JS集合去重和排序

定义类型和集合

export interface DeptRoomLocItem {

    DeptId: number;

    DeptName: string;

    RoomId: number;

    RoomName: string;

    RoomNO: string;

    RoomLocId: number;

    RoomLocName: string;

    RoomLocNO: string;

}

const allDeptRoomLocs:DeptRoomLocItem[]=[…..]

 方式一：用 filter + findIndex

const newRoomOptions = allDeptRoomLocs
    .filter(x => x.DeptId === deptId)
    .filter((item, index, arr) => arr.findIndex(x => x.RoomNO === item.RoomNO) === index)
    .sort((s1, s2) => s1.RoomNO.localeCompare(s2.RoomNO))
    .map(x => ({
        label: `${x.RoomName}(NO.:${x.RoomNO})`,
        value: x.RoomId,
    }));

方式二：用 Map（性能更好） 

const newRoomOptions = […new Map(
    allDeptRoomLocs
        .filter(x => x.DeptId === deptId)
        .map(x => [x.RoomNO, x])  // 以 RoomNO 为 key
).values()]
    .sort((s1, s2) => s1.RoomNO.localeCompare(s2.RoomNO))
    .map(x => ({
        label: `${x.RoomName}(NO.:${x.RoomNO})`,
        value: x.RoomId,
    }));

方式三：用 reduce 

const newRoomOptions = allDeptRoomLocs
    .filter(x => x.DeptId === deptId)
    .reduce((acc, cur) => {
        if (!acc.some(x => x.RoomNO === cur.RoomNO)) {
            acc.push(cur);
        }
        return acc;
    }, [] as typeof allDeptRoomLocs)
    .sort((s1, s2) => s1.RoomNO.localeCompare(s2.RoomNO))
    .map(x => ({
        label: `${x.RoomName}(NO.:${x.RoomNO})`,
        value: x.RoomId,
    }));

推荐方式二，数据量大时性能最好。