在javascript中删除重复的数组数组

Question

我想对数组数组进行重复数据删除。重复数组是匹配元素索引子集的数组。在这种情况下，比如说 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()可以给我这样的结果？我已经尝试了以下功能，但这只是一个开始。它离给我想要的结果还差得远。但它让我知道我正在尝试做什么。/** * 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, );  });}我研究了一些可能会有所帮助的 lodash 函数，_.filter但_.some到目前为止，我似乎无法找到产生所需结果的结构。

Answer 1

您可以在迭代行时根据列中的值创建Set 。您可以选择仅为指定列创建集合，例如在您的情况下为 1 和 3。然后，在遍历每一行时，检查该行中的任何指定列是否具有已经在相应集合中的值，如果存在，则丢弃该行。

（在手机上，无法输入实际代码。我猜代码也很简单）

Answer 2

这可能不是最有效的算法，但我会做类似的事情

function getDeduplicated(unDeduplicated, idxs) {

  const result = [];

  const used = new Set();

  unDeduplicated.forEach(arr => {

    const vals = idxs.map(i => arr[i]).join();

    if (!used.has(vals)) {

      result.push(arr);

      used.add(vals);

    }

  });

  return result;

}

Answer 3

如果我明白你想做什么，我知道，但这是我所做的

list = [

  [ 11, 12, 13, 14, 15, ],

  [ 21, 22, 23, 24, 25, ],

  [ 21, 58, 49, 57, 28, ],

  [ 31, 88, 33, 88, 35, ],

  [ 41, 42, 43, 44, 45, ],

  [ 51, 88, 53, 88, 55, ],

  [ 41, 77, 16, 29, 37, ],

];

el_list = []  // Auxiliar to save all unique numbers

res_list = list.reduce(

    (_list, row) => {

        // console.log(_list)

        this_rows_el = []  // Auxiliar to save this row's elements

        _list.push(row.reduce(

            (keep_row, el) => {

                // console.log(keep_row, this_rows_el, el)

                if(keep_row && el_list.indexOf(el)==-1 ){

                    el_list.push(el)

                    this_rows_el.push(el)

                    return true

                }else if(this_rows_el.indexOf(el)!=-1) return true  // Bypass repeated elements in this row

                else return false

            }, true) ? row : null)  // To get only duplicated rows (...) ? null : row )

        return _list

    }, []

)

console.log(res_list)

Answer 4

不是最有效的，但这将删除多个重复数组的副本

const unDeduplicated = [ [ 11, 12, 13, 14, 15, ], [ 21, 22, 23, 24, 25, ], [ 31, 88, 33, 99, 35, ], [ 41, 33, 43, 44, 45, ], [ 51, 88, 53, 99, 55, ]]

const unDeduplicated1 = [

  [ 11, 12, 13, 14, 15, ],

  [ 21, 22, 23, 24, 25, ],// duplicate in indices: 1, 3 with row index 3

  [ 31, 88, 33, 99, 35, ], // duplicate in indices: 1, 3 with row index 4

  [ 21, 22, 43, 24, 45, ],// duplicate in indices: 1, 3 // delete this

  [ 51, 88, 53, 99, 55, ], // duplicate in indices: 1, 3 // delete this row from result

];

function getDeduplicated(arr, arind) {

  for (let i = 0; i < arr.length; i++) {

    for (let j = 1 + i; j < arr.length; j++) {

      if (arr[j].includes(arr[i][arind[0]]) && arr[j].includes(arr[i][arind[1]])) {

        arr.splice(j, 1)

        i--

      } else continue

    }

  }

  return arr

}

const deduplicated = getDeduplicated(unDeduplicated, [1, 3]);

const deduplicated2 = getDeduplicated(unDeduplicated1, [1, 3]);

console.log(deduplicated)

console.log("#####################")

console.log(deduplicated2)

Answer 5

这是相当简洁的。它使用嵌套过滤器。它也适用于任意数量的重复项，只保留第一个。

init = [

  [ 11, 12, 13, 14, 15],

  [ 21, 22, 23, 24, 25],

  [ 31, 88, 33, 99, 35],

  [ 41, 42, 43, 44, 45],

  [ 51, 88, 53, 99, 55],

];

var deDuplicate = function(array, indices){

var res = array.filter(

  (elem) => !array.some(

  (el) =>

  array.indexOf(el) < array.indexOf(elem) && //check that we don't discard the first dupe

  el.filter((i) => indices.includes(el.indexOf(i))).every((l,index) => l === elem.filter((j) => indices.includes(elem.indexOf(j)))[index])

//check if the requested indexes are the same.

// Made a bit nasty by the fact that you can't compare arrays with ===

  )

);

return(res);

}

console.log(deDuplicate(init,[1,3]));