Pretty file size in Ruby? - ruby

Question

I'm trying to make a method that converts an integer that represents bytes to a string with a 'prettied up' format.
Here's my half-working attempt:
class Integer
def to_filesize
{
'B' => 1024,
'KB' => 1024 * 1024,
'MB' => 1024 * 1024 * 1024,
'GB' => 1024 * 1024 * 1024 * 1024,
'TB' => 1024 * 1024 * 1024 * 1024 * 1024
}.each_pair { |e, s| return "#{s / self}#{e}" if self < s }
end
end
What am I doing wrong?

Answer 1

If you use it with Rails - what about standard Rails number helper?
http://api.rubyonrails.org/classes/ActionView/Helpers/NumberHelper.html#method-i-number_to_human_size
number_to_human_size(number, options = {})
?

Answer 2

How about the Filesize gem ? It seems to be able to convert from bytes (and other formats) into pretty printed values:
example:
Filesize.from("12502343 B").pretty # => "11.92 MiB"
http://rubygems.org/gems/filesize

Answer 3

I agree with #David that it's probably best to use an existing solution, but to answer your question about what you're doing wrong:
The primary error is dividing s by self rather than the other way around.
You really want to divide by the previous s, so divide s by 1024.
Doing integer arithmetic will give you confusing results, so convert to float.
Perhaps round the answer.
So:
class Integer
def to_filesize
{
'B' => 1024,
'KB' => 1024 * 1024,
'MB' => 1024 * 1024 * 1024,
'GB' => 1024 * 1024 * 1024 * 1024,
'TB' => 1024 * 1024 * 1024 * 1024 * 1024
}.each_pair { |e, s| return "#{(self.to_f / (s / 1024)).round(2)}#{e}" if self < s }
end
end
lets you:
1.to_filesize
# => "1.0B"
1020.to_filesize
# => "1020.0B"
1024.to_filesize
# => "1.0KB"
1048576.to_filesize
# => "1.0MB"
Again, I don't recommend actually doing that, but it seems worth correcting the bugs.

Answer 4

This is my solution:
def filesize(size)
units = %w[B KiB MiB GiB TiB Pib EiB ZiB]
return '0.0 B' if size == 0
exp = (Math.log(size) / Math.log(1024)).to_i
exp += 1 if (size.to_f / 1024 ** exp >= 1024 - 0.05)
exp = units.size - 1 if exp > units.size - 1
'%.1f %s' % [size.to_f / 1024 ** exp, units[exp]]
end
Compared to other solutions it's simpler, more efficient, and generates a more proper output.
Format
All other methods have the problem that they report 1023.95 bytes wrong. Moreover to_filesize simply errors out with big numbers (it returns an array).
- method: [ filesize, Filesize, number_to_human, to_filesize ]
- 0 B: [ 0.0 B, 0.00 B, 0 Bytes, 0.0B ]
- 1 B: [ 1.0 B, 1.00 B, 1 Byte, 1.0B ]
- 10 B: [ 10.0 B, 10.00 B, 10 Bytes, 10.0B ]
- 1000 B: [ 1000.0 B, 1000.00 B, 1000 Bytes, 1000.0B ]
- 1 KiB: [ 1.0 KiB, 1.00 KiB, 1 KB, 1.0KB ]
- 1.5 KiB: [ 1.5 KiB, 1.50 KiB, 1.5 KB, 1.5KB ]
- 10 KiB: [ 10.0 KiB, 10.00 KiB, 10 KB, 10.0KB ]
- 1000 KiB: [ 1000.0 KiB, 1000.00 KiB, 1000 KB, 1000.0KB ]
- 1 MiB: [ 1.0 MiB, 1.00 MiB, 1 MB, 1.0MB ]
- 1 GiB: [ 1.0 GiB, 1.00 GiB, 1 GB, 1.0GB ]
- 1023.95 GiB: [ 1.0 TiB, 1023.95 GiB, 1020 GB, 1023.95GB ]
- 1 TiB: [ 1.0 TiB, 1.00 TiB, 1 TB, 1.0TB ]
- 1 EiB: [ 1.0 EiB, 1.00 EiB, 1 EB, ERROR ]
- 1 ZiB: [ 1.0 ZiB, 1.00 ZiB, 1020 EB, ERROR ]
- 1 YiB: [ 1024.0 ZiB, 1024.00 ZiB, 1050000 EB, ERROR ]
Performance
Also, it has the best performance (seconds to process 1 million numbers):
- filesize: 2.15
- Filesize: 15.53
- number_to_human: 139.63
- to_filesize: 2.41

Answer 5

Here is a method using log10:
def number_format(d)
e = Math.log10(d).to_i / 3
return '%.3f' % (d / 1000 ** e) + ['', ' k', ' M', ' G'][e]
end
s = number_format(9012345678.0)
puts s == '9.012 G'
https://ruby-doc.org/core/Math.html#method-c-log10

Answer 6

You get points for adding a method to Integer, but this seems more File specific, so I would suggest monkeying around with File, say by adding a method to File called .prettysize().
But here is an alternative solution that uses iteration, and avoids printing single bytes as float :-)
def format_mb(size)
conv = [ 'b', 'kb', 'mb', 'gb', 'tb', 'pb', 'eb' ];
scale = 1024;
ndx=1
if( size < 2*(scale**ndx) ) then
return "#{(size)} #{conv[ndx-1]}"
end
size=size.to_f
[2,3,4,5,6,7].each do |ndx|
if( size < 2*(scale**ndx) ) then
return "#{'%.3f' % (size/(scale**(ndx-1)))} #{conv[ndx-1]}"
end
end
ndx=7
return "#{'%.3f' % (size/(scale**(ndx-1)))} #{conv[ndx-1]}"
end

Answer 7

#Darshan Computing's solution is only partial here. Since the hash keys are not guaranteed to be ordered this approach will not work reliably. You could fix this by doing something like this inside the to_filesize method,
conv={
1024=>'B',
1024*1024=>'KB',
...
}
conv.keys.sort.each { |s|
next if self >= s
e=conv[s]
return "#{(self.to_f / (s / 1024)).round(2)}#{e}" if self < s }
}
This is what I ended up doing for a similar method inside Float,
class Float
def to_human
conv={
1024=>'B',
1024*1024=>'KB',
1024*1024*1024=>'MB',
1024*1024*1024*1024=>'GB',
1024*1024*1024*1024*1024=>'TB',
1024*1024*1024*1024*1024*1024=>'PB',
1024*1024*1024*1024*1024*1024*1024=>'EB'
}
conv.keys.sort.each { |mult|
next if self >= mult
suffix=conv[mult]
return "%.2f %s" % [ self / (mult / 1024), suffix ]
}
end
end

Answer 8

FileSize may be dead, but now there is ByteSize.
require 'bytesize'
ByteSize.new(1210000000) #=> (1.21 GB)
ByteSize.new(1210000000).to_s #=> 1.21 GB