如何按shop与status分组合并数组并实现求和与统计？

核心思路 #

利用Array.reduce()遍历数组，以shop + status的组合作为唯一分组键，在累加器中维护每个分组的数值总和与计数，最后将累加器对象转换为数组得到最终结果。

示例实现 #

假设输入数据 #

const inputData = [
  { shop: "Shop A", status: "Paid", rent_total: 100, latefee_total: 10, amount_total: 110 },
  { shop: "Shop A", status: "Paid", rent_total: 200, latefee_total: 20, amount_total: 220 },
  { shop: "Shop A", status: "Unpaid", rent_total: 150, latefee_total: 15, amount_total: 165 },
  { shop: "Shop B", status: "Paid", rent_total: 300, latefee_total: 0, amount_total: 300 },
];

实现代码 #

// 分组处理逻辑
const groupedData = Object.values(inputData.reduce((acc, current) => {
  // 生成唯一分组标识：拼接shop和status
  const groupKey = `${current.shop}-${current.status}`;

  // 若分组不存在，初始化结构
  if (!acc[groupKey]) {
    acc[groupKey] = {
      shop: current.shop,
      status: current.status,
      rent_total: 0,
      latefee_total: 0,
      amount_total: 0,
      total_shops: 0
    };
  }

  // 累加数值字段
  acc[groupKey].rent_total += current.rent_total;
  acc[groupKey].latefee_total += current.latefee_total;
  acc[groupKey].amount_total += current.amount_total;
  // 统计该分组下的条目数量（即total_shops）
  acc[groupKey].total_shops += 1;

  return acc;
}, {}));

console.log(groupedData);

输出结果 #

[
  {
    shop: "Shop A",
    status: "Paid",
    rent_total: 300,
    latefee_total: 30,
    amount_total: 330,
    total_shops: 2
  },
  {
    shop: "Shop A",
    status: "Unpaid",
    rent_total: 150,
    latefee_total: 15,
    amount_total: 165,
    total_shops: 1
  },
  {
    shop: "Shop B",
    status: "Paid",
    rent_total: 300,
    latefee_total: 0,
    amount_total: 300,
    total_shops: 1
  }
]

自定义调整说明 #

如果total_shops需要统计分组内的唯一店铺数（而非条目数，例如存在同名不同店铺的场景），可以改用Set来维护：

const groupedData = Object.values(inputData.reduce((acc, current) => {
  const groupKey = `${current.shop}-${current.status}`;

  if (!acc[groupKey]) {
    acc[groupKey] = {
      shop: current.shop,
      status: current.status,
      rent_total: 0,
      latefee_total: 0,
      amount_total: 0,
      _uniqueShops: new Set()
    };
  }

  acc[groupKey].rent_total += current.rent_total;
  acc[groupKey].latefee_total += current.latefee_total;
  acc[groupKey].amount_total += current.amount_total;
  acc[groupKey]._uniqueShops.add(current.shop); // 存储唯一店铺标识

  return acc;
}, {})).map(group => ({
  ...group,
  total_shops: group._uniqueShops.size,
  _uniqueShops: undefined // 移除临时字段
}));

内容的提问来源于stack exchange，提问作者Abdul Razique