arrays - 有没有办法在没有注入方法的情况下将此数组转换为哈希？

Question

animals = [['dogs', 4], ['cats', 3], ['dogs', 7]]

将动物转换为：

{'dogs' => 11, 'cats' => 3}

Answer 1

您可以使用each_with_object：

=> array =  [['dogs', 4], ['cats', 3], ['dogs', 7]]
=> array.each_with_object(Hash.new(0)) do |(pet, n), accum| 
=>   accum[pet] += n
=> end
#> {'dogs' => 11, 'cats' => 3}

Answer 2

我使用了 Enumerable#group_by。更好的方法是使用计数哈希，@Зелёный 已经这样做了。

animals = [['dogs', 4], ['cats', 3], ['dogs', 7]]

animals.group_by(&:first).tap { |h| h.keys.each { |k| h[k] = h[k].transpose[1].sum } }
  #=> {"dogs"=>11, "cats"=>3}

Answer 3

data = [['dogs', 4], ['cats', 3], ['dogs', 7]]
data.dup
    .group_by(&:shift)
    .map { |k, v| [k, v.flatten.reduce(:+)] }
    .to_h

与Hash#merge：

data.reduce({}) do |acc, e|
  acc.merge([e].to_h) { |_, v1, v2| v1 + v2 }
end

data.each_with_object({}) do |e, acc|
  acc.merge!([e].to_h) { |_, v1, v2| v1 + v2 }
end

Answer 4

这是通过遍历每个数组元素完成的另一种方法：

animals = [['dogs', 4], ['cats', 3], ['dogs', 7]]

result = Hash.new(0)

animals.each do |animal|
    result[animal[0]] += animal[1].to_i
end

p result

Answer 5

to_h如果您使用的是 ruby​​ <= 2.1，则可以使用该方法。

例如：

animals = [['dogs', 4], ['cats', 3], ['dogs', 7]]
animals.group_by(&:first).map { |k,v| [k,v.transpose.last.reduce(:+)]}.to_h # return {"dogs"=>11, "cats"=>3}