Question

假设我有以下数组：

a = (1..10).to_a

是否有一个内置的ruby（或rails）功能能够将阵列分成完全N个大致相等的部分，同时保持顺序？

我正在寻找类似的东西：

a.bucketize(3)
=> [[1,2,3,4],[5,6,7],[8,9,10]]
a.bucketize(5)
=> [[1,2],[3,4],[5,6],[7,8],[9,10]]

提示：each_slice不会这样做。

另外，我知道我可以自己编写此函数并打开Array类或Enumerable模块。

感谢。

Answer 1

我会这样做：

ary = (1..10).to_a

ary.each_slice((ary.length.to_f/3).ceil).to_a
=> [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10]]

ary.each_slice((ary.length.to_f/5).ceil).to_a
=> [[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]]

这并不完美，但确实很接近：

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

ary.each_slice((ary.length.to_f/2).ceil).to_a
=> [[1, 2, 3, 4, 5], [6, 7, 8, 9]]

ary.each_slice((ary.length.to_f/3).ceil).to_a
=> [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

ary.each_slice((ary.length.to_f/4).ceil).to_a
=> [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Answer 2

使用功能方法可以最好地解决这类任务。这是一个尾递归函数实现（除了在数组上有效累积的不可避免的<<之外）：

class Array
  def bucketize(n, index = 0, acc = [])
    return acc if n <= 0 || size <= index
    n0 = ((size - index).to_f / n).ceil
    bucketize(n - 1, index + n0, acc << self[index, n0])
  end
end

(1..9).to_a.bucketize(3)
#=> [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
(1..10).to_a.bucketize(3)
#=> [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
(1..11).to_a.bucketize(3)
#=> [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11]]

Answer 3

这是我最终做的事情：

class Array
  def bucketize(n)
    return [] if (buckets = n.to_i) <= 0
    j = length / buckets.to_f
    result = each_with_index.chunk { |_, i| (i / j).floor }.map { |_, v| v.map(&:first) }
    result << [] until result.length == buckets
    result
  end
end

示例：

a = (1..10).to_a
=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
a.bucketize(1)
=> [[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]]
a.bucketize(2)
=> [[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]]
a.bucketize(3)
=> [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
a.bucketize(4)
=> [[1, 2, 3], [4, 5], [6, 7, 8], [9, 10]]
...
a.bucketize(9)
=> [[1, 2], [3], [4], [5], [6], [7], [8], [9], [10]]
...
a.bucketize(11)
=> [[1], [2], [3], [4], [5], [6], [7], [8], [9], [10], []]

Ruby“bucketize”一个数组

3 个答案: