找出“扁平化索引”来自“堆叠索引”数组的

发布于 2024-10-13 07:05:08 字数 660 浏览 0 评论 0原文

我试图提出一个方程，从“堆叠索引”中以数学方式确定数组的“扁平索引”。请观察以下 Ruby 示例。

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

        [[ 9,10,11,12],
         [13,14,15,16]]]

在此示例中，matx 是一个三维矩阵，元素 7 位于 matx[0][1][2]。然而，在下一个示例中：

matx.flatten!  # => [1, 2, 3, 4, 5, 6, 7, 8, 
               #     9, 10, 11, 12, 13, 14, 15, 16]

现在元素 7 位于 matx[6]。

所以本质上，我正在寻找一种方法，在给定矩阵的维度和特定元素的索引集的情况下，从堆叠矩阵转换为展平矩阵。 ~~Reverse 也很棒，但我认为获得该结果的方法与获得此结果的方法类似（但本质上是相反的）。~~ 我意识到 reverse 实际上并不是一个函数，因为没有一种必然能够区分 5 映射到 [2,3] 还是 [3,2] 等的方法。所以我不打算研究这个。

原文

I'm trying to come up with an equation to mathematically determine the "flattened index" of an array from the "stacked index." Observe the following example in Ruby.

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

        [[ 9,10,11,12],
         [13,14,15,16]]]

In this example, matx is a three dimensional matrix, and the element 7 is located at matx[0][1][2]. However, in the next example:

matx.flatten!  # => [1, 2, 3, 4, 5, 6, 7, 8, 
               #     9, 10, 11, 12, 13, 14, 15, 16]

Now the element 7 is located at matx[6].

So essentially, I'm looking for a way to, given the dimensions of the matrix and the set of indices for the particular element, convert from the stacked matrix to the flattened matrix. ~~Reverse would be awesome, too, but I figure the way to get that is similar (but essentially reversed) to the method of obtaining this result.~~ I realized that reverse is not actually a function, because there's no way to necessarily tell the difference as to whether 5 maps to [2,3] or [3,2], etc. So I'm not going to look into that one.

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末蓝 2024-10-20 07:05:08

class Index
  def initialize *dims
    @dims = dims.reverse
  end

  def if_flat *subs
    raise unless @dims && @dims.size == subs.size
    res = 0
    subs.reverse.each_with_index { |s, i| res += s * @dims[0...i].inject(1) { |m, e| m * e }}
    res
  end
end
puts Index.new(2, 2, 4).if_flat 0, 1, 2

class Index
  def initialize *dims
    @dims = dims.reverse
  end

  def if_flat *subs
    raise unless @dims && @dims.size == subs.size
    res = 0
    subs.reverse.each_with_index { |s, i| res += s * @dims[0...i].inject(1) { |m, e| m * e }}
    res
  end
end
puts Index.new(2, 2, 4).if_flat 0, 1, 2

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