Group array elements in groups of two

ruby-talk
1

A better way to do this? :

arr= [1, 2, 3, 4, 5, 6, 7, 8]
new= []
while !arr.empty?
  elem1, elem2= arr.pop, arr.pop
  new << [elem2, elem1];
end
new.reverse!

new= [[1, 2], [3, 4], [5, 6], [7, 8]]

Thanks!

···

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2

Alle lunedì 17 settembre 2007, Emmanuel Oga ha scritto:

A better way to do this? :

arr= [1, 2, 3, 4, 5, 6, 7, 8]
new=
while !arr.empty?
  elem1, elem2= arr.pop, arr.pop
  new << [elem2, elem1];
end
new.reverse!

new= [[1, 2], [3, 4], [5, 6], [7, 8]]

Thanks!

Use each_slice:
  
  require 'enumerator'
  arr = [1, 2, 3, 4, 5, 6, 7, 8]
  new =
  arr.each_slice(2){|s| new << s}

or each_slice, enum_for an inject:

  require 'enumerator'
  arr = [1, 2, 3, 4, 5, 6, 7, 8]
  new = arr.enum_for(:each_slice , 2).inject(){|res, c| res << c}

I hope this helps

Stefano

3

irb(main):012:0> require 'enumerator'
=> true
irb(main):013:0> new =
=>
irb(main):014:0> [1,2,3,4,5,6].each_slice(2) do |slice|
irb(main):015:1* new << slice
irb(main):016:1> end
=> nil
irb(main):017:0> new
=> [[1, 2], [3, 4], [5, 6]]

···

On 17/09/2007, Emmanuel Oga <oga_emmanuel_oga@yahoo.com.ar> wrote:

A better way to do this? :

arr= [1, 2, 3, 4, 5, 6, 7, 8]
new=
while !arr.empty?
  elem1, elem2= arr.pop, arr.pop
  new << [elem2, elem1];
end
new.reverse!

new= [[1, 2], [3, 4], [5, 6], [7, 8]]

Thanks!

4

A better way to do this? :

arr= [1, 2, 3, 4, 5, 6, 7, 8]
new=
while !arr.empty?
  elem1, elem2= arr.pop, arr.pop
  new << [elem2, elem1];
end
new.reverse!

new= [[1, 2], [3, 4], [5, 6], [7, 8]]

>> a = (1..8).to_a
=> [1, 2, 3, 4, 5, 6, 7, 8]
>> require "enumerator"
=> true
>> a.enum_slice(2).to_a
=> [[1, 2], [3, 4], [5, 6], [7, 8]]

Hope that helps.

James Edward Gray II

···

On Sep 17, 2007, at 10:18 AM, Emmanuel Oga wrote:

5

This question has come up several times.

require 'enumerator'
    ==>true
[1, 2, 3, 4, 5, 6, 7, 8].enum_slice(2).to_a
    ==>[[1, 2], [3, 4], [5, 6], [7, 8]]

Without 'require':

f=nil
    ==>nil
[1, 2, 3, 4, 5, 6, 7, 8].partition{f=!f}.transpose
    ==>[[1, 2], [3, 4], [5, 6], [7, 8]]

···

On Sep 17, 10:18 am, Emmanuel Oga <oga_emmanuel_...@yahoo.com.ar> wrote:

A better way to do this? :

arr= [1, 2, 3, 4, 5, 6, 7, 8]
new=
while !arr.empty?
  elem1, elem2= arr.pop, arr.pop
  new << [elem2, elem1];
end
new.reverse!

new= [[1, 2], [3, 4], [5, 6], [7, 8]]

6

Hi,

···

In message "Re: group array elements in groups of two" on Tue, 18 Sep 2007 00:18:09 +0900, Emmanuel Oga <oga_emmanuel_oga@yahoo.com.ar> writes:

A better way to do this? :

arr= [1, 2, 3, 4, 5, 6, 7, 8]
new=
while !arr.empty?
elem1, elem2= arr.pop, arr.pop
new << [elem2, elem1];
end
new.reverse!

new= [[1, 2], [3, 4], [5, 6], [7, 8]]

require 'enumerator'
arr= [1, 2, 3, 4, 5, 6, 7, 8]
new=arr.to_enum(:each_slice, 2).to_a

7

Stefano Crocco wrote:

  require 'enumerator'
  arr = [1, 2, 3, 4, 5, 6, 7, 8]
  new = arr.enum_for(:each_slice , 2).inject(){|res, c| res << c}

You can simply replace the inject with a call to to_a:
new = arr.enum_for(:each_slice , 2).to_a

···

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8

very usefull. Thank you!

···

--
Posted via http://www.ruby-forum.com/.

9

Nirvana at last! I won a round of golf with
Matz!

YM: .to_enum(:each_slice, 2).to_a
WJ: .enum_slice(2).to_a

···

On Sep 17, 10:35 am, Yukihiro Matsumoto <m...@ruby-lang.org> wrote:

Hi,

In message "Re: group array elements in groups of two" > on Tue, 18 Sep 2007 00:18:09 +0900, Emmanuel Oga <oga_emmanuel_...@yahoo.com.ar> writes:

>A better way to do this? :
>
>arr= [1, 2, 3, 4, 5, 6, 7, 8]
>new=
>while !arr.empty?
> elem1, elem2= arr.pop, arr.pop
> new << [elem2, elem1];
>end
>new.reverse!
>
>new= [[1, 2], [3, 4], [5, 6], [7, 8]]

require 'enumerator'
arr= [1, 2, 3, 4, 5, 6, 7, 8]
new=arr.to_enum(:each_slice, 2).to_a

10

Perhaps, but you both sliced your shots.

···

On Sep 17, 10:07 am, William James <w_a_x_...@yahoo.com> wrote:

Nirvana at last! I won a round of golf with
Matz!

YM: .to_enum(:each_slice, 2).to_a
WJ: .enum_slice(2).to_a

11

Hi,

···

In message "Re: group array elements in groups of two" on Tue, 18 Sep 2007 01:10:10 +0900, William James <w_a_x_man@yahoo.com> writes:

Nirvana at last! I won a round of golf with
Matz!

YM: .to_enum(:each_slice, 2).to_a
WJ: .enum_slice(2).to_a

Well done, but your version does not work on 1.8. :wink:

              matz.

12

William James wrote:

[...]
Nirvana at last! I won a round of golf with
Matz!

YM: new=arr.to_enum(:each_slice, 2).to_a
WJ: new=arr.enum_slice(2).to_a

  SK: new=*arr.enum_slice(2)

cheers

Simon

13

Yukihiro Matsumoto wrote:

Hi,

>Nirvana at last! I won a round of golf with
>Matz!
>
>YM: .to_enum(:each_slice, 2).to_a
>WJ: .enum_slice(2).to_a

Well done, but your version does not work on 1.8. :wink:

              matz.

Will the world stop turning or am i just trapped in a parallel universe?

$ ruby -v -renumerator -e "p [1,2,3,4,5,6,7,8,9,0].enum_slice(2).to_a"
ruby 1.8.4 (2005-12-24) [i386-cygwin]
[[1, 2], [3, 4], [5, 6], [7, 8], [9, 0]]

cheers

Simon

···

In message "Re: group array elements in groups of two" > on Tue, 18 Sep 2007 01:10:10 +0900, William James <w_a_x_man@yahoo.com> writes: