在 Ruby 中，如何根据键中单词的数量对散列中的元素进行排序？

Question

例如：

{"job" => 1, "big job" => 2, "super job" => 1, "work" => 2, "super big job" => 1}

我想变成这样

{"super big job" => 1, "big job" => 2, "super job" => 1, "job" => 1, , "work" => 2}

即三字键在前，二字键在后，一字键在最后。我知道我可以使用 key.split(' ').length 函数来计算单词数，但是我如何使用它来实际修改哈希中元素的顺序。

Answer 1

你可以这样做：

a = {"job" => 1, "big job" => 2, "super job" => 1, "work" => 2, "super big job" => 1}
Hash[a.sort_by { |k, v| k.split.size }.reverse]
# => {"super big job"=>1, "super job"=>1, "big job"=>2, "work"=>2, "job"=>1}

Answer 2

x = {"job" => 1, "big job" => 2, "super job" => 1, "work" => 2, "super big job" => 1}

p  x.sort_by{|x,|-x.split.size}.to_h #=> {"super big job"=>1, "big job"=>2, "super job"=>1, "work"=>2, "job"=>1}

在 Ruby 中，如何根据键中单词的数量对散列中的元素进行排序？

In Ruby, how do I sort the elements in a hash by the quantity of words in the keys?

ruby

hash