Question

我想对数组数组进行重复数据删除。复制数组是与元素索引的子集匹配的数组。在这种情况下，例如，索引[1]和索引[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()可以给我带来这样的结果？

我尝试了以下功能，但这只是一个开始。而且这还不能给我理想的结果。但这可以让我了解我要做什么。

/**
 * 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lodash函数，它们可能会像_.filter和_.some一样有用，但到目前为止，我似乎找不到能够产生所需结果的结构。

Answer 1

遍历行时，可以根据列中的值创建Set。您可以选择仅为指定的列创建集合，例如您的情况下为1和3。然后，在遍历每一行时，请检查该行中是否有任何指定的列具有已在对应集中的值，并且是否将其丢弃。

（在电话上，不能键入实际代码。我想代码也很简单）

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

如果我对自己的工作了解得很好，那么这就是我所做的

mix.options({
    extractVueStyles: true,
    globalVueStyles: 'resources/sass/' + process.env.MIX_THEME + '/_variables.scss'
});

Answer 4

效率不是最高，但这将删除多个重复数组的重复

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

这很简洁。它使用嵌套过滤器。它也适用于任意数量的重复项，仅保留第一个重复项。

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]));

在JavaScript中对数组进行重复数据删除

5 个答案: