从打字稿中的对象列表中查找键的唯一值的有效方式

我正在开发一个具有

1. 一组过滤器
2. 并记录在表格中

表格中的列对应于过滤器。过滤器随时都包含来自相应列的唯一值作为选项。

对于一条记录，一列可以包含多个值（即，来自相应过滤器的多个选项）

当用户从过滤器中选择一个选项时，表中的记录将被过滤，过滤后的结果（根据用户的选择）将显示给用户。

一旦得出过滤记录集，就可以通过为每一列查找唯一值来从过滤记录集中得出每个过滤器的唯一值。

Filter对象的键对应于Record对象的列，我有一个记录列表和一个过滤器列表。我想遍历这两个列表，并找到每个键的列的唯一值。

我正在使用以下逻辑从邮件中查找过滤器的唯一选项。

export function filterOptionsService(records: Record[], filters: RecordFilter[]): RecordFilter[] {
    const newFilters: RecordFilter[] = filters.map(filter => {

        //logic to find all values for a column from the set of records 
        const filterOptions = records.reduce((uniqueList, record) => uniqueList.concat(record[filter.key]), []);

        //logic to find unique values for a column from the set of records
        //which act as options of corresponding filters.
        const uniqueOptions = uniqBy(filterOptions, (opt) => filter.valueFunction ? filter.valueFunction(opt) : opt);

        const filterOptions: FilterOption[] = uniqueOptions.map(value => {
            return {
                label: filter.labelFunction ? filter.labelFunction(value) : value,
                value: filter.valueFunction ? filter.valueFunction(value) : value,
            };
        });
        filter.options = orderBy(dropListOptions, 'label');

        //here is my logic to find the count of each option, present in the filtered records
        filter.options = filter.options.map(option => ({
            ...option,
            count: filter.valueFunction
                ? filterOptions.filter(value => filter.valueFunction(value) === option.value).length
                : filterOptions.filter(value => value === option.value).length
        }));
        return filter;
    });
    return newFilters;
}

interface Filter {
    key: string;
    labelFunction?: Function;
    valueFunction?: Function;
    order: number;
    multiple: boolean;
    options: DropListOption[];
    values: DropListOption[];
}


interface Record {
    column1: string;
    column2: string;
    column3: string | string[];
    .
    .
    columnN: string;
}

下面的逻辑在我的代码中花费的时间最多。8k记录大约需要7秒。

const filterOptions = records.reduce((uniqueList, record) => 
                              uniqueList.concat(record[filter.key]), []); 

我无法使我的代码性能更好。请您提出我要去哪里的建议。

这是typescrip操场上的示例代码 https：//www.typescriptlang。组织/播放？＃代码/ FASwdgLgpgTgZgQwMZQAQDEQBtowPIAOEIA9mKgN7Co201YIBGUWAXKgM4QzgDmA3NTq0AbgiwBXKOwRgAnoOG1kxEVAD87RiRJYosxUtQEEcrCQQATTalkKhSpCQmQbYCQFtmMQzQC + -LTA1AD0IRwIHgR6qGJ6HKhwJDCoANZQcrYwaAQwJASwEHIAjAA0xnkFMEUATKigkLCIKKgASlBOMJaYOLCUDsLpcuxcPGAC9UaoDMxY6C5IxGQ282CLpGC + wnFSq + vL7HtLmwN0yZaw7O5esFt0HhI4INHSqNq6 + IDT + ccch9i4QjHADaAF07qJxFI-hgAbAgRswVsAkFgGEKvlqiAoAkSHA2h1zqgkAwOAkENlOJE0AgEkNUHiCZ1unCUiTaRw0SFUBAABYgBICzggKJYEBwbGWVDZKo4yAIY6oDyZFQScTE0kJXm0pV2JXJHKVQrYzngXDNNCtEgAd36dFy + UKJRG3D4SIGDqqtRdYwEwACwWATjAXFQ4BAxHEVutMOjYNQAF5UMDTlNhBRPU7iqxgQAiMhQXPlXN87KF0GlTPVOQ1Vi5phIXN + UqptO0DNG6vZvMStRF1C5sDgcuVzve3MXOBNlttqYdx1dnP5sCF4ulqCrge9zeD4e5itV8cNputtPg4JBMIASVsHgZYCwKrJIF4Q-GsShOJ5JDSGQxXsyJIUg4EVonFEAkAVDYgzIUNw0jOZWVjQkuh6XB4yTFNZyUCghjrQ8Sn7c5LjKJVHmIF5WEQLAOCgcofg2P4wXKHYcXYMFm1PNNcIyfCxxrIiukuGpygeJ5KOo2j6KIRicwrD9JDY5NQT8GguKMc9gglXp8Bk2CAGVYBECCoAACngkAoxtDhSgs8Q0NgDgAEpBGCdEb21NRUGKAAWRJWQZRgACsOggTl0WtXkoHIDyEC8gAOVJpRtQKQsWBI + TQWYUuySwJBaTEICgsgeQQdIEgANgAWgAdk4QkwEsDgADpgDgBZFW0wE9JDQyYGMlBTLyGN2DjeSusc0aUJZHSwSc-pTgmmAWo8BACFMpbEwAPjtIwwnMXgIO-fzGtsLAsAU6FEmSWxiV0TxyDgPI70y + qIAZfFsmZBJT2DUMloRWDE2SmNmty-KzNMlwQAARykAAZAUIHKYb5oTHbUGhuGoERrhmuDSCICGm1gSW5qhlBJzyjmwIuN + 2CPmag7TIAcgBnqOBZ0p2d + FzsNbPwXP9QMgA