ruby - 如果数组中存在键，如何从另一个哈希中获取哈希

Question

我有一个哈希，说：

hash1 ={1=>"a",2=>"b",3=>"c",4=>"d"}

和一个数组，说：

arr=[2,3]

我必须找到一个结果哈希，如：

hash2={2="b",3=>"c"}

也就是说，生成的散列必须只包含给定数组中存在键的键值对。没有循环可以做到这一点吗？

Answer 1

以下将做你想要的，但会破坏原来的hash1.

hash2 = hash1.keep_if {|k,v| arr.include? k}

以下将做你想做的，但保持hash1原来的样子。

hash2 = hash1.select {|k,v| arr.include? k}

Answer 2

  hash1.select {|k,v| arr.member? k}  # {2=>"b", 3=>"c"}

Answer 3

你在寻找这样的东西吗？

1.9.3p392 :001 > hash1 ={1=>"a",2=>"b",3=>"c",4=>"d"}
 => {1=>"a", 2=>"b", 3=>"c", 4=>"d"} 
1.9.3p392 :002 > arr=[2,3]
 => [2, 3] 
1.9.3p392 :003 > hash2 = hash1.keep_if{|key, value| arr.include?(key)}
 => {2=>"b", 3=>"c"} 

我知道你说没有循环，但这已经尽可能接近了

Answer 4

到目前为止所有答案的基准：

require 'fruity'

hash1 = Hash[[*1..10000].zip[*1.10000]]
arr   = 1.upto(1000).select(&:odd?)

compare do
  keep_if_include do
    hash2 = hash1.dup.keep_if {|k,v| arr.include? k}
  end

  values_at_zip do
    hash2 = Hash[arr.zip(hash1.dup.values_at(*arr))]
  end

  select_member do
    hash2 = hash1.dup.select {|k,v| arr.member? k}
  end
end

结果：

Running each test 4096 times. Test will take about 17 seconds.
select_member is faster than keep_if_include by 2x ± 0.1
keep_if_include is faster than values_at_zip by 410x ± 100.0

Answer 5

如果您使用的是导轨，请尝试

hash2 = hash1.sliec(*arr)

如果你只是使用红宝石，试试这个

hash2 = hash1.select {|k, v| arr.include?(k) }