how to sort this array

L

Li Chen

Hi,

I have an array containing files names. How do I sort them so that I can
get the expected results?

Thanks,

Li

#############
files=[
"c:/ruby/self/2004/20.txt",
"c:/ruby/self/2004/3.txt",
"c:/ruby/self/2004/2.txt",
"c:/ruby/self/2004/10.txt",
"c:/ruby/self/2004/1.txt"
]

expected results:
[
"c:/ruby/self/2004/1.txt",
"c:/ruby/self/2004/2.txt",
"c:/ruby/self/2004/3.txt",
"c:/ruby/self/2004/10.txt",
]
"c:/ruby/self/2004/20.txt",
 
S

Stefan Rusterholz

Li said:
Hi,

I have an array containing files names. How do I sort them so that I can
get the expected results?

Thanks,

Li

#############
files=[
"c:/ruby/self/2004/20.txt",
"c:/ruby/self/2004/3.txt",
"c:/ruby/self/2004/2.txt",
"c:/ruby/self/2004/10.txt",
"c:/ruby/self/2004/1.txt"
]

expected results:
[
"c:/ruby/self/2004/1.txt",
"c:/ruby/self/2004/2.txt",
"c:/ruby/self/2004/3.txt",
"c:/ruby/self/2004/10.txt",
]
"c:/ruby/self/2004/20.txt",

You get that result because in "2" <=> "10" the comparison is byte by
byte and then "2" is > "1", so it aborts there with "2" being > than
"10".
What you want is called natural sorting. Googling for natsort and ruby
you should get a few results.

Regards
Stefan
 
L

Li Chen

here is my code:

C:\Users\Alex>irb
irb(main):001:0> files=[
irb(main):002:1* "c:/ruby/self/2004/20.txt",
irb(main):003:1* "c:/ruby/self/2004/3.txt",
irb(main):004:1* "c:/ruby/self/2004/2.txt",
irb(main):005:1* "c:/ruby/self/2004/10.txt",
irb(main):006:1* "c:/ruby/self/2004/1.txt"
irb(main):007:1> ]
=> ["c:/ruby/self/2004/20.txt", "c:/ruby/self/2004/3.txt",
"c:/ruby/self/2004/2.txt", "c:/ruby/self/
2004/10.txt", "c:/ruby/self/2004/1.txt"]
irb(main):008:0>
irb(main):009:0*
irb(main):010:0* files=files.sort_by do|s|
irb(main):011:1* s.split(/\//)[-1].split(/\./)[0].to_i
irb(main):012:1> end
=> ["c:/ruby/self/2004/1.txt", "c:/ruby/self/2004/2.txt",
"c:/ruby/self/2004/3.txt", "c:/ruby/self/2
004/10.txt", "c:/ruby/self/2004/20.txt"]
irb(main):013:0>

I am not sure if it is the Rubyish way.

Li
 
S

Simon Krahnke

* Li Chen said:
files=[
"c:/ruby/self/2004/20.txt",
"c:/ruby/self/2004/3.txt",
"c:/ruby/self/2004/2.txt",
"c:/ruby/self/2004/10.txt",
"c:/ruby/self/2004/1.txt"
]

expected results:
[
"c:/ruby/self/2004/1.txt",
"c:/ruby/self/2004/2.txt",
"c:/ruby/self/2004/3.txt",
"c:/ruby/self/2004/10.txt",
"c:/ruby/self/2004/20.txt",
]

files.sort_by { | fn | fn.match(/(\d+)\.txt$/)[1].to_i }

mfg, simon .... hth
 
G

Gregory Seidman

Hi,

I have an array containing files names. How do I sort them so that I can
get the expected results?

Thanks,

Li

#############
files=[
"c:/ruby/self/2004/20.txt",
"c:/ruby/self/2004/3.txt",
"c:/ruby/self/2004/2.txt",
"c:/ruby/self/2004/10.txt",
"c:/ruby/self/2004/1.txt"
]

expected results:
[
"c:/ruby/self/2004/1.txt",
"c:/ruby/self/2004/2.txt",
"c:/ruby/self/2004/3.txt",
"c:/ruby/self/2004/10.txt",
]
"c:/ruby/self/2004/20.txt",

files = files.sort_by { |f| f[/\/(\d+)\.[^\/]*\Z/, 1].to_i }

There are several things to explain here:

- The #sort_by method calls the block on each element to produce a set of
surrogate values to use as sorting keys.

- The regular expression captures the series of digits between the last
slash in a string and the following dot.

- The [] method can be called on String in many ways, including passing a
RegExp and an index. That form returns the capture of the appropriate
index from matching the RegExp or nil if the RegExp does not match.

- Calling #to_i on the result means that it will either parse the captured
sequence of digits as an integer or, if the RegExp fails, return 0.

--Greg
 
R

Rob Biedenharn

* Li Chen said:
files=[
"c:/ruby/self/2004/20.txt",
"c:/ruby/self/2004/3.txt",
"c:/ruby/self/2004/2.txt",
"c:/ruby/self/2004/10.txt",
"c:/ruby/self/2004/1.txt"
]

expected results:
[
"c:/ruby/self/2004/1.txt",
"c:/ruby/self/2004/2.txt",
"c:/ruby/self/2004/3.txt",
"c:/ruby/self/2004/10.txt",
"c:/ruby/self/2004/20.txt",
]

files.sort_by { | fn | fn.match(/(\d+)\.txt$/)[1].to_i }

mfg, simon .... hth


Since #to_i will stop at a non-digit, you could get simpler:

files.sort_by {|fn| fn[%r{.*/([^/]+)},1].to_i }

I tend to use %r{ } when the Regexp deals with / characters.

-Rob

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

botp

expected results:
[ "c:/ruby/self/2004/1.txt",
"c:/ruby/self/2004/2.txt",
"c:/ruby/self/2004/3.txt",
"c:/ruby/self/2004/10.txt",
"c:/ruby/self/2004/20.txt",
]

apology in advance for dupl email. i sent an email an hrs back but it
seems it got blackholed :)

anyway, if you want basename numeric sorting, then just do

eg,

files.sort_by{|f| File.basename(f).to_i}
 
R

Robert Klemme

2008/10/31 botp said:
expected results:
[ "c:/ruby/self/2004/1.txt",
"c:/ruby/self/2004/2.txt",
"c:/ruby/self/2004/3.txt",
"c:/ruby/self/2004/10.txt",
"c:/ruby/self/2004/20.txt",
]

apology in advance for dupl email. i sent an email an hrs back but it
seems it got blackholed :)

anyway, if you want basename numeric sorting, then just do

eg,

files.sort_by{|f| File.basename(f).to_i}

Very elegant! Here's another variant

irb(main):013:0> puts files.sort_by {|f| f[/\d+(?=\.txt$)/].to_i}
c:/ruby/self/2004/1.txt
c:/ruby/self/2004/2.txt
c:/ruby/self/2004/3.txt
c:/ruby/self/2004/10.txt
c:/ruby/self/2004/20.txt
=> nil

Kind regards

robert
 
R

Robert Klemme

And here's a variant for cases where there are multiple subdirectories:

irb(main):010:0> puts files.sort_by {|f| f.scan(/\d+/).map{|x|x.to_i}}
c:/ruby/self/2004/1.txt
c:/ruby/self/2004/2.txt
c:/ruby/self/2004/3.txt
c:/ruby/self/2004/10.txt
c:/ruby/self/2004/20.txt

Kind regards

robert
 

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