在哈希中交换键和值

时间:2012-06-12 00:46:15

标签: ruby hashmap

在Ruby中,如何在Hash上交换键和值?

假设我有以下哈希:

{:a=>:one, :b=>:two, :c=>:three}

我想转变为:

{:one=>:a, :two=>:b, :three=>:c}

使用地图似乎相当乏味。是否有更短的解决方案?

6 个答案:

答案 0 :(得分:247)

Ruby有一个哈希的辅助方法,可以让你将哈希处理为反转。

{a: 1, b: 2, c: 3}.key(1)
=> :a

如果你想保留反向哈希,那么Hash#invert应该适用于大多数情况。

{a: 1, b: 2, c: 3}.invert
=> {1=>:a, 2=>:b, 3=>:c}

<强> BUT ...

如果您有重复值,invert将丢弃除最后一个值之外的所有值。同样,key只会返回第一个匹配项。

{a: 1, b: 2, c: 2}.key(2)
=> :b

{a: 1, b: 2, c: 2}.invert
=> {1=>:a, 2=>:c}

所以..如果你的值是唯一的,你可以使用Hash#invert,如果没有,那么你可以将所有值保存为数组,如下所示:

class Hash
  # like invert but not lossy
  # {"one"=>1,"two"=>2, "1"=>1, "2"=>2}.inverse => {1=>["one", "1"], 2=>["two", "2"]} 
  def safe_invert
    each_with_object({}) do |(key,value),out| 
      out[value] ||= []
      out[value] << key
    end
  end
end

注意:此代码包含测试现在为here

或者简而言之......

class Hash
  def safe_invert
    self.each_with_object({}){|(k,v),o|(o[v]||=[])<<k}
  end
end

答案 1 :(得分:61)

你打赌有一个!在Ruby中总是有一种更简单的方法!

这很简单,只需使用Hash#invert

{a: :one, b: :two, c: :three}.invert
=> {:one=>:a, :two=>:b, :three=>:c}

Etvoilà!

答案 2 :(得分:2)

files = {
  'Input.txt' => 'Randy',
  'Code.py' => 'Stan',
  'Output.txt' => 'Randy'
}

h = Hash.new{|h,k| h[k] = []} # Create hash that defaults unknown keys to empty an empty list
files.map {|k,v| h[v]<< k} #append each key to the list at a known value
puts h

这也将处理重复值。

答案 3 :(得分:1)

# this doesn't looks quite as elegant as the other solutions here,
# but if you call inverse twice, it will preserve the elements of the original hash

# true inversion of Ruby Hash / preserves all elements in original hash
# e.g. hash.inverse.inverse ~ h

class Hash

  def inverse
    i = Hash.new
    self.each_pair{ |k,v|
      if (v.class == Array)
        v.each{ |x|
          i[x] = i.has_key?(x) ? [k,i[x]].flatten : k
        }
      else
        i[v] = i.has_key?(v) ? [k,i[v]].flatten : k
      end
    }
    return i
  end

end

Hash#inverse为您提供:

 h = {a: 1, b: 2, c: 2}
 h.inverse
  => {1=>:a, 2=>[:c, :b]}
 h.inverse.inverse
  => {:a=>1, :c=>2, :b=>2}  # order might not be preserved
 h.inverse.inverse == h
  => true                   # true-ish because order might change

而内置的invert方法刚刚破解:

 h.invert
  => {1=>:a, 2=>:c}    # FAIL
 h.invert.invert == h 
  => false             # FAIL

答案 4 :(得分:1)

使用数组

input = {:key1=>"value1", :key2=>"value2", :key3=>"value3", :key4=>"value4", :key5=>"value5"}
output = Hash[input.to_a.map{|m| m.reverse}]

使用哈希

input = {:key1=>"value1", :key2=>"value2", :key3=>"value3", :key4=>"value4", :key5=>"value5"}
output = input.invert

答案 5 :(得分:0)

如果你有一个哈希键,那么键是唯一的,你可以使用Hash#invert

> {a: 1, b: 2, c: 3}.invert
=> {1=>:a, 2=>:b, 3=>:c} 

如果您使用非唯一键,那将无效,但是,只保留最后看到的键:

> {a: 1, b: 2, c: 3, d: 3, e: 2, f: 1}.invert
=> {1=>:f, 2=>:e, 3=>:d}

如果你有一个非唯一键的哈希,你可以这样做:

> hash={a: 1, b: 2, c: 3, d: 3, e: 2, f: 1}
> hash.each_with_object(Hash.new { |h,k| h[k]=[] }) {|(k,v), h| 
            h[v] << k
            }     
=> {1=>[:a, :f], 2=>[:b, :e], 3=>[:c, :d]}

如果散列的值已经是数组,则可以执行以下操作:

> hash={ "A" => [14, 15, 16], "B" => [17, 15], "C" => [35, 15] }
> hash.each_with_object(Hash.new { |h,k| h[k]=[] }) {|(k,v), h| 
            v.map {|t| h[t] << k}
            }   
=> {14=>["A"], 15=>["A", "B", "C"], 16=>["A"], 17=>["B"], 35=>["C"]}