# Grouping elements of an array?

Hello,

What’s the best way to do the following?

[1,2,3,4,5,6,7,8,9] => [[1,2,3],[4,5,6],[7,8,9]]

I want to transform the array into an array of arrays with fixed number
of elements.
I think there should be a method but I can’t find it.

I can iterate the elements using a counter to divide but it’s so
boring.

Sam

class Array
def group_in(how_many)
ret_array = []
new_array = []
self.each {|elem| how_many.times { new_array << self.shift };
ret_array << new_array; new_array = []; }
ret_array
end
end

x = [1,2,3,4,5,6,7,8,9]
x = x.group_in(3)
require ‘pp’
pp x

[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

That should do it. Not sure if it’s the best way, but it will work.

–Jeremy

On 1/8/07, Sam K. [email protected] wrote:

I can iterate the elements using a counter to divide but it’s so
boring.

Sam

My free Ruby e-book:
http://www.humblelittlerubybook.com/book/

My blogs:

http://www.rubyinpractice.com/

On 08/01/07, Sam K. [email protected] wrote:

I can iterate the elements using a counter to divide but it’s so
boring.

Sam

requre ‘enumerator’

[1,2,3,4,5,6,7,8,9].enum_slice(3).inject([]){|array,slice| array <<
slice}

Farrel

FYI. B is what you expect:
A=[1,2,3,4,5,6,7,8,9];B=[];
3.times {B<<A.slice!(0…2);}

Farrel L. wrote:

requre ‘enumerator’

[1,2,3,4,5,6,7,8,9].enum_slice(3).inject([]){|array,slice| array << slice}

Farrel

requre ‘enumerator’
[ 1, 2, 3, 4, 5, 6, 7, 8 ].enum_slice(3).to_aj

Without “require”:
[ 1, 2, 3, 4, 5, 6, 7, 8 ].inject([[]]){|a,e|
a << [] if a.last.size==3; a.last << e; a}

Hi William,

William J. wrote:

Without “require”:
[ 1, 2, 3, 4, 5, 6, 7, 8 ].inject([[]]){|a,e|
a << [] if a.last.size==3; a.last << e; a}

This looks really clever.
I like it.

Sam

Farrel L. wrote:

On 08/01/07, Sam K. [email protected] wrote:

What’s the best way to do the following?

[1,2,3,4,5,6,7,8,9] => [[1,2,3],[4,5,6],[7,8,9]]

[1,2,3,4,5,6,7,8,9].enum_slice(3).inject([]){|array,slice| array << slice}

Alternately,

[1,2,3,4,5,6,7,8,9].enum_for(:each_slice,3).to_a
=> [[1, 2, 3], [4, 5, 6], [7, 8, 9]]