JS/TS实现：统计对象数组指定属性值的出现次数并生成数组

需求 #

遍历对象数组，提取target属性的所有唯一值，生成一个包含[属性值, 出现次数]格式的数组。

原始数据 #

[
  { name: "doesnt matter", target: "A", other: "doesnt matter" },
  { name: "doesnt matter", target: "B", other: "doesnt matter" },
  { name: "doesnt matter", target: "C", other: "doesnt matter" },
  { name: "doesnt matter", target: "A", other: "doesnt matter" },
  { name: "doesnt matter", target: "B", other: "doesnt matter" },
  { name: "doesnt matter", target: "C", other: "doesnt matter" },
  { name: "doesnt matter", target: "A", other: "doesnt matter" },
  { name: "doesnt matter", target: "B", other: "doesnt matter" },
  { name: "doesnt matter", target: "A", other: "doesnt matter" }
]

预期输出 #

const newArray = [['A', 4], ['B', 3], ['C', 2]]

尝试的代码（未达预期） #

arr.reduce((accumulator: any, currentOrder: any) => {
  if (!accumulator[currentOrder[property]]) { 
    accumulator[currentOrder[property]] = [];
  }
  accumulator[currentOrder[property]].push(currentOrder);
  return accumulator;
}, []);

问题分析 #

这段代码有两个核心问题：

• 用数组作为累加器，但数组的键是数字索引，无法通过字符串（如"A"）直接存取属性，导致统计逻辑错误；
• 把整个currentOrder对象push进数组，最终得到的是分组后的对象数组，而非次数统计。

解决方案 #

完整实现代码 #

const arr = [
  { name: "doesnt matter", target: "A", other: "doesnt matter" },
  { name: "doesnt matter", target: "B", other: "doesnt matter" },
  { name: "doesnt matter", target: "C", other: "doesnt matter" },
  { name: "doesnt matter", target: "A", other: "doesnt matter" },
  { name: "doesnt matter", target: "B", other: "doesnt matter" },
  { name: "doesnt matter", target: "C", other: "doesnt matter" },
  { name: "doesnt matter", target: "A", other: "doesnt matter" },
  { name: "doesnt matter", target: "B", other: "doesnt matter" },
  { name: "doesnt matter", target: "A", other: "doesnt matter" }
];

// 1. 用对象做累加器统计每个target值的出现次数
const countObj = arr.reduce((acc, item) => {
  const targetVal = item.target;
  acc[targetVal] = (acc[targetVal] || 0) + 1;
  return acc;
}, {} as Record<string, number>);

// 2. 将统计对象转为目标二维数组格式
const newArray = Object.entries(countObj);

console.log(newArray); // 输出: [['A', 4], ['B', 3], ['C', 2]]

简化一行版 #

const newArray = Object.entries(arr.reduce((acc, {target}) => (acc[target] = (acc[target] || 0) + 1, acc), {} as Record<string, number>));

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