ruby-on-rails - 我们可以写一个没有哈希类的哈希吗？

Question

为了这个问题，我们假设我们在 Ruby 中没有 hash 类。

是否可以在不使用 Ruby 中的 hash class 的情况下创建 hash？或者，如果我们没有散列概念，我们可能会称它为关联数组。

Answer 1

虽然可以使用数组来模拟哈希的行为，但性能会受到影响:

require 'fruity'

CHARS = ('a' .. 'z').to_a
ARY = CHARS.zip(CHARS)
HASH = ARY.to_h

compare do 
  _assoc_a { ARY.assoc('a') }
  _hash_a  { HASH['a']      }
  _hash_m  { HASH['m']      }
  _assoc_m { ARY.assoc('m') }
  _assoc_z { ARY.assoc('z') }
  _hash_z  { HASH['z']      }
end
# >> Running each test 32768 times. Test will take about 3 seconds.
# >> _hash_m is similar to _hash_z (results differ: m vs z)
# >> _hash_z is similar to _hash_a (results differ: z vs a)
# >> _hash_a is faster than _assoc_a by 2x ± 1.0 (results differ: a vs ["a", "a"])
# >> _assoc_a is faster than _assoc_m by 5x ± 1.0 (results differ: ["a", "a"] vs ["m", "m"])
# >> _assoc_m is faster than _assoc_z by 2x ± 0.1 (results differ: ["m", "m"] vs ["z", "z"])

进行反向查找，从第二个元素到第一个，或者从值到键:

compare do
  _rassoc_a      { ARY.rassoc('a')  }
  _hash_rassoc_a { HASH.rassoc('a') }
  _rassoc_m      { ARY.rassoc('m')  }
  _hash_rassoc_m { HASH.rassoc('m') }
  _rassoc_z      { ARY.rassoc('z')  }
  _hash_rassoc_z { HASH.rassoc('z') }
end
# >> Running each test 32768 times. Test will take about 4 seconds.
# >> _rassoc_a is faster than _hash_rassoc_a by 2x ± 1.0
# >> _hash_rassoc_a is faster than _rassoc_m by 3x ± 1.0 (results differ: ["a", "a"] vs ["m", "m"])
# >> _rassoc_m is similar to _hash_rassoc_m
# >> _hash_rassoc_m is similar to _rassoc_z (results differ: ["m", "m"] vs ["z", "z"])
# >> _rassoc_z is similar to _hash_rassoc_z

对我来说，将散列反向查找到 key 会更慢，因为散列并未针对此类事情进行优化。如果我需要这样做，我会尝试创建一个单独的哈希来反转键和值，然后使用它来进行反向查找:

hash = {'a' => 1, 'b' => 2}
reversed_hash = hash.invert # => {1=>"a", 2=>"b"}

如果值是一个数组或值不是唯一的，这会中断，有时生成的键使用起来不是很方便。