.6 KiB'
>>> humanize.naturalsize(1000000, gnu=True)
'976.6K'
Reference:
One such library is hurry.filesize.
>>> from hurry.filesize import alternative
>>> size(1, system=alternative)
'1 byte'
>>> size(10, system=alternative)
'10 bytes'
>>> size(1024, system=alternative)
'1 KB'
The HumanFriendly project helps with this.
import humanfriendly
humanfriendly.format_size(1024)
The above code will give 1KB as answer.
Examples can be found here.
What you're about to find below is by no means the most performant or shortest solution among the ones already posted. Instead, it focuses on one particular issue that many of the other answers miss.
Namely the case when input like 999_995 is given:
Python 3.6.1 ...
...
>>> value = 999_995
>>> base = 1000
>>> math.log(value, base)
1.999999276174054
which, being truncated to the nearest integer and applied back to the input gives
>>> order = int(math.log(value, base))
>>> value/base**order
999.995
This seems to be exactly what we'd expect until we're required to control output precision. And this is when things start to get a bit difficult.
With the precision set to 2 digits we get:
>>> round(value/base**order, 2)
1000 # K
instead of 1M.
How can we counter that?
Of course, we can check for it explicitly:
if round(value/base**order, 2) == base:
order += 1
But can we do better? Can we get to know which way the order should be cut before we do the final step?
It turns out we can.
Assuming 0.5 decimal rounding rule, the above if condition translates into:
resulting in
def abbreviate(value, base=1000, precision=2, suffixes=None):
if suffixes is None:
suffixes = ['', 'K', 'M', 'B', 'T']
if value == 0:
return f'{0}{suffixes[0]}'
order_max = len(suffixes) - 1
order = log(abs(value), base)
order_corr = order - int(order) >= log(base - 0.5/10**precision, base)
order = min(int(order) + order_corr, order_max)
factored = round(value/base**order, precision)
return f'{factored:,g}{suffixes[order]}'
giving
>>> abbreviate(999_994)
'999.99K'
>>> abbreviate(999_995)
'1M'
>>> abbreviate(999_995, precision=3)
'999.995K'
>>> abbreviate(2042, base=1024)
'1.99K'
>>> abbreviate(2043, base=1024)
'2K'
To get the file size in a human readable form, I created this function:
import os
def get_size(path):
size = os.path.getsize(path)
if size < 1024:
return f"{size} bytes"
elif size < 1024*1024:
return f"{round(size/1024, 2)} KB"
elif size < 1024*1024*1024:
return f"{round(size/(1024*1024), 2)} MB"
elif size < 1024*1024*1024*1024:
return f"{round(size/(1024*1024*1024), 2)} GB"
>>> get_size("a.txt")
1.4KB
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
This feature if available in Boltons which is a very handy library to have for most projects.
>>> bytes2human(128991)
'126K'
>>> bytes2human(100001221)
'95M'
>>> bytes2human(0, 2)
'0.00B'
Referencing Sridhar Ratnakumar's answer, updated to:
def formatSize(sizeInBytes, decimalNum=1, isUnitWithI=False, sizeUnitSeperator=""):
"""format size to human readable string"""
# https://en.wikipedia.org/wiki/Binary_prefix#Specific_units_of_IEC_60027-2_A.2_and_ISO.2FIEC_80000
# K=kilo, M=mega, G=giga, T=tera, P=peta, E=exa, Z=zetta, Y=yotta
sizeUnitList = ['','K','M','G','T','P','E','Z']
largestUnit = 'Y'
if isUnitWithI:
sizeUnitListWithI = []
for curIdx, eachUnit in enumerate(sizeUnitList):
unitWithI = eachUnit
if curIdx >= 1:
unitWithI += 'i'
sizeUnitListWithI.append(unitWithI)
# sizeUnitListWithI = ['','Ki','Mi','Gi','Ti','Pi','Ei','Zi']
sizeUnitList = sizeUnitListWithI
largestUnit += 'i'
suffix = "B"
decimalFormat = "." + str(decimalNum) + "f" # ".1f"
finalFormat = "%" + decimalFormat + sizeUnitSeperator + "%s%s" # "%.1f%s%s"
sizeNum = sizeInBytes
for sizeUnit in sizeUnitList:
if abs(sizeNum) < 1024.0:
return finalFormat % (sizeNum, sizeUnit, suffix)
sizeNum /= 1024.0
return finalFormat % (sizeNum, largestUnit, suffix)
and example output is:
def testKb():
kbSize = 3746
kbStr = formatSize(kbSize)
print("%s -> %s" % (kbSize, kbStr))
def testI():
iSize = 87533
iStr = formatSize(iSize, isUnitWithI=True)
print("%s -> %s" % (iSize, iStr))
def testSeparator():
seperatorSize = 98654
seperatorStr = formatSize(seperatorSize, sizeUnitSeperator=" ")
print("%s -> %s" % (seperatorSize, seperatorStr))
def testBytes():
bytesSize = 352
bytesStr = formatSize(bytesSize)
print("%s -> %s" % (bytesSize, bytesStr))
def testMb():
mbSize = 76383285
mbStr = formatSize(mbSize, decimalNum=2)
print("%s -> %s" % (mbSize, mbStr))
def testTb():
tbSize = 763832854988542
tbStr = formatSize(tbSize, decimalNum=2)
print("%s -> %s" % (tbSize, tbStr))
def testPb():
pbSize = 763832854988542665
pbStr = formatSize(pbSize, decimalNum=4)
print("%s -> %s" % (pbSize, pbStr))
def demoFormatSize():
testKb()
testI()
testSeparator()
testBytes()
testMb()
testTb()
testPb()
# 3746 -> 3.7KB
# 87533 -> 85.5KiB
# 98654 -> 96.3 KB
# 352 -> 352.0B
# 76383285 -> 72.84MB
# 763832854988542 -> 694.70TB
# 763832854988542665 -> 678.4199PB
Here is an option using while:
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')
This solution might also appeal to you, depending on how your mind works:
from pathlib import Path
def get_size(path = Path('.')):
""" Gets file size, or total directory size """
if path.is_file():
size = path.stat().st_size
elif path.is_dir():
size = sum(file.stat().st_size for file in path.glob('*.*'))
return size
def format_size(path, unit="MB"):
""" Converts integers to common size units used in computing """
bit_shift = {"B": 0,
"kb": 7,
"KB": 10,
"mb": 17,
"MB": 20,
"gb": 27,
"GB": 30,
"TB": 40,}
return "{:,.0f}".format(get_size(path) / float(1 << bit_shift[unit])) + " " + unit
# Tests and test results
>>> get_size("d:\\media\\bags of fun.avi")
'38 MB'
>>> get_size("d:\\media\\bags of fun.avi","KB")
'38,763 KB'
>>> get_size("d:\\media\\bags of fun.avi","kb")
'310,104 kb'
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