javascript - 按值对对象的属性进行排序并保留值

Question

我创建了一个包含重复数据的食物名称数组。通过使用:

listFoodnames.forEach(x => {
  counts[x] = (counts[x] || 0) + 1;
});

我可以计算数组中每个项目存在的次数。

使用:

Object.keys(counts).sort((a, b) => counts[b]-counts[a])

我可以将项目的数量从大到小排序，但是，我丢失了我想保留的实际数量。

我该如何解决这个问题？

$('document').ready(function() {
  var listFoodnames = ["Batteries", "Bread", "Milk", "Bread", "Milk", "Milk"]
  var counts = {};

  listFoodnames.forEach(x => {
    counts[x] = (counts[x] || 0) + 1;
  });

  console.log(counts) //LOGS EACH ITEM AS PROPERTY AND OCCURRENCES AS VALUE
  
  var sortedCounts = Object.keys(counts).sort((a, b) => counts[b]-counts[a])
  
  console.log(sortedCounts) //LOSES VALUES
})

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Answer 1

不幸的是，根据规范，不能保证对象属性是有序的。您可以尝试使用对象数组。

另请注意，当您将数组中的项目组合成单个对象时，reduce 比 forEach 更合适。

const listFoodnames = ["Batteries", "Bread", "Milk", "Bread", "Milk", "Milk"];
const counts =
  Object.values(
    listFoodnames.reduce((a, foodName) => {
      if (!a[foodName]) a[foodName] = { foodName, count: 0 };
      a[foodName].count++;
      return a;
    }, {})
  );
counts.sort((a, b) => a.count - b.count);
console.log(counts);