JavaScript如何从数组获取去重结果，合并同分组非空属性生成新数组

采用「哈希表分组+单次遍历合并」的逻辑实现，时间复杂度为O(n)，为该场景下的最优解法，完整代码如下：

const array = [
    [{"name":"Joe", "age":null, "date":"1959-08-10", "home":"ny"}],
    [{"name":"Bill", "age":null, "date":"1956-08-10", "home":"tx"}],
    [{"name":"Joe", "age":"17", "date":null, "home":"ny"}],
    [{"name":null, "age":"17", "date":"1956-08-10", "home":"tx"}],
    [{"name":"Joe", "age":"17", "date":"1959-08-10", "home":null}]
];

const groupMap = new Map();

// 处理带有效home值的条目，按home分组合并非空属性
array.forEach(item => {
    const curr = item[0];
    const homeKey = curr.home;
    if (!homeKey) return;

    if (!groupMap.has(homeKey)) {
        groupMap.set(homeKey, { name: null, age: null, date: null, home: homeKey });
    }
    const target = groupMap.get(homeKey);
    Object.keys(target).forEach(key => {
        if (curr[key] !== null) target[key] = curr[key];
    });
});

// 处理home为null的条目，匹配同名分组补充属性
array.forEach(item => {
    const curr = item[0];
    if (curr.home) return;
    for (const [_, target] of groupMap) {
        if (curr.name === target.name) {
            Object.keys(target).forEach(key => {
                if (curr[key] !== null && target[key] === null) target[key] = curr[key];
            });
            break;
        }
    }
});

// 转换为要求的输出格式
const new_array = Array.from(groupMap.values()).map(i => [i]);

执行后输出的new_array完全符合预期：

[
    [{"name":"Joe", "age":"17", "date":"1959-08-10", "home":"ny"}],
    [{"name":"Bill", "age":"17", "date":"1956-08-10", "home":"tx"}]
]

方案说明 #

• 用Map做分组存储，读写性能高于普通对象，分组操作时间复杂度为O(1)
• 仅需要两次遍历原数组，整体时间复杂度为O(n)，远高于多次循环筛选的写法
• 逻辑分层清晰，先处理明确分组的条目，再补全无有效home的条目，避免逻辑混乱

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