javascript - 合并对象使用 javascript 中的加法保留它的值

Question

我的 js 对象数组结构如下，

items= [
    {
      discount: 27.6,
      name: 'Floy Vandervort',
      price: 230,
      quantity: 3,
      taxable: 662.4
    },
    {
      discount: 122.88,
      name: 'Adriel Abshire II',
      price: 256,
      quantity: 6,
      taxable: 1413.12
    },
    {
      discount: 159.66,
      name: 'Tabitha Stroman',
      price: 887,
      quantity: 2,
      taxable: 1614.34
    },
    {
      discount: 27.6,
      name: 'Floy Vandervort',
      price: 230,
      quantity: 3,
      taxable: 662.4
    },
    {
      discount: 122.88,
      name: 'Adriel Abshire II',
      price: 256,
      quantity: 6,
      taxable: 1413.12
    },
    {
      discount: 159.66,
      name: 'Tabitha Stroman',
      price: 887,
      quantity: 2,
      taxable: 1614.34
    },
    {
      discount: 27.6,
      name: 'Floy Vandervort',
      price: 230,
      quantity: 3,
      taxable: 662.4
    },
    {
      discount: 122.88,
      name: 'Adriel Abshire II',
      price: 256,
      quantity: 6,
      taxable: 1413.12
    },
    {
      discount: 159.66,
      name: 'Tabitha Stroman',
      price: 887,
      quantity: 2,
      taxable: 1614.34
    }
  ]

我想避免基于 name 属性的对象重复。所以我决定通过保留其评估来合并它们，如下所示，

用例

考虑属性 name，这里在上面的数组 Floy Vandervort 中重复了 3 次。要将其转换为单个对象，请通过加法保留值将它们合并为单个对象。因此，除了 price 属性外，discount、quantity 和 taxable 属性应该通过加法合并。

我正在寻找一个最佳解决方案，我通过迭代原始数组并将合并的对象推送到另一个数组来实现。我想消除复杂性，可能吗？如果是，如何？这是我正在使用的功能

function(items) {
  let filtered = [];
  items.forEach((item) => {
    if (!isContains(filtered, item)) {
      filtered.push(item);
    } else {
      index = filtered.findIndex((x) => x.name === item.name);
      filtered[index].discount += item.discount;
      filtered[index].quantity += item.quantity;
      filtered[index].taxable += item.taxable;
    }
  });

  return filtered;
}

function isContains(items, ob) {
  items.forEach((item) => {
    if (item.name === ob.name) {
      return true;
    }
  });
  return false;
}

Answer 1

在 forEach 回调中返回无效 - isContains 将始终返回 false。最好使用按名称索引的对象或 Map，以将计算复杂度降低一个数量级 - 然后您可以获取该对象的值以获取所需的数组:

const items=[{discount:27.6,name:"Floy Vandervort",price:230,quantity:3,taxable:662.4},{discount:122.88,name:"Adriel Abshire II",price:256,quantity:6,taxable:1413.12},{discount:159.66,name:"Tabitha Stroman",price:887,quantity:2,taxable:1614.34},{discount:27.6,name:"Floy Vandervort",price:230,quantity:3,taxable:662.4},{discount:122.88,name:"Adriel Abshire II",price:256,quantity:6,taxable:1413.12},{discount:159.66,name:"Tabitha Stroman",price:887,quantity:2,taxable:1614.34},{discount:27.6,name:"Floy Vandervort",price:230,quantity:3,taxable:662.4},{discount:122.88,name:"Adriel Abshire II",price:256,quantity:6,taxable:1413.12},{discount:159.66,name:"Tabitha Stroman",price:887,quantity:2,taxable:1614.34}];

function squish(items) {
  const squishedItemsByName = items.reduce((a, { name, ...props }) => {
    if (!a[name]) {
      a[name] = { name };
    }
    Object.entries(props).forEach(([prop, val]) => {
      a[name][prop] = (a[name][prop] || 0) + val;
    });
    return a;
  }, {});
  return Object.values(squishedItemsByName);
}

console.log(squish(items));