Sort and list the files from a directory in numeric order

Question

This is my folder structure.

/home/files/encounters 9-22-11-0.jpg .. /home/files/encounters 9-22-11-[n].jpg

puts Dir.glob("/home/files/*.jpg")[0]

When i execute the above code, it displayed the sixth file (index 5 => /home/files/encounters 9-22-11-5.jpg), but actually i need the output as first file(index 0 => /home/files/encounters 9-22-11-0.jpg)

How can i sort the files as user defined sorting order?. like

When i tried..
..[0] => /home/files/encounters 9-22-11-5.jpg
..[1] => /home/files/encounters 9-22-11-21.jpg
..[2] => /home/files/encounters 9-22-11-39.jpg

But, I need
..[0] => /home/files/encounters 9-22-11-0.jpg
..[1] => /home/files/encounters 9-22-11-1.jpg
..[2] => /home/files/encounters 9-22-11-2.jpg

Additional information, sorting is also not working.

f = Dir.glob("/home/files/*.jpg").sort
f[0] => /home/files/encounters 9-22-11-0.jpg
f[0] => /home/files/encounters 9-22-11-1.jpg
f[0] => /home/files/encounters 9-22-11-10.jpg
f[0] => /home/files/encounters 9-22-11-11.jpg

Jonas Elfström · Answer 1 · 2011-10-19T11:59:52.313

14

puts Dir.glob("/home/files/*.jpg").sort

Would work if you had a format like 11-09-22-05.jpg instead of 9-22-11-5.jpg. You could try to sort them as a number instead.

Dir.glob("/home/files/*.jpg").sort_by {|s| s.gsub("-","").to_i }

But as it seems you have month-day-year-number I guess that the correct way to sort is a bit more complicated than that.

arr=%w[9-22-12-33.jpg 9-22-11-5.jpg 9-22-10-99.jpg 12-24-11-1.jpg]
arr.sort_by do |s|
  t = s.split("-").map(&:to_i)
  [t[2], t[0], t[1], t[3]]
end

It works by reformatting 9-22-11-5.jpg to an array containing [11, 9, 22, 5] and then sorts by that instead.

edited Oct 19 '11 at 11:59

answered Oct 19 '11 at 10:37

Jonas Elfström

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Why format it to a string? Sorting by an array would have the same semantics: `arr.sort_by do |s| t = ...; [t[2], t[0], t[1], t[3]]; end` – qerub Oct 19 '11 at 11:52
@Qerub Thanks, I totally forgot that you can do that. – Jonas Elfström Oct 19 '11 at 11:58
@JonasElfström: Happy to remind you! (Feel free to upvote my comment! :]) – qerub Oct 19 '11 at 12:24
1

I'd use `s.scan(/\d+/).last(4).map(&:to_i)` instead of split, to allow for arbitrary prefixes in the filenames. – Lars Haugseth Oct 19 '11 at 14:58

score 1 · Answer 2 · answered Oct 19 '11 at 10:43

1

If possible, I'd create the files with a fixed width numeric format instead:

encounters 9-22-11-05.jpg
encounters 9-22-11-11.jpg
encounters 9-22-11-99.jpg

If that's impossible, you can extract the last numeric part and use that in a custom sort criterion:

a = Dir.glob("*.jpg")
r = Regexp.new(".*-([0-9]+).jpg")
b = a.sort do |f1, f2|
  n1 = r.match(f1)[1].to_i
  n2 = r.match(f2)[1].to_i
  n1 <=> n2
end
puts b

This extracts the last numeric part from each filename (using a regular expression) and sorts by this.
If you have files belonging to different dates, you'll have to modify this so it sorts by their "base" names plus the numeric part.

answered Oct 19 '11 at 10:43

Frank Schmitt

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Thanks @Frank Schmitt. It's working for me. Can you please tell me "n1 <=> n2"?. – Mr. Black Oct 19 '11 at 10:58
It's the spaceship operator. http://stackoverflow.com/questions/827649/ruby-spaceship-operator – Jonas Elfström Oct 19 '11 at 11:30
Be aware that this solution only sorts on the last number. It will take no notice of the, presumed, date before that. – Jonas Elfström Oct 19 '11 at 11:31

Sort and list the files from a directory in numeric order

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