如何根据对象中的键值对将数组映射到新数组？答案

【问题标题】：How can you map an array to a new array based on the key-value pairs in an object?如何根据对象中的键值对将数组映射到新数组？
【发布时间】：2018-02-22 18:29:31
【问题描述】：

我正在编写一个可以为测试评分的脚本。答案是口头的，每个单词对应一个从 1 到 5 的特定数值。

我创建了一个记录这些通信的对象：

const answerValues = {
    "Consistently": 5,
    "Often": 4,
    "Sometimes": 3,
    "Rarely": 2,
    "Never": 1
  }

答案以如下结构的数组形式给出：

const answers = [
  ["Consistently, "Often", "Sometimes", "Rarely", "Never"],
  ["Often, "Sometimes", "Consistently", "Never", "Rarely"],
  ["Sometimes, "Rarely", "Consistently", "Rarely", "Often"]
]

我需要做的是将原始的answers 映射到一个新数组，而不是这样：

const answerResults = [
  [5, 4, 3, 2, 1],
  [4, 3, 5, 1, 2],
  [3, 2, 5, 1, 4]
]

我似乎无法让它工作；任何帮助将不胜感激。

PS 如果需要，我可以将answers 数组更改为一个对象，如果这样会更容易。

【问题讨论】：

标签： javascript arrays ecmascript-6

【解决方案1】：

使用map

var output = answers.map( s => s.map( t => answerValues[t] ) )

演示

var answerValues = {
  "Consistently": 5,
  "Often": 4,
  "Sometimes": 3,
  "Rarely": 2,
  "Never": 1
};
var answers = [
  ["Consistently", "Often", "Sometimes", "Rarely", "Never"],
  ["Often", "Sometimes", "Consistently", "Never", "Rarely"],
  ["Sometimes", "Rarely", "Consistently", "Rarely", "Often"]
];
var output = answers.map(s => s.map(t => answerValues[t]));
console.log(output);

说明

- use `map` to iterate `answers`,
-   use `map` for each `s` in `answers` and iterate the values
-     *replace* each value `t` with its `answerValues[t]`

【讨论】：

【解决方案2】：

您可以将外部数组与内部数组的对象的值进行映射。

const
    answerValues = { Consistently: 5, Often: 4, Sometimes: 3, Rarely: 2, Never: 1 },
    answers = [["Consistently", "Often", "Sometimes", "Rarely", "Never"],["Often", "Sometimes", "Consistently", "Never", "Rarely"], ["Sometimes", "Rarely", "Consistently", "Rarely", "Often"]],
    result = answers.map(a => a.map(k => answerValues[k]));

console.log(result);

【讨论】：

【解决方案3】：

简单使用嵌套映射，然后使用 answerValues 映射返回响应

const answers = [
  ["Consistently", "Often", "Sometimes", "Rarely", "Never"],
  ["Often", "Sometimes", "Consistently", "Never", "Rarely"],
  ["Sometimes", "Rarely", "Consistently", "Rarely", "Often"]
]

const answerValues = {
    "Consistently": 5,
    "Often": 4,
    "Sometimes": 3,
    "Rarely": 2,
    "Never": 1
}

const res = answers.map(answer => {
    return answer.map(resp => answerValues[resp]);
})
console.log(res)

【讨论】：