有沒有辦法將這個數組轉換成哈希沒有注入方法？

Question

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

轉換動物：

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

Answer 1

這是通過每個數組元素迭代完成的另一種方法：

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 2

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 3

你如果您使用r，可以使用to_h方法uby < = 2.1。

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} 

Answer 4

我用Enumerable#group_by。更好的方法是使用計數散列，@Зелёный完成了。

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 5

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

隨着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