如何根据键名合并2个数组并根据合并后的值排序？ [英] How to merge 2 arrays based on a key name and sort based on the merged value?

问题描述

假设我有两个列表

const listA = [{"apple":100}, {"banana": 50}, {"pearl": 10}, {"cherry": 5}, {"kiwi": 3}]
const listB = [{"peach": 30}, {"apple": 15}, {"kiwi": 10}, {"mango": 5}]

问题是怎么办我将两个列表合并为一个堆栈，并以数字增量将同一项目合并，然后按数量排序？我的意思是最终结果应该是->

The question is how do I merge two list into one stack up the same item with number increment and sort by the qty? I mean the final result should be ->

const listMerged = [{"apple":115}, {"banana": 50} , {"peach": 30}, {"kiwi": 13}, {"pearl": 10}, {"cherry": 5}, {"mango": 5}]

我知道它会是这样的：

sortListDesc(list) {

    return obj.sort(function (l1,l2) {
      return l2< l1 ? -1
           : l2 >l1 ? 1
           : 0
    })
  }

但是

推荐答案

您可以使用 reduce 和 排序 和 Object.values 像这样：

You can use reduce, and sort and Object.values like this:

const listA = [{"apple":100}, {"banana": 50}, {"pearl": 10}, {"cherry": 5}, {"kiwi": 3}]
, listB = [{"peach": 30}, {"apple": 15}, {"kiwi": 10}, {"mango": 5}]

let merged = Object.values(listA.concat(listB).reduce((acc, a) => {
  const [k, v] = Object.entries(a)[0];
  (acc[k] = acc[k] || {[k]: 0})[k] += v;
  return acc;
}, {}));

merged.sort((a, b) => Object.values(b)[0] - Object.values(a)[0]);
console.log(merged);

或者，

使用 reduce 创建一个对象，其中所有水果为键，单个总和为值。然后使用 Object。条目 ，排序和地图，例如：

Using reduce, create an object with all the fruits as keys and individual sum as values. Then use Object.entries, sort and map like this:

const listA = [{"apple":100}, {"banana": 50}, {"pearl": 10}, {"cherry": 5}, {"kiwi": 3}]
, listB = [{"peach": 30}, {"apple": 15}, {"kiwi": 10}, {"mango": 5}]

let merged2 = listA.concat(listB).reduce((acc, a) => {
  const [k, v] = Object.entries(a)[0];
  acc[k] = (acc[k] || 0) + v;
  return acc;
}, {});

const final = Object.entries(merged2)
  .sort(([, v1], [, v2]) => v2 - v1)
  .map(([k, v]) => ({[k]: v}))

console.log(final);

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