将所有数字字符串递归转换为Ruby哈希中的整数 [英] Recursively convert all numeric strings to integers in a Ruby hash

问题描述

我有一个随机大小的哈希，它的值可能像"100"，我想将其转换为整数.我知道我可以使用value.to_i if value.to_i.to_s == value来做到这一点，但是考虑到值可以是字符串，也可以是(哈希或字符串的)数组，也可以是其他值，因此我不确定如何在哈希中递归地执行此操作.哈希.

I have a hash of a random size, which may have values like "100", which I would like to convert to integers. I know I can do this using value.to_i if value.to_i.to_s == value, but I'm not sure how would I do that recursively in my hash, considering that a value can be either a string, or an array (of hashes or of strings), or another hash.

推荐答案

这是一个非常简单的递归实现(尽管必须同时处理数组和哈希值会增加一些棘手的问题).

This is a pretty straightforward recursive implementation (though having to handle both arrays and hashes adds a little trickiness).

def fixnumify obj
  if obj.respond_to? :to_i
    # If we can cast it to a Fixnum, do it.
    obj.to_i

  elsif obj.is_a? Array
    # If it's an Array, use Enumerable#map to recursively call this method
    # on each item.
    obj.map {|item| fixnumify item }

  elsif obj.is_a? Hash
    # If it's a Hash, recursively call this method on each value.
    obj.merge( obj ) {|k, val| fixnumify val }

  else
    # If for some reason we run into something else, just return
    # it unmodified; alternatively you could throw an exception.
    obj

  end
end

而且，嘿，它甚至可以工作:

And, hey, it even works:

hsh = { :a => '1',
        :b => '2',
        :c => { :d => '3',
                :e => [ 4, '5', { :f => '6' } ]
              },
        :g => 7,
        :h => [],
        :i => {}
      }

fixnumify hsh
# => {:a=>1, :b=>2, :c=>{:d=>3, :e=>[4, 5, {:f=>6}]}, :g=>7, :h=>[], :i=>{}}

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