Sorting numbers as strings

J

Jack Bauer

I'm trying to sort some strings containing numbers. The strings
themselves can't be changed (they're items being pulled from a DB.) This
is an example of some of the things I need to sort. First is how I
wanted them sorted:

FastEthernet0/1
FastEthernet0/10
FastEthernet0/11
FastEthernet0/12
FastEthernet0/13
FastEthernet0/2
FastEthernet0/3
FastEthernet0/4
FastEthernet0/5
FastEthernet0/6
FastEthernet0/7
FastEthernet0/8
FastEthernet0/9
...

I need to get it like this:
FastEthernet0/1
FastEthernet0/2
FastEthernet0/3
FastEthernet0/4
FastEthernet0/5
FastEthernet0/6
FastEthernet0/7
FastEthernet0/8
FastEthernet0/9
FastEthernet0/10
FastEthernet0/11
FastEthernet0/12
FastEthernet0/13

Then they could turn into FastEthernet1/1, etc. Also, the name doesn't
really matter as it could have FastEthernet, another name, or none at
all and just be 0/1 or whatever.

I'm guessing something like a regex to strip non-numbers so
FastEthernet0/1 becomes 01 then sort numerically, but that wouldn't help
if I have FastEthernet0/1, GigabitEthernet0/1, and FastEthernet0/2 since
it would come out in an incorrect order (01, 01, 02 instead of 01, 02,
01 because of the F and G alphabetical sort.)

Any ideas of the best way to go about this?
 
J

Jack Bauer

Jack said:
First is how I wanted them sorted:



Sorry I meant to say that the first version is how they're CURRENTLY
being sorted and the second version is how I WANT them sorted.
 
D

Douglas Seifert

[Note: parts of this message were removed to make it a legal post.]

Maybe something like this:

irb> a = ["FastEthernet0/1", "FastEthernet0/10", "FastEthernet0/11",
"FastEthernet0/12", "FastEthernet0/13", "FastEthernet0/2",
"FastEthernet0/3", "FastEthernet0/4", "FastEthernet0/5", "FastEthernet0/6",
"FastEthernet0/7", "FastEthernet0/8", "FastEthernet0/9"]
irb> a.sort_by{|s| n,i = s.split('/'); [n, i.to_i]}
=> ["FastEthernet0/1", "FastEthernet0/2", "FastEthernet0/3",
"FastEthernet0/4", "FastEthernet0/5", "FastEthernet0/6", "FastEthernet0/7",
"FastEthernet0/8", "FastEthernet0/9", "FastEthernet0/10",
"FastEthernet0/11", "FastEthernet0/12", "FastEthernet0/13"]
 
R

Robert Klemme

2009/5/18 Jack Bauer said:
I'm trying to sort some strings containing numbers. The strings
themselves can't be changed (they're items being pulled from a DB.) This
is an example of some of the things I need to sort. First is how I
wanted them sorted:

FastEthernet0/1
FastEthernet0/10
FastEthernet0/11
FastEthernet0/12
FastEthernet0/13
FastEthernet0/2
FastEthernet0/3
FastEthernet0/4
FastEthernet0/5
FastEthernet0/6
FastEthernet0/7
FastEthernet0/8
FastEthernet0/9
...

I need to get it like this:
FastEthernet0/1
FastEthernet0/2
FastEthernet0/3
FastEthernet0/4
FastEthernet0/5
FastEthernet0/6
FastEthernet0/7
FastEthernet0/8
FastEthernet0/9
FastEthernet0/10
FastEthernet0/11
FastEthernet0/12
FastEthernet0/13

Then they could turn into FastEthernet1/1, etc. Also, the name doesn't
really matter as it could have FastEthernet, another name, or none at
all and just be 0/1 or whatever.

I'm guessing something like a regex to strip non-numbers so
FastEthernet0/1 becomes 01 then sort numerically, but that wouldn't help
if I have FastEthernet0/1, GigabitEthernet0/1, and FastEthernet0/2 since
it would come out in an incorrect order (01, 01, 02 instead of 01, 02,
01 because of the F and G alphabetical sort.)

Any ideas of the best way to go about this?

16:39:42 Temp$ ruby19 srt.rb
FastEthernet0/1
FastEthernet0/2
FastEthernet0/3
FastEthernet0/4
FastEthernet0/5
FastEthernet0/6
FastEthernet0/7
FastEthernet0/8
FastEthernet0/9
FastEthernet0/10
FastEthernet0/11
FastEthernet0/12
FastEthernet0/13
16:40:24 Temp$ cat srt.rb
puts DATA.sort_by {|s| s.scan(/\d+/).map {|x| x.to_i} }
__END__
FastEthernet0/1
FastEthernet0/10
FastEthernet0/11
FastEthernet0/12
FastEthernet0/13
FastEthernet0/2
FastEthernet0/3
FastEthernet0/4
FastEthernet0/5
FastEthernet0/6
FastEthernet0/7
FastEthernet0/8
FastEthernet0/9
16:40:28 Temp$

Kind regards

robert
 
J

Jack Bauer

Douglas said:
Maybe something like this:
irb> a.sort_by{|s| n,i = s.split('/'); [n, i.to_i]}

Thanks to both of you guys. The one I quoted above looks similar to what
I was doing, I just thought it wasn't working because I'm using
pagination and it's ordering each page rather than the whole result and
then paginating.
 
R

Rob Biedenharn

16:40:24 Temp$ cat srt.rb
puts DATA.sort_by {|s| s.scan(/\d+/).map {|x| x.to_i} }
Kind regards

robert

If you do need to consider things like "FastEthernet..." v.
"GigabitEthernet...", then perhaps you want something like this which
breaks the original string into digits and non-digits and converts the
digits to be integers.

puts DATA.sort_by {|s| s.scan(/\d+|\D+/).map {|x| x =~ /\d+/ ?
x.to_i : x } }

-Rob

Rob Biedenharn http://agileconsultingllc.com
(e-mail address removed)
 
R

Robert Klemme

If you do need to consider things like "FastEthernet..." v.
"GigabitEthernet...", then perhaps you want something like this which
breaks the original string into digits and non-digits and converts the
digits to be integers.

puts DATA.sort_by {|s| s.scan(/\d+|\D+/).map {|x| x =~ /\d+/ ?
x.to_i : x } }

Good point! Here's an interesting variant:

puts DATA.sort_by {|s| s.scan(/(\d+)|\D+/).map { $1 ? $1.to_i : $& } }

Kind regards

robert
 
R

Rob Biedenharn

Good point! Here's an interesting variant:

puts DATA.sort_by {|s| s.scan(/(\d+)|\D+/).map { $1 ? $1.to_i : $& } }

Kind regards

robert

Interesting, but "wrong" in that it doesn't sort the way the OP wanted.
Actually, since all the regexp applications have been applied by #scan
before the #map happens, the values of $1 and $& are effectively
constants and no sorting happens at all.

However, that did inspire me to make my version a little better.

puts DATA.sort_by {|s| s.scan(/(\d+)|(\D+)/).map {|(n,s)| s ||
n.to_i } }

I'd rather make the variables local than invoke the Perlish Regexp
globals (even if they did were assigned in the block the way you
expected). It could be even more readable if (n,s) were replaced with
(digits,nondigits), but it looks OK to me with n and s.

-Rob

Rob Biedenharn http://agileconsultingllc.com
(e-mail address removed)
 
R

Robert Klemme

Interesting, but "wrong" in that it doesn't sort the way the OP wanted.
Actually, since all the regexp applications have been applied by #scan
before the #map happens, the values of $1 and $& are effectively
constants and no sorting happens at all.

Aaargh! Yes, you are completely right.
However, that did inspire me to make my version a little better.

puts DATA.sort_by {|s| s.scan(/(\d+)|(\D+)/).map {|(n,s)| s ||
n.to_i } }

I'd rather make the variables local than invoke the Perlish Regexp
globals (even if they did were assigned in the block the way you
expected). It could be even more readable if (n,s) were replaced with
(digits,nondigits), but it looks OK to me with n and s.

Absolutely agree, I try to use local variables whenever possible.
Although I recently learned that $1 etc. are local to the current stack
frame! I did not knew that before and it certainly makes their use a
lot safer.

Thanks for catching my mistake!

Kind regards

robert
 
J

Jack Bauer

You guys are great. I went with Bob's (pick one, hah) since it ended up
being moderately faster than the method I was originally using.
 
J

Johan Holmberg

You guys are great. I went with Bob's (pick one, hah) since it ended up
being moderately faster than the method I was originally using.
--

Here is a variant similar to the other solutions you already got:

puts arr.sort_by {|s| s.split(/(\d+)/).each_with_index.map {|x,i|
i.odd? ? x.to_i : x }}

Works with Ruby 1.9.

/Johan Holmberg
 
R

Rob Biedenharn

Here is a variant similar to the other solutions you already got:

puts arr.sort_by {|s| s.split(/(\d+)/).each_with_index.map {|x,i|
i.odd? ? x.to_i : x }}

Works with Ruby 1.9.

/Johan Holmberg


Are you trying to say that you believe some part of a solution that
Jack already accepted won't work in Ruby 1.9? (Because they should
work fine.)

-Rob

Rob Biedenharn http://agileconsultingllc.com
(e-mail address removed)
 
J

Johan Holmberg

Are you trying to say that you believe some part of a solution that Jack
already accepted won't work in Ruby 1.9? (Because they should work fine.)

-Rob

No, not at all. I tried to say that I believed *my* variant only
worked in 1.9 (and maybe 1.8.7). (I should have said "Only works with
Ruby 1.9").

Sorry for the misunderstanding.

/Johan Holmberg
 

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