JavaScript合并动态键对象数组 仅对键名含_abs的字段值求和

现有如下结构的对象数组：

[
  {"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2, ...},
  {"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2, ...},
  {"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2, ...},
  {"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2, ...}
]

处理规则：

• 每个对象可包含任意n组键值对，每类指标都对应带_perc和_abs后缀的两个字段
• 数组归约为单个对象：所有键名带_abs后缀的字段值累加求和，所有_perc后缀字段统一赋值为"-"
  预期输出格式：

{"A_abs": 16, "A_perc": "-", "B_abs": 16, "B_perc": "-", "C_abs": x, "C_perc": "-", ...}

原有参考的通用reduce求和逻辑会对所有字段无差别累加，无法区分_abs和_perc字段，不符合需求，原代码如下：

var data = [{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2},{"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2}],
    result = data.reduce((r, o) => (Object.entries(o).forEach(([k, v]) => r[k] = (r[k] || 0) + v), r), {});

console.log(result);

只需要在遍历键值对时增加后缀判断逻辑即可，无需提前预知字段数量、字段前缀，自动适配所有符合后缀规则的字段：

const data = [
  {"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2},
  {"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2},
  {"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2},
  {"A_abs":4,"A_perc":2, "B_abs":4,"B_perc":2}
];

const result = data.reduce((acc, cur) => {
  Object.entries(cur).forEach(([key, val]) => {
    if (key.endsWith('_abs')) {
      // 仅_abs后缀字段执行累加
      acc[key] = (acc[key] || 0) + val;
    } else if (key.endsWith('_perc')) {
      // _perc字段固定赋值为"-"
      acc[key] = "-";
    }
  });
  return acc;
}, {});

console.log(result);
// 运行输出：{A_abs: 16, A_perc: "-", B_abs: 16, B_perc: "-"}

该逻辑兼容性强，后续新增任意指标（比如C、D、E类的_abs/_perc字段）都不需要修改代码，会自动完成累加和赋值。

内容的提问来源于stack exchange，提问作者toffler