使用javascript中的加法合并对象以保留其值答案

【问题标题】：Merge Object preserving it's value using addition in javascript使用javascript中的加法合并对象以保留其值
【发布时间】：2020-01-25 06:31:28
【问题描述】：

我的 js 对象数组的结构如下所示，

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，在上面的数组中 Floy Vandervort 重复 3 次。要将其转换为单个对象，请通过添加保留值将它们合并为单个对象。因此，discount、quantity 和 taxable 属性应通过加法合并，但 price 属性除外。

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，有可能吗？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;
}

【问题讨论】：

@CertainPerformance 我刚刚进行了编辑并放置了我正在使用的代码。
@CertainPerformance 这就是我想要的，谢谢。

标签： javascript arrays optimization duplicates array-merge

【解决方案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您想要的数组：

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));

【讨论】：

【解决方案2】：

试试

let h = {};
items.forEach(x=> h[x.name]= !h[x.name] ? x : {
      discount: h[x.name].discount+x.discount,
      name: x.name,
      price: x.price,
      quantity: h[x.name].quantity+x.quantity,
      taxable: h[x.name].taxable+x.taxable
    });

let result = Object.keys(h).map(k=> h[k]);

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
    }
  ]

let h = {};
items.forEach(x=> h[x.name]= !h[x.name] ? x : {
      discount: h[x.name].discount+x.discount,
      name: x.name,
      price: x.price,
      quantity: h[x.name].quantity+x.quantity,
      taxable: h[x.name].taxable+x.taxable
    });
    
let result = Object.keys(h).map(k=> h[k]);

console.log(result);

【讨论】：

【解决方案3】：

执行此操作的许多方法之一可能如下：

const uniqueItems = Object.values(
    items.reduce((hashMap, item) => { 
        if (!hashMap[item.name]) {
            hashMap[item.name] = { ...item };
        } else {
            // You should use a safest way to add 
            hashMap[item.name].discount += item.discount;
            hashMap[item.name].price += item.price;
            hashMap[item.name].quantity += item.quantity;
            hashMap[item.name].taxable += item.taxable;
        }

        return hashMap;
    }, {})
)

【讨论】：