我想對數(shù)組數(shù)組進(jìn)行重復(fù)數(shù)據(jù)刪除。重復(fù)數(shù)組是匹配元素索引子集的數(shù)組。在這種情況下，比如說 index[1]和 index [3]。const unDeduplicated = [  [ 11, 12, 13, 14, 15, ],  [ 21, 22, 23, 24, 25, ],  [ 31, 88, 33, 99, 35, ], // duplicate in indices: 1, 3 with row index 4  [ 41, 42, 43, 44, 45, ],  [ 51, 88, 53, 99, 55, ], // duplicate in indices: 1, 3 // delete this row from result];const deduplicated = getDeduplicated( unDeduplicated, [ 1, 3, ], );console.log( deduplicated );// expected result:// [//   [ 11, 12, 13, 14, 15, ],//   [ 21, 22, 23, 24, 25, ],//   [ 31, 88, 33, 99, 35, ],//   [ 41, 42, 43, 44, 45, ],//   // this row was omitted from result because it was duplicated at indices 1 and 3 with row index 2// ]什么功能getDeduplicated()可以給我這樣的結(jié)果？我已經(jīng)嘗試了以下功能，但這只是一個開始。它離給我想要的結(jié)果還差得遠(yuǎn)。但它讓我知道我正在嘗試做什么。/** * Returns deduplicated array as a data grid ([][] -> 2D array) * @param { [][] } unDedupedDataGrid The original data grid to be deduplicated to include only unque rows as defined by the indices2compare. * @param { Number[] } indices2compare An array of indices to compare for each array element. * If every element at each index for a given row is duplicated elsewhere in the array, * then the array element is considered a duplicate * @returns { [][] } */const getDeduplicated = ( unDedupedDataGrid, indices2compare, ) => {  let deduped = [];  unDedupedDataGrid.forEach( row => {    const matchedArray = a.filter( row => row[1] === 88 && row[3] === 99 );    const matchedArrayLength = matchedArray.length;    if( matchedArrayLength ) return;    deduped.push( row, );  });}我研究了一些可能會有所幫助的 lodash 函數(shù)，_.filter但_.some到目前為止，我似乎無法找到產(chǎn)生所需結(jié)果的結(jié)構(gòu)。