Lodash 过滤器对对象进行分组答案

【问题标题】：Lodash filter to group objectsLodash 过滤器对对象进行分组
【发布时间】：2023-01-25 01:13:26
【问题描述】：

我有一个数组，我喜欢按 ID 分组，但将数组中的所有不同元素推送到结果中。

let data = [
    {   id: "1", elements: ["a","b"], files: []    },
    {   id: "1", elements: ["a","b"], files: []    },
    {   id: "2", elements: ["a","b","c"],
        files: [
            { name: "test1",
              extension: "pdf"
            },
            { name: "test2",
              extension: "pdf"
            }
        ]
    },
    {   id: "2", elements: ["a","b","c"],
        files: [
            { name: "test3",
              extension: "png"
            },
            { name: "test4",
              extension: "png"
            },
            { name: "test5",
              extension: "pdf"
            }
        ]
    },
    {   id: "2", elements: ["a","b","c"], files: []
    }
];

我想要一个结果

result = 
[
    {
        "id": "1", "elements": [ "a", "b" ], "files": []    },
    {
        "id": "2", "elements": [ "a", "b", "c" ],
        "files": [
            { "name": "test1", "extension": "pdf" },
            { "name": "test2", "extension": "pdf" },
            { "name": "test3", "extension": "png" },
            { "name": "test4", "extension": "png" },
            { "name": "test5", "extension": "pdf" },
        ]
    }
]

lodash 或任何其他进程如何获得所需的输出

【问题讨论】：

标签： javascript lodash

【解决方案1】：

您可以使用Array.reduce() 将项目按id 分组。

这会为每个 id 创建一个具有属性的对象，然后我们可以使用 Object.values() 将结果作为数组获取。

对于 elements 数组，我们还将使用 Set 来确保我们不会重复元素。例如，我们只想要 ['a','b'] 而不是 ['a','b','a','b']。

let data = [ {   id: "1", elements: ["a","b"], files: []    }, {   id: "1", elements: ["a","b"], files: []    }, {   id: "2", elements: ["a","b","c"], files: [ { name: "test1", extension: "pdf" }, { name: "test2", extension: "pdf" } ] }, {   id: "2", elements: ["a","b","c"], files: [ { name: "test3", extension: "png" }, { name: "test4", extension: "png" }, { name: "test5", extension: "pdf" } ] }, {   id: "2", elements: ["a","b","c"], files: [] } ];

const result = Object.values(data.reduce((acc, { id, elements, files }) => { 
    acc[id] = acc[id] || { id, elements: [], files: []};
    // Use a Set to avoid duplicates...
    acc[id].elements = [...new Set([...elements, ...acc[id].elements])];
    acc[id].files.push(...files);
    return acc;
}, {}));

console.log('Result:', result)

.as-console-wrapper { max-height: 100% !important; }

【讨论】：