使用JavaScript合并二维数组重复记录：按首两元素拼接值聚合

问题描述 #

给定以下二维数组：

let array = [
    ["a23","3000","0","ES08"],
    ["a23","3000","0","ES93"],
    ["a23","3000","0","ES71"],
    ["12w","3007","0","ES34"],
    ["12w","3007","0","ES56"],
    ["1v8","3008","0","ES08"]
];

重复判定规则：当数组元素的第0位与第1位拼接值（array[i][0] + array[i][1]）相同时，视为重复记录。

需求：将重复记录合并为单条，把对应元素的最后一列值拼接在一起，最终得到仅包含唯一记录的数组，预期结果如下：

[
    ["a23","3000","0","ES08ES93ES71"],
    ["12w","3007","0","ES34ES56"],
    ["1v8","3008","0","ES08"]
]

实现方案 #

使用Map结构跟踪唯一键（拼接后的第0+第1位），遍历数组时合并重复项的最后一列值，最后将Map的值转换为数组即可。

完整代码 #

let array = [
    ["a23","3000","0","ES08"],
    ["a23","3000","0","ES93"],
    ["a23","3000","0","ES71"],
    ["12w","3007","0","ES34"],
    ["12w","3007","0","ES56"],
    ["1v8","3008","0","ES08"]
];

// 初始化Map存储唯一键与合并后的元素
const mergedMap = new Map();

array.forEach(item => {
    const key = item[0] + item[1];
    if (!mergedMap.has(key)) {
        // 首次遇到该键，复制前3项并初始化最后一项
        mergedMap.set(key, [...item.slice(0, 3), item[3]]);
    } else {
        // 已存在该键，拼接最后一列值
        const existingItem = mergedMap.get(key);
        existingItem[3] += item[3];
    }
});

// 将Map的值转换为数组，得到最终结果
const mergedArray = Array.from(mergedMap.values());

console.log(mergedArray);

逻辑说明 #

1. 键生成：用每个元素的第0位和第1位拼接成唯一键，确保重复项被精准识别。
2. Map存储：首次遇到键时，复制元素的前3项并将最后一项初始化为当前值；后续遇到相同键时，直接拼接最后一列的内容。
3. 结果转换：通过Array.from()将Map的values集合转换为目标数组，得到去重合并后的结果。

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