Python format size application (converting B to KB, MB, GB, TB)

Question

I am trying to write an application to convert bytes to kb to mb to gb to tb. Here\'s what I have so far:

def size_format(b):
    if b < 1000:
              r         


        

Answer 1

Yet another humanbytes version, with no loops/if..else, in python3 syntax.

Test numbers stolen from @whereisalext's answer.

Mind you, it's still a sketch, e.g. if the numbers are large enough it will traceback.

import math as m


MULTIPLES = ["B", "k{}B", "M{}B", "G{}B", "T{}B", "P{}B", "E{}B", "Z{}B", "Y{}B"]


def humanbytes(i, binary=False, precision=2):
    base = 1024 if binary else 1000
    multiple = m.trunc(m.log2(i) / m.log2(base))
    value = i / m.pow(base, multiple)
    suffix = MULTIPLES[multiple].format("i" if binary else "")
    return f"{value:.{precision}f} {suffix}"


if __name__ == "__main__":
    sizes = [
        1, 1024, 500000, 1048576, 50000000, 1073741824, 5000000000,
        1099511627776, 5000000000000]

    for i in sizes:
        print(f"{i} == {humanbytes(i)}, {humanbytes(i, binary=True)}")

Results:

1 == 1.00 B, 1.00 B
1024 == 1.02 kB, 1.00 kiB
500000 == 500.00 kB, 488.28 kiB
1048576 == 1.05 MB, 1.00 MiB
50000000 == 50.00 MB, 47.68 MiB
1073741824 == 1.07 GB, 1.00 GiB
5000000000 == 5.00 GB, 4.66 GiB
1099511627776 == 1.10 TB, 1.00 TiB
5000000000000 == 5.00 TB, 4.55 TiB

Update:

As pointed out in comments (and as noted originally: "Mind you, it's still a sketch"), this code is slow and buggy. Please see @mitch-mcmabers 's answer.

Update 2: I was also lying about having no ifs.

Answer 2

When you divide the value you're using an integer divide, since both values are integers. You need to convert one of them to float first:

return '%.1f' % float(b)/1000 + 'KB'

or even just

return '%.1f' % b/1000.0 + 'KB'

Answer 3

def humanbytes(B):
   'Return the given bytes as a human friendly KB, MB, GB, or TB string'
   B = float(B)
   KB = float(1024)
   MB = float(KB ** 2) # 1,048,576
   GB = float(KB ** 3) # 1,073,741,824
   TB = float(KB ** 4) # 1,099,511,627,776

   if B < KB:
      return '{0} {1}'.format(B,'Bytes' if 0 == B > 1 else 'Byte')
   elif KB <= B < MB:
      return '{0:.2f} KB'.format(B/KB)
   elif MB <= B < GB:
      return '{0:.2f} MB'.format(B/MB)
   elif GB <= B < TB:
      return '{0:.2f} GB'.format(B/GB)
   elif TB <= B:
      return '{0:.2f} TB'.format(B/TB)

tests = [1, 1024, 500000, 1048576, 50000000, 1073741824, 5000000000, 1099511627776, 5000000000000]

for t in tests: print '{0} == {1}'.format(t,humanbytes(t))

Output:

1 == 1.0 Byte
1024 == 1.00 KB
500000 == 488.28 KB
1048576 == 1.00 MB
50000000 == 47.68 MB
1073741824 == 1.00 GB
5000000000 == 4.66 GB
1099511627776 == 1.00 TB
5000000000000 == 4.55 TB

and for future me here it is in Perl too:

sub humanbytes {
   my $B = shift;
   my $KB = 1024;
   my $MB = $KB ** 2; # 1,048,576
   my $GB = $KB ** 3; # 1,073,741,824
   my $TB = $KB ** 4; # 1,099,511,627,776

   if ($B < $KB) {
      return "$B " . (($B == 0 || $B > 1) ? 'Bytes' : 'Byte');
   } elsif ($B >= $KB && $B < $MB) {
      return sprintf('%0.02f',$B/$KB) . ' KB';
   } elsif ($B >= $MB && $B < $GB) {
      return sprintf('%0.02f',$B/$MB) . ' MB';
   } elsif ($B >= $GB && $B < $TB) {
      return sprintf('%0.02f',$B/$GB) . ' GB';
   } elsif ($B >= $TB) {
      return sprintf('%0.02f',$B/$TB) . ' TB';
   }
}

Answer 4

This is a compact version that converts B (bytes) to any higher order such MB, GB without using a lot of if...else in python. I use bit-wise to deal with this. Also it allows to return a float output if you trigger the parameter return_output in the function as True:

import math

def bytes_conversion(number, return_float=False):

    def _conversion(number, return_float=False):

        length_number = int(math.log10(number))

        if return_float:

           length_number = int(math.log10(number))
           return length_number // 3, '%.2f' % (int(number)/(1 << (length_number//3) *10))

        return length_number // 3, int(number) >> (length_number//3) * 10

    unit_dict = {
        0: "B",  1: "kB",
        2: "MB", 3: "GB",
        4: "TB", 5: "PB",
        6: "EB"
    }

    if return_float:

        num_length, number = _conversion(number, return_float=return_float)

    else:
        num_length, number = _conversion(number)

    return "%s %s" % (number, unit_dict[num_length])

#Example usage:
#print(bytes_conversion(491266116, return_float=True))

This is only a few of my posts in StackOverflow. Please let me know if I have any errors or violations.

Answer 5

An output with no decimal places:

>>> format_file_size(12345678)
'11 MiB, 792 KiB, 334 bytes'

format_file_size(
    def format_file_size(fsize):
        result = []
        units = {s: u for s, u in zip(reversed([2 ** n for n in range(0, 40, 10)]), ['GiB', 'MiB', 'KiB', 'bytes'])}
        for s, u in units.items():
            t = fsize // s
            if t > 0:
                result.append('{} {}'.format(t, u))
            fsize = fsize % s
        return ', '.join(result) or '0 bytes'

Answer 6

Fixed version of Bryan_Rch's answer:

def format_bytes(size):
    # 2**10 = 1024
    power = 2**10
    n = 0
    power_labels = {0 : '', 1: 'kilo', 2: 'mega', 3: 'giga', 4: 'tera'}
    while size > power:
        size /= power
        n += 1
    return size, power_labels[n]+'bytes'