数组转换为散列红宝石，而preserving数组索引值

问题描述

我有一个数组 X 在它值的数量。下面的阵列只有4个，但我需要code是动态的，仅具有四个数组对象不依赖。

I have an array that has X number of values in it. The following array only has 4, but I need the code to be dynamic and not reliant on only having four array objects.

数组= [成人，家庭，单，孩子]

我想阵列转换为一个哈希看起来像这样：

I want to convert array to a hash that looks like this:

哈希= {0 =＆GT; 成人，1 =＆GT; 家庭，2 =＆GT; '单'，3 =＆GT; '儿童'}

的哈希应该有尽可能多的键/值对作为阵列具有目的和值应在0和递增1为每个对象开始

The hash should have as many key/value pairs as the array has objects, and the values should start at 0 and increment by 1 for each object.

推荐答案

使用<$c$c>Enumerable#each_with_index:

Hash[array.each_with_index.map { |value, index| [index, value] }]
# => {0=>"Adult", 1=>"Family", 2=>"Single", 3=>"Child"}

由于@hirolau评论， each_with_index.map 也可以写成 map.with_index 。

Hash[array.map.with_index { |value, index| [index, value] }]
# => {0=>"Adult", 1=>"Family", 2=>"Single", 3=>"Child"}

更新

Alterantive使用：

Alterantive that use Hash#invert:

Hash[array.map.with_index{|*x|x}].invert
# => {0=>"Adult", 1=>"Family", 2=>"Single", 3=>"Child"}
Hash[[*array.map.with_index]].invert
# => {0=>"Adult", 1=>"Family", 2=>"Single", 3=>"Child"}

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