使用對象的兩個屬性減少單個對象數組

Question

我有一個對象數組，我想將它縮減為基於兩個鍵值對的另一個對象數組。

const original =
    [
        {
            key1: 'a',
            key2: 'AA',
            value: 1
        },
        {
            key1: 'a',
            key2: 'AA',
            anotherValue: 2
        },
        {
            key1: 'b',
            key2: 'BB',
            value: 1
        },
        {
            key1: 'a',
            key2: 'AA',
            yetAnother: 3
        },
        {
            key1: 'b',
            key2: 'BB',
            anotherValue: 4
        },
        {
            key1: 'c',
            key2: 'CC',
            value: 1
        }
    ];

應該改成：

const result =
    [
        {
            key1: 'a',
            key2: 'AA',
            value: 1,
            anotherValue: 2,
            yetAnother: 3
        },
        {
            key1: 'b',
            key2: 'BB',
            value: 1,
            anotherValue: 4
        },
        {
            key1: 'c',
            key2: 'CC',
            value: 1
        },
    ];

厭倦了使用 map 和 reduce 甚至 lodash。 然而，我所有的嘗試都是徒勞的。

Answer 1

使用Array.reduce ，您可以按key1和key2值對對對象進行分組。 變量groupBy對象值包含您想要的結果，您只能使用Object.values(groupBy)生成值。

 const original = [ { key1: 'a', key2: 'AA', value: 1 }, { key1: 'a', key2: 'AA', anotherValue: 2 }, { key1: 'b', key2: 'BB', value: 1 }, { key1: 'a', key2: 'AA', yetAnother: 3 }, { key1: 'b', key2: 'BB', anotherValue: 4 }, { key1: 'c', key2: 'CC', value: 1 } ]; const groupBy = original.reduce((acc, cur) => { const { key1, key2, ...rest } = cur; const key = key1 + "_" + key2; acc[key] ? acc[key] = { ...acc[key], ...rest } : acc[key] = cur; return acc; }, {}); const result = Object.values(groupBy); console.log(result);

Answer 2

使用Array.prototype.reduce()構建一個條目Map ，由任何key*屬性的組合鍵控，同時合並其他屬性。

然后您可以將Map值轉換為新數組

 const original = [{"key1":"a","key2":"AA","value":1},{"key1":"a","key2":"AA","anotherValue":2},{"key1":"b","key2":"BB","value":1},{"key1":"a","key2":"AA","yetAnother":3},{"key1":"b","key2":"BB","anotherValue":4},{"key1":"c","key2":"CC","value":1}] const t1 = performance.now() const result = [...original.reduce((map, obj) => { // create a key from all the "key*" properties const key = Object.keys(obj) .filter(key => key.startsWith("key")) .sort() .map(key => obj[key]).join(":") // looks like "a:AA", "b:BB", etc // merge with previously found matches const entry = { ...(map.get(key) ?? {}), ...obj } // collect the merged object return map.set(key, entry) }, new Map()).values()] const t2 = performance.now() console.info(result, `in ${t2 - t1}ms`)

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