Ruby on rails 转换多组结果集的哈希结果

这是

Product.group（[：name，：category]）.order（[：name，：category]）.count的输出

：

{
  ["Product A", "Category 2"]=>42,
  ["Product A", "Category 3"]=>83,
  ["Product A", "Category 4"]=>47,
  ["Product B", "Category 2"]=>1,
  ["Product B", "Category 3"]=>4,
  ["Product B", "Category 4"]=>10,
  ["Product C", "Category 3"]=>2,
  ["Product C", "Category 4"]=>4,
  ["Product D", "Category 1"]=>6,
  ["Product D", "Category 2"]=>13,
  ["Product D", "Category 3"]=>57,
  ["Product D", "Category 4"]=>27
}

每种产品可分为1-4类。我需要有0的计数，以及为最终的转换

所需转换数组列：产品名称、类别1计数、类别2计数、类别3计数和类别4计数：

[
  ["Product A",  0, 42, 83, 47],
  ["Product B",  0,  1,  4, 10],
  ["Product C",  0,  0,  2,  4],
  ["Product D",  6, 13, 57, 27]
]

任意类别名称的另一个选项：

CATEGORY_NAMES = ["Category 1", "Category 2", "Category 3", "Category 4"]

groups.group_by { |(p,_),_| p }.map do |product, prod_groups| 
  sub_groups = Hash[prod_groups.map { |(p, c), v| [c, v] }]
  sub_groups.default = 0
  [product, *CATEGORY_NAMES.map { |c| sub_groups[c] }]
end
# => [["Product A", 0, 42, 83, 47], ["Product B", 0, 1, 4, 10], ["Product C", 0, 0, 2, 4], ["Product D", 6, 13, 57, 27]] 

---

很好地使用了解构绑定。

CATEGORY_NAMES = ["Category 1", "Category 2", "Category 3", "Category 4"]

groups.group_by { |(p,_),_| p }.map do |product, prod_groups| 
  sub_groups = Hash[prod_groups.map { |(p, c), v| [c, v] }]
  sub_groups.default = 0
  [product, *CATEGORY_NAMES.map { |c| sub_groups[c] }]
end
# => [["Product A", 0, 42, 83, 47], ["Product B", 0, 1, 4, 10], ["Product C", 0, 0, 2, 4], ["Product D", 6, 13, 57, 27]]