Group By 和 SUM 的 ES6 实现答案

【问题标题】：ES6 Implementation of Group By and SUMGroup By 和 SUM 的 ES6 实现
【发布时间】：2018-03-26 06:04:22
【问题描述】：

我有一组具有以下属性的对象：material_no、material_name、qty。

let data = [
  { material_no: '1001', material_name: 'Material 1', qty: 100 },
  { material_no: '1001', material_name: 'Material 1', qty: 50 },
  { material_no: '1002', material_name: 'Material 2', qty: 44 },
  { material_no: '1003', material_name: 'Material 3', qty: 125 },
  { material_no: '1002', material_name: 'Material 2', qty: 59 },
  { material_no: '1004', material_name: 'Material 4', qty: 999 },
  { material_no: '1005', material_name: 'Material 5', qty: 80 },
  { material_no: '1005', material_name: 'Material 5', qty: 66 }
]

如何返回按其 material_no/material_name 分组的对象数组以及具有相同 material_no/material_name 的数量之和？

[
  { material_no: '1001', material_name: 'Material 1', qty: 150 },
  { material_no: '1002', material_name: 'Material 2', qty: 103 },
  { material_no: '1003', material_name: 'Material 3', qty: 125 },
  { material_no: '1004', material_name: 'Material 4', qty: 999 },
  { material_no: '1005', material_name: 'Material 5', qty: 146 }
]

【问题讨论】：

到目前为止你有没有尝试过？能否提供示例数据和预期输出？
@nem035 我添加了示例和预期的输出。我在必须选择是否应该使用 filter、map、reduce 或 Object.assign 之间陷入困境。

标签： javascript ecmascript-6

【解决方案1】：

您可以将数组缩减为 Map，该数组将项目总数存储为 material_no。然后将地图值提取到数组中。

let data = [
  { material_no: '1001', material_name: 'Material 1', qty: 100 },
  { material_no: '1001', material_name: 'Material 1', qty: 50 },
  { material_no: '1002', material_name: 'Material 2', qty: 44 },
  { material_no: '1003', material_name: 'Material 3', qty: 125 },
  { material_no: '1002', material_name: 'Material 2', qty: 59 },
  { material_no: '1004', material_name: 'Material 4', qty: 999 },
  { material_no: '1005', material_name: 'Material 5', qty: 80 },
  { material_no: '1005', material_name: 'Material 5', qty: 66 }
]

const sums = [
  ...data.reduce(
    (map, item) => {
      const { material_no: key, qty } = item;
      const prev = map.get(key);
      
      if(prev) {
        prev.qty += qty
      } else {
        map.set(key, Object.assign({}, item))
      }
      
      return map
    },
    new Map()
  ).values()
]

console.log(sums)

【讨论】：

【解决方案2】：

你可以像这样使用.reduce()：

let data = [{ material_no: '1001', material_name: 'Material 1', qty: 100 },{ material_no: '1001', material_name: 'Material 1', qty: 50 },{ material_no: '1002', material_name: 'Material 2', qty: 44 },{ material_no: '1003', material_name: 'Material 3', qty: 125 },{ material_no: '1002', material_name: 'Material 2', qty: 59 },{ material_no: '1004', material_name: 'Material 4', qty: 999 },{ material_no: '1005', material_name: 'Material 5', qty: 80 },{ material_no: '1005', material_name: 'Material 5', qty: 66 }];

let result = Object.values(
               data.reduce((a, c) => (
                 a[c.material_no] = a[c.material_no] ?
                 (a[c.material_no].qty += c.qty, a[c.material_no]) :
                 c, a), {}
               )
             );

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

有用的资源：

【讨论】：

这太美了！

【解决方案3】：

data
    .sort((a, b) => a.material_no - b.material_no)
    .map(({material_no, material_name, qty}, index, arr) => ({
        material_no,
        material_name,
        qty: qty + (material_no === (arr[index - 1] || {}).material_no && arr[index - 1].qty)
    }))
    .filter(({material_no}, index, arr) => material_no !== (arr[index + 1] || {}).material_no);

它很慢，只是为了好玩:)

【讨论】：

【解决方案4】：

let data = [
  { material_no: '1001', material_name: 'Material 1', qty: 100 },
  { material_no: '1001', material_name: 'Material 1', qty: 50 },
  { material_no: '1002', material_name: 'Material 2', qty: 44 },
  { material_no: '1003', material_name: 'Material 3', qty: 125 },
  { material_no: '1002', material_name: 'Material 2', qty: 59 },
  { material_no: '1004', material_name: 'Material 4', qty: 999 },
  { material_no: '1005', material_name: 'Material 5', qty: 80 },
  { material_no: '1005', material_name: 'Material 5', qty: 66 }
];

let accumulation = data.reduce((total, val, index)=>{
  let foundItemIndex = total.findIndex((obj)=>obj.material_no == val.material_no);
  if(foundItemIndex < 0) total.push(val) 
  else total[foundItemIndex].qty += val.qty;
  return total;
}, []);

console.log(accumulation);

【讨论】：