如果我有一个Hash
,比如有子灰,它们嵌套在Hash
:值的数组中
{"school" => "School name", "teachers" => [{"name" => "Schmith", "age" => 34},{"name" => "McRight", "age" => 51}]}
def get_keys(hash)
(hash.keys + hash.values.grep(Hash){|sub_hash| get_keys(sub_hash)}).flatten
end
使用前面的函数,我只能获得["school","teachers"]
,但我还需要哈希的密钥,它们位于"教师"值的数组中:["school","teachers","name","age","name","age"]
最好我想消除结果中的重复:["school","teachers","name","age"]
def get_keys(object)
if object.is_a? Hash
(object.keys + get_keys(object.values)).flatten.uniq
elsif object.is_a? Array
object.collect{|value| get_keys value}
else
[]
end
end
我认为这很清楚,你可以随心所欲地筑巢。如果对象不是数组或散列,那么就没有键。如果它是一个数组,它将查找可能在所有元素中的键如果它是一个散列,它将返回它的密钥,加上散列值上的散列的密钥。然后应用flatten
来删除可能的空数组,应用uniq
来删除重复的值。
使用数组,它将获得内部的密钥
您可以使用Enumerable#each_with_object
执行类似的操作,当值为Array时,为数组中的每个对象递归调用相同的函数。
hash = {"school" => "School name", "teachers" => [{"name" => "Schmith", "age" => 34},{"name" => "McRight", "age" => 51}]}
def get_keys(hash)
hash.each_with_object([]) do |(k, v), arr|
arr << k
if v.is_a? Array
arr << v.map { |hash2| get_keys(hash2) }.uniq
end
end
end
puts get_keys(hash)
=> ["school", "teachers", "name", "age"]