需要合并具有相同键但具有不同值的对象

Question

我想找到最短，最漂亮的方法来转换具有相同类别键值的对象数组：

[{
  "label": "Apple",
  "category": "Fruits"
}, {
  "label": "Orange",
  "category": "Fruits"
}, {
  "label": "Potato",
  "category": "Vegetables"
}, {
  "label": "Tomato",
  "category": "Vegetables"
}, {
  "label": "Cherry",
  "category": "Berries"
}]

到具有相同类别的分组标签的那个：

[{
  "label": ["Apple", "Orange"],
  "category": "Fruits"
}, {
  "label": ["Potato", "Tomato"],
  "category": "Vegetables"
}, {
  "label": ["Cherry"],
  "category": "Berries"
}]

Answer 1

您可以将对象用作哈希表并对类别进行分组。

 var data = [{ "label": "Apple", "category": "Fruits" }, { "label": "Orange", "category": "Fruits" }, { "label": "Potato", "category": "Vegetables" }, { "label": "Tomato", "category": "Vegetables" }, { "label": "Cherry", "category": "Berries" }], grouped = []; data.forEach(function (a) { if (!this[a.category]) { this[a.category] = { label: [], category: a.category }; grouped.push(this[a.category]); } this[a.category].label.push(a.label); }, Object.create(null)); console.log(grouped); 

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Answer 2

这是我如何使用lodash做到这一点：

_(coll)
  .groupBy('category')
  .map((v, k) => ({
    category: k,
    label: _.map(v, 'label')
  }))
  .value()

基本上， groupBy（）创建一个具有唯一类别作为键的对象。 然后， map（）将此对象转换回数组，其中每个项目都具有您需要的结构。

Answer 3

这是我尝试解决您的问题

 var result = []; var input = [{ "label": "Apple", "category": "Fruits" }, { "label": "Orange", "category": "Fruits" }, { "label": "Potato", "category": "Vegetables" }, { "label": "Tomato", "category": "Vegetables" }, { "label": "Cherry", "category": "Berries" }]; var cat = []; for(i = 0; i < input.length; i++) { if(!cat[input[i].category]) { cat[input[i].category] = {category: input[i].category, label:[input[i].label]}; result.push(cat[input[i].category]); } else { cat[input[i].category].label.push(input[i].label); } } console.log(result);