Splitting an array into arrays x
I have an array:
arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
I want to split arr1
into x slices where each slice is as complete and equal as possible.
arr2 = arr1.foo(3)
# => [1, 2, 3, 4][5, 6, 7][8, 9, 10]
each_slice
does the opposite of what I want, separating the array from groups of x elements.
arr2 = arr1.each_slice(3)
# => [1, 2, 3][4, 5, 6][7, 8, 9][10]
If possible, I want to do this without using special methods like in_groups
.
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3 answers
Another approach:
def in_groups(array, n)
a = array.dup
n.downto(1).map { |i| a.pop(a.size / i) }.reverse
end
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
in_groups(arr, 1) #=> [[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]]
in_groups(arr, 2) #=> [[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]]
in_groups(arr, 3) #=> [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
in_groups(arr, 4) #=> [[1, 2, 3], [4, 5, 6], [7, 8], [9, 10]]
in_groups(arr, 5) #=> [[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]]
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You can use recursion:
def in_groups(arr, n)
return [arr] if n == 1
len = arr.size/n
[arr[0,len]].concat in_groups(arr[len..-1], n-1)
end
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
in_groups(arr, 1) #=> [[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]]
in_groups(arr, 2) #=> [[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]]
in_groups(arr, 3) #=> [[1, 2, 3], [4, 5, 6], [7, 8, 9, 10]]
in_groups(arr, 4) #=> [[1, 2], [3, 4], [5, 6, 7], [8, 9, 10]]
in_groups(arr, 5) #=> [[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]]
in_groups(arr, 9) #=> [[1], [2], [3], [4], [5], [6], [7], [8], [9, 10]]
in_groups(arr, 10) #=> [[1], [2], [3], [4], [5], [6], [7], [8], [9], [10]]
in_groups(arr, 11) #=> [[], [1], [2], [3], [4], [5], [6], [7], [8], [9], [10]]
Edit 1: first for the largest groups, click on .reverse
or replace the antepult line like this:
len = (arr.size.to_f/n).ceil
Edit 2: Below is a small variation on @undur's answer, maybe easier to follow for those with "B" and "C" brain types:
class Array
def in_groups(n)
size_small, nbr_large = count.divmod(n)
size_large, nbr_small = size_small+1, n-nbr_large
nbr_for_large = nbr_large * size_large
self[0, nbr_for_large].each_slice(size_large).to_a.concat(
self[nbr_for_large..-1].each_slice(size_small).to_a)
end
end
(1..10).to_a.in_groups(3)
#=> [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
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