javascript - 当对象可以具有不同类型时如何按键对对象数组进行分组

Question

我有一个像这样的对象数组:

const input = [
    {
        "name" : "car",
        "sign" : "+",
        "options" : "benz"
    },
    {
        "name" : "bike",
        "sign" : "+",
        "options" : "pulsar"
    },
    {
        "name" : "bike",
        "sign" : "+",
        "options" : "enfield"
    },
    {
        "name" : "car",
        "sign" : "+",
        "options" : ["toyota","hyundai","benz"]
    },
    {
        "name" : "",
        "sign" : "",
        "options" : "" 
    },
    {
        "name" : "car",
        "sign" : "+",
        "options" : ["audi", "ford"]
    }
]

输入结构是这样的:每个对象有3个键值对。

1. 全部可以为空

2. 它可以有有效值:

   a.选项可以是字符串

   b.或者它可以是一个数组

我不想在输出中添加值为“”(空字符串)的对象，输出应该是这样的:

[
    {"car" : [ "benz", "toyota","hyundai", "audi", "ford"]},
    {"bike" : ["pulsar","enfield" ]}
]

其他各种答案没有显示当对象值可以是数组或字符串时如何处理，所以我正在写一个新问题。

我不确定是否必须使用reduce或map。 非常感谢任何帮助，提前致谢。

Answer 1

这是一个相当标准的“分组依据”操作，按一个属性进行分组并将另一个属性累加到一个数组中。

为了容纳字符串和数组，您使用 concat()将 options 值添加到累加器中。 Concat 接受单个值或值数组，并将后者展平到目标数组中。为了避免重复，您可以使用 Set如果 name 和 options 都为 true，下面的代码片段只会向累加器添加元素。

const input = [
  { "name": "", "sign": "+", "options": "benz" }, // blank 'name' property
  { "name": "bike", "sign": "+", "options": "" }, // blank 'options' property
  { "name": "bike", "sign": "+", "options": "enfield" },
  { "name": "car", "sign": "+", "options": ["toyota", "hyundai"] },
  { "name": "car", "sign": "", "options": "ford" },
  { "name": "car", "sign": "+", "options": ["audi", "ford"] }]

const result = Object.values(
  input.reduce((acc, { name, options }) => {
    if (name && options) {
      const category = (acc[name] || (acc[name] = { [name]: [] }));
      category[name] = [...new Set(category[name].concat(options))];
    }
    return acc;
  }, {}));

console.log(result)

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或者，为了避免在每次迭代时创建新的 Set，您可以累积到一个 Set 中，然后映射 reduce() 调用的结果，并将每个集合转换为数组。

const input = [
  { "name": "", "sign": "+", "options": "benz" }, // blank 'name' property
  { "name": "bike", "sign": "+", "options": "" }, // blank 'options' property
  { "name": "bike", "sign": "+", "options": "enfield" },
  { "name": "car", "sign": "+", "options": ["toyota", "hyundai"] },
  { "name": "car", "sign": "", "options": "ford" },
  { "name": "car", "sign": "+", "options": ["audi", "ford"] }]

const result = Object.entries(
  input.reduce((acc, { name, options }) => {
    if (name && options) {
      acc[name] || (acc[name] = new Set);
      [].concat(options).forEach(option => acc[name].add(option));
    }
    return acc;
  }, {}))
  .map(([name, options]) => ({ [name]: [...options] }));

console.log(result);

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由于您似乎无法使用 ES6 迭代器，因此这里有一个选项，仅使用 Array#includes() 检查以避免重复。

const input = [
  { "name": "", "sign": "+", "options": "benz" }, // blank 'name' property
  { "name": "bike", "sign": "+", "options": "" }, // blank 'options' property
  { "name": "bike", "sign": "+", "options": "enfield" },
  { "name": "car", "sign": "+", "options": ["toyota", "hyundai"] },
  { "name": "car", "sign": "", "options": "ford" },
  { "name": "car", "sign": "+", "options": ["audi", "ford"] }]

const result = Object.values(
  input.reduce((acc, { name, options }) => {
    if (name && options) {
      const category = (acc[name] || (acc[name] = { [name]: [] }));
      [].concat(options).forEach(option => {
        if (!category[name].includes(option)) category[name].push(option)
      });
    }
    return acc;
  }, {}));

console.log(result);

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ES3

var input = [
  { "name": "", "sign": "+", "options": "benz" },
  { "name": "bike", "sign": "+", "options": "" },
  { "name": "bike", "sign": "+", "options": "enfield" },
  { "name": "car", "sign": "+", "options": ["toyota", "hyundai"] },
  { "name": "car", "sign": "", "options": "ford" },
  { "name": "car", "sign": "+", "options": ["audi", "ford"] }
];

var temp = {};

for (var i = 0; i < input.length; i++) {
  var name_ = input[i].name;
  var options_ = input[i].options;

  if (name_ !== undefined && name_ !== "" && options_ !== undefined && options_ !== "") {

    if (!temp.hasOwnProperty(name_)) {
      temp[name_] = {};
      temp[name_][name_] = [];
    }

    var options_ = [].concat(options_);

    for (var j = 0; j < options_.length; j++) {
      if (!temp[name_][name_].includes(options_[j])) {
        temp[name_][name_].push(options_[j]);
      }
    }

  }
}

var result = [];

for (var key_ in temp) {
  if (temp.hasOwnProperty(key_)) {
    result.push(temp[key_]);
  }
}

console.log(result);

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