获取数组元素的索引比O（n）快 [英] Get index of array element faster than O(n)

问题描述

鉴于我有一个巨大的数组，以及它的值。我想得到数组中的值的索引。有没有其他方法，而是调用数组#index 来获取它？问题来自需要保持真正庞大的数组并且调用数组#index 的次数很多。

Given I have a HUGE array, and a value from it. I want to get index of the value in array. Is there any other way, rather then call Array#index to get it? The problem comes from the need of keeping really huge array and calling Array#index enormous amount of times.

之后几次尝试我发现通过使用（值，索引）字段而不是值本身存储结构，缓存内部元素内的索引给了一个巨大的步骤在性能上（20倍胜利）。

After a couple of tries I found that caching indexes inside elements by storing structs with (value, index) fields instead of the value itself gives a huge step in performance (20x times win).

我仍然想知道是否有更方便的方法来查找en元素的索引而不进行缓存（或者有一个很好的缓存技术可以提升性能）。

Still I wonder if there's a more convenient way of finding index of en element without caching (or there's a good caching technique that will boost up the performance).

推荐答案

将数组转换为哈希值。然后寻找密钥。

Convert the array into a hash. Then look for the key.

array = ['a', 'b', 'c']
hash = Hash[array.map.with_index.to_a]    # => {"a"=>0, "b"=>1, "c"=>2}
hash['b'] # => 1

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