我有一個(gè)結(jié)構(gòu)如下的對象?const data = [? ? { academicYearId: 1, classLevelId: 1, subjectId: 1, ...},? ? { academicYearId: 1, classLevelId: 1, subjectId: 2, ...},? ? { academicYearId: 1, classLevelId: 1, subjectId: 3, ...},? ? ,,,?]我需要?jiǎng)?chuàng)建一個(gè)將返回唯一列的函數(shù)，例如? ?const uniqueColumns = ( val, columns)=> {? ? ?//? ?}? ?const val = [? ? { id: 1, name: 'n1', val: 1 },? ? { id: 1, name: 'n1', val: 2 },? ? { id: 2, name: 'n2', val: 1 },? ? { id: 3, name: 'n2', val: 2 }? ?]? ?let result = uniqueColumns(val)? ?console.log(val)? ?/**?? ? * Expected? ? * [{ id: 1, name: 'n1'}, { id: 2, name: 'n2'}, { id: 3, name: 'n2'}]? ? */}我試圖查看如何從 JavaScript 中的對象數(shù)組中獲取不同的值？我已經(jīng)設(shè)法想出了下面的? ?const uniqueColumns = (val, columns) =>?? ? ?([...new Set(? ? ? ?val.map(item =>?? ? ? ? ?columns.reduce((prev, next) =>?? ? ? ? ? ?({[next]: item[next], ...prev}), {})? ? ? ?).map(item => JSON.stringify(item)))? ? ?].map(item => JSON.parse(item)))? ?? ?const val = [? ? { id: 1, name: 'n1', val: 1 },? ? { id: 1, name: 'n1', val: 2 },? ? { id: 2, name: 'n2', val: 1 },? ? { id: 3, name: 'n2', val: 2 }? ?]? ?const result = uniqueColumns(val, ['id', 'name'])? ?console.log(result)我想問的是是否有更好的方法而不是必須將對象轉(zhuǎn)換為字符串然后再轉(zhuǎn)換回對象來實(shí)現(xiàn)這一點(diǎn)