Question

我想找到一种最短，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"
}]

来自同一类别的分组标签：

[{
  "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。

＆＃13;

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);

＆＃13;

.as-console-wrapper { max-height: 100% !important; top: 0; }

＆＃13;

Answer 2

以下是我将如何使用lodash：

_(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

以下是我尝试解决您的问题

＆＃13;

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);

＆＃13;

需要合并具有相同键但具有不同值的对象

3 个答案: