Split array into sub-arrays based on value

Question

I was looking for an Array equivalent String#split in Ruby Core, and was surprised to find that it did not exist. Is there a more elegant way than the following to split an array into sub-arrays based on a value?

class Array
  def split( split_on=nil )
    inject([[]]) do |a,v|
      a.tap{
        if block_given? ? yield(v) : v==split_on
          a << []
        else
          a.last << v
        end
      }
    end.tap{ |a| a.pop if a.last.empty? }
  end
end

p (1..9 ).to_a.split{ |i| i%3==0 },
  (1..10).to_a.split{ |i| i%3==0 }
#=> [[1, 2], [4, 5], [7, 8]]
#=> [[1, 2], [4, 5], [7, 8], [10]]

Edit: For those interested, the "real-world" problem which sparked this request can be seen in this answer, where I've used @fd's answer below for the implementation.

Well, in Python you could convert it into a string (values separated by commas or something), split that, and then go back to a list. Dunno if that's an option in Ruby. — Rafe Kettler
– Rafe Kettler, Commented Jan 26, 2011 at 0:24
@Rafe It would be, but only if the contents were only strings. Even then, that could hardly be considered elegant. :p — Phrogz
– Phrogz, Commented Jan 26, 2011 at 0:38
@Phrogz if they were numbers it'd work fine too. You'd just do ','.join([str(x) for x in list_of_nums]), then split on whatever, then rejoin and split on commas. Functional, yes, elegant, eh no. — Rafe Kettler
– Rafe Kettler, Commented Jan 26, 2011 at 0:45
@Rafe Perhaps I should also accept answers for most roundabout hack. To/from YAML, anyone? :) — Phrogz
– Phrogz, Commented Jan 26, 2011 at 1:22
FYI: I don't see anything in your solution that requires self to be an Array. You could pull that method up into Enumerable, since you only depend on self responding to inject. (Incidentally, that also would allow you to get rid of the to_a in your two testcases.) — Jörg W Mittag
– Jörg W Mittag, Commented Jan 26, 2011 at 11:16

scaryguy · Accepted Answer · 2013-09-21 17:33:45Z

20

Sometimes partition is a good way to do things like that:

(1..6).partition { |v| v.even? } 
#=> [[2, 4, 6], [1, 3, 5]]

answered Sep 21, 2013 at 17:33

scaryguy

7,9603 gold badges42 silver badges53 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Lloeki Over a year ago

Irrelevant to the question: the author wants to split delimited running sequences.

Phrogz · Accepted Answer · 2013-01-16 19:22:06Z

14

I tried golfing it a bit, still not a single method though:

(1..9).chunk{|i|i%3==0}.reject{|sep,ans| sep}.map{|sep,ans| ans}

Or faster:

(1..9).chunk{|i|i%3==0 || nil}.map{|sep,ans| sep&&ans}.compact

Also, Enumerable#chunk seems to be Ruby 1.9+, but it is very close to what you want.

For example, the raw output would be:

(1..9).chunk{ |i|i%3==0 }.to_a                                       
=> [[false, [1, 2]], [true, [3]], [false, [4, 5]], [true, [6]], [false, [7, 8]], [true, [9]]]

(The to_a is to make irb print something nice, since chunk gives you an enumerator rather than an Array)

Edit: Note that the above elegant solutions are 2-3x slower than the fastest implementation:

module Enumerable
  def split_by
    result = [a=[]]
    each{ |o| yield(o) ? (result << a=[]) : (a << o) }
    result.pop if a.empty?
    result
  end
end

edited Jan 16, 2013 at 19:22

Phrogz

304k115 gold badges669 silver badges758 bronze badges

answered Jan 26, 2011 at 1:30

Mike Tunnicliffe

10.8k3 gold badges34 silver badges46 bronze badges

6 Comments

Phrogz Over a year ago

Nice! I hadn't seen chunk before. For the record, it's 1.9.2+, but that's wholly acceptable to me.

Mike Tunnicliffe Over a year ago

Here's a link to the doc: ruby-doc.org/core/classes/Enumerable.html#M001523

Phrogz Over a year ago

Not surprisingly (due to the extra iterations needed for reject/map) chunk is a good bit slower; I've added a benchmarking 'answer' collecting implementations.

SwiftMango Over a year ago

(1..10).chunk{|n| n % 3 == 0 ? :_separator : :keep}.map{|_,v| v}

Mike Tunnicliffe Over a year ago

(1..10).chuck{|n| n%3==0 || nil}.map{|_,v| v}

|

Phrogz · Accepted Answer · 2013-01-16 19:21:46Z

5

Here are benchmarks aggregating the answers (I'll not be accepting this answer):

require 'benchmark'
a = *(1..5000); N = 1000
Benchmark.bmbm do |x|
  %w[ split_with_inject split_with_inject_no_tap split_with_each
      split_with_chunk split_with_chunk2 split_with_chunk3 ].each do |method|
    x.report( method ){ N.times{ a.send(method){ |i| i%3==0 || i%5==0 } } }
  end
end
#=>                                user     system      total        real
#=> split_with_inject          1.857000   0.015000   1.872000 (  1.879188)
#=> split_with_inject_no_tap   1.357000   0.000000   1.357000 (  1.353135)
#=> split_with_each            1.123000   0.000000   1.123000 (  1.123113)
#=> split_with_chunk           3.962000   0.000000   3.962000 (  3.984398)
#=> split_with_chunk2          3.682000   0.000000   3.682000 (  3.687369)
#=> split_with_chunk3          2.278000   0.000000   2.278000 (  2.281228)

The implementations being tested (on Ruby 1.9.2):

class Array
  def split_with_inject
    inject([[]]) do |a,v|
      a.tap{ yield(v) ? (a << []) : (a.last << v) }
    end.tap{ |a| a.pop if a.last.empty? }
  end

  def split_with_inject_no_tap
    result = inject([[]]) do |a,v|
      yield(v) ? (a << []) : (a.last << v)
      a
    end
    result.pop if result.last.empty?
    result
  end

  def split_with_each
    result = [a=[]]
    each{ |o| yield(o) ? (result << a=[]) : (a << o) }
    result.pop if a.empty?
    result
  end

  def split_with_chunk
    chunk{ |o| !!yield(o) }.reject{ |b,a| b }.map{ |b,a| a }
  end

  def split_with_chunk2
    chunk{ |o| !!yield(o) }.map{ |b,a| b ? nil : a }.compact
  end

  def split_with_chunk3
    chunk{ |o| yield(o) || nil }.map{ |b,a| b && a }.compact
  end
end

edited Jan 16, 2013 at 19:21

answered Jan 26, 2011 at 4:49

Phrogz

304k115 gold badges669 silver badges758 bronze badges

1 Comment

Confusion Over a year ago

A bit late too the party, but: these methods aren't entirely comparable, because the results of these methods aren't all the same. The first three return something similar to what String#split returns (including empty arrays when two subsequent separators are found), while split_with_chunk and split_with_chunk2 never return empty arrays and while split_with_chunk3 still contains the 'grouping' value of chunk.

kurumi · Accepted Answer · 2011-01-26 01:35:44Z

1

Other Enumerable methods you might want to consider is each_slice or each_cons

I don't know how general you want it to be, here's one way

>> (1..9).each_slice(3) {|a| p a.size>1?a[0..-2]:a}
[1, 2]
[4, 5]
[7, 8]
=> nil
>> (1..10).each_slice(3) {|a| p a.size>1?a[0..-2]:a}
[1, 2]
[4, 5]
[7, 8]
[10]

edited Jan 26, 2011 at 1:35

answered Jan 26, 2011 at 1:27

kurumi

25.7k5 gold badges47 silver badges52 bronze badges

1 Comment

Phrogz Over a year ago

Only for my specific mod 3 example, but not in general.

Community · Accepted Answer · 2017-05-23 12:34:36Z

1

here is another one (with a benchmark comparing it to the fastest split_with_each here https://stackoverflow.com/a/4801483/410102):

require 'benchmark'

class Array
  def split_with_each
    result = [a=[]]
    each{ |o| yield(o) ? (result << a=[]) : (a << o) }
    result.pop if a.empty?
    result
  end

  def split_with_each_2
    u, v = [], []
    each{ |x| (yield x) ? (u << x) : (v << x) }
    [u, v]
  end
end

a = *(1..5000); N = 1000
Benchmark.bmbm do |x|
  %w[ split_with_each split_with_each_2 ].each do |method|
    x.report( method ){ N.times{ a.send(method){ |i| i%3==0 || i%5==0 } } }
  end
end

                        user     system      total        real
split_with_each     2.730000   0.000000   2.730000 (  2.742135)
split_with_each_2   2.270000   0.040000   2.310000 (  2.309600)

edited May 23, 2017 at 12:34

CommunityBot

11 silver badge

answered Dec 18, 2012 at 23:16

akonsu

29.7k39 gold badges126 silver badges204 bronze badges

1 Comment

Lloeki Over a year ago

This is like Array#partition, not String#split.

Collectives™ on Stack Overflow

Split array into sub-arrays based on value

5 Answers 5

1 Comment

6 Comments

1 Comment

1 Comment

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

1 Comment

6 Comments

1 Comment

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related