如何在TypeScript中根据嵌套对象的_id字段筛选唯一对象？

给定如下TypeScript对象数组，需要根据group.application._id字段筛选出唯一对象，移除重复项（示例中第二个和第三个元素的group.application._id相同，需保留第二个、移除第三个）：

const myObjArray: any = [
    {
        "group": {
            "name": "SAM-admin",
            "active": true,
            "application": {
                "_id": "g4fac238hma10kjo3mv473vbs",
                "name": "Prism",
                "owners": "",
                "url": "",
                "description": "",
                "active": "",
                "created_by": "",
                "updated_by": ""
            },
            "created_by": "",
            "updated_by": "",
        },
        "user": "g4fac238hma10kjo3mv473vbs",
        "role": "g4fac238hma10kjo3mv473vbs",
        "isAdmin": true,
        "active": true,
        "created_by": "Abhilash.Shajan1@gmail.com",
        "updated_by": "",
    },
    {
        "group": {
            "name": "SAM-super-admin",
            "active": true,
            "application": {
                "_id": "g4fac238hma10kjo3mv473asc",
                "name": "Utopia",
                "owners": "",
                "url": "",
                "description": "",
                "active": "",
                "created_by": "",
                "updated_by": ""
            },
            "created_by": "",
            "updated_by": "",
        },
        "user": "g4fac238hma10kjo3mv473asc",
        "role": "g4fac238hma10kjo3mv473asc",
        "isAdmin": true,
        "active": true,
        "created_by": "Hima.Thomas@gmail.com",
        "updated_by": "",
    },
    {
        "group": {
            "name": "SAM-guest",
            "active": true,
            "application": {
                "_id": "g4fac238hma10kjo3mv473asc",
                "name": "Utopia",
                "owners": "",
                "url": "",
                "description": "",
                "active": "",
                "created_by": "",
                "updated_by": ""
            },
            "created_by": "",
            "updated_by": "",
        },
        "user": "g4fac238hma10kjo3mv473asc",
        "role": "g4fac238hma10kjo3mv473asc",
        "isAdmin": true,
        "active": true,
        "created_by": "Hima.Thomas@gmail.com",
        "updated_by": "",
    }
];

最简单的实现方式 #

推荐用Array.filter()结合临时对象追踪已出现的ID，一次遍历完成去重，性能更优：

// 记录已出现的application _id
const seenAppIds: Record<string, boolean> = {};
// 过滤出唯一对象
const uniqueArray = myObjArray.filter(item => {
    const appId = item.group.application._id;
    if (!seenAppIds[appId]) {
        seenAppIds[appId] = true;
        return true;
    }
    return false;
});

如果追求更简洁的写法，也可以用Array.reduce()：

const uniqueArray = myObjArray.reduce((acc: any[], current) => {
    const appId = current.group.application._id;
    // 检查结果数组中是否已有该ID的对象
    if (!acc.some(item => item.group.application._id === appId)) {
        acc.push(current);
    }
    return acc;
}, []);

说明 #

• filter+对象记录的时间复杂度是O(n)，适合处理较大数组；reduce+some的时间复杂度是O(n²)，小数据量下使用更简洁。
• 两种方式都会保留每个group.application._id第一次出现的对象，满足需求中移除第三个重复对象的要求。

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