Question

我想知道合并2个对象数组itemsA和itemsB的最佳方法是什么。合并后的数据应位于mergedList中。

标准：

source='STAPLE'项不应在合并数组中重复。
包含任何其他来源的项目可能会重复。例如，name: 'Ball'项source: 'USER'可能会存在两次。

itemsA有6件商品，itemsB有7件商品，mergedList应该有11件商品

let itemsA = [
{name: 'Milk', source: 'STAPLE'},
{name: 'Bread', source: 'AD'},
{name: 'Egg', source: 'STAPLE'},
{name: 'Ball', source: 'USER'},
{name: 'Pasta', source: 'STAPLE'},
{name: 'Coke', source: 'AD'}];
let itemsB = [
{name: 'Milk', source: 'USER'},
{name: 'Bread', source: 'AD'},
{name: 'Egg', source: 'STAPLE'},
{name: 'Ball', source: 'USER'},
{name: 'Mango', source: 'USER'},
{name: 'Pasta', source: 'STAPLE'},
{name: 'Coke', source: 'USER'}]

mergedList应该等于

let mergedList = [
{name: 'Milk', source: 'STAPLE'},
{name: 'Bread', source: 'AD'},
{name: 'Egg', source: 'STAPLE'},
{name: 'Ball', source: 'USER'},
{name: 'Pasta', source: 'STAPLE'},
{name: 'Coke', source: 'AD'}] 
{name: 'Milk', source: 'USER'},
{name: 'Bread', source: 'AD'},
{name: 'Ball', source: 'USER'},
{name: 'Mango', source: 'USER'},
{name: 'Coke', source: 'USER'}]   ];

Answer 1

function merge(itemsA, itemsB) {
  let merged = [];
  itemsA.concat(itemsB).reduce((stapleSet, obj) => (obj.source != "STAPLE") ?
  (merged.push(obj), stapleSet) : 
  (stapleSet.has(obj.name) || merged.push(obj), 
  stapleSet.add(obj.name), stapleSet), new Set());

  return merged;
}

获得两个阵列。
创建合并数组以将对象推送到。
将两个项目数组连接在一起。
减少连接数组 - 如果对象源是主要的而不在stapleSet中，则将对象的名称添加到stapleSet - 设置对象只允许每个条目中的一个 - 然后将对象推送到merged数组。否则将对象推送到merged数组。
从函数返回合并的数组。

＆＃13;

let itemsA = [
{name: 'Milk', source: 'STAPLE'},
{name: 'Bread', source: 'AD'},
{name: 'Egg', source: 'STAPLE'},
{name: 'Ball', source: 'USER'},
{name: 'Pasta', source: 'STAPLE'},
{name: 'Coke', source: 'AD'}];
let itemsB = [
{name: 'Milk', source: 'USER'},
{name: 'Bread', source: 'AD'},
{name: 'Egg', source: 'STAPLE'},
{name: 'Ball', source: 'USER'},
{name: 'Mango', source: 'USER'},
{name: 'Pasta', source: 'STAPLE'},
{name: 'Coke', source: 'USER'}];

    function merge(itemsA, itemsB) {
      let merged = [];
      itemsA.concat(itemsB).reduce((stapleSet, obj) => (obj.source != "STAPLE") ?
      (merged.push(obj), stapleSet) : 
      (stapleSet.has(obj.name) || merged.push(obj), 
      stapleSet.add(obj.name), stapleSet), new Set());

      return merged;
    }
    console.log( merge(itemsA, itemsB) );

＆＃13;

修改：OP的其他格式

function merge(itemsA, itemsB) {
  let merged = [];
  itemsA.concat(itemsB).reduce((stapleSet, obj) => {
    if (obj.source != "STAPLE") {
      merged.push(obj);
      return stapleSet;
    } else {
      if (stapleSet.has(obj.name)) {
        return stapleSet;
      } else {
        merged.push(obj);
        stapleSet.add(obj.name);
        return stapleSet;
      }
    }
  }, new Set());
  return merged;
}

如何使用ES6从对象数组中删除重复项？

1 个答案: