比较javascript中常见值的未指定数量的数组

Question

I would like to know how to compare two or more -- potentially unlimited -- arrays for common values and push these values into a new array efficiently. Below I have a function that will accept unlimited arguments, but I am uncertain if this is a good place to begin. PHP appears to have a method that can do what I want called array_intersect. Does javascript offer something similar?

Note: I have found examples of how this can be done with two or so arrays, but I have not found examples of how such approaches might be applied to an unspecified number of arrays as of yet. Therefore I do not see this as a duplicate question.

To further clarify, the arrays might be filled with anything. Letters, numbers, symbols, words, you name it, it might be there.

 var sampleOne = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]; var sampleTwo = [5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]; function FindDirectRelation() { for(var i = 0; i < arguments.length; ++i) { console.log(arguments[i]); }; }; var directRelation = FindDirectRelation(sampleOne, sampleTwo); 

I am still a coding novice, so please ensure that everything is explained in a way that is simple enough for me to understand.

Answer 1

using an existing intersect that works with 2 arrays, we can chain together a common sub-set using the built-in reduce() method on an array of arrays that need intersected:

function intersect(a, b) {
  var aa = {};
  a.forEach(function(v) { aa[v]=1; });
  return b.filter(function(v) { return v in aa; });
}

var r1=[1,2,3], 
r2=[1,3,4,5], 
r3=[5,1,3];

alert([r1, r2, r3].reduce(intersect)) // shows: 1,3

if you define "intersect" as just being in more than one array (not every), then it's more complex...

Answer 2

Check to make sure the elements in the first array are also in the remaining arrays:

function multi_intersect(a) {
  var other_arrays = Array.prototype.slice.call(arguments, 1);

  return a . filter(function(elt) {
    return other_arrays.every(function(an) {
      return an.indexOf(elt) !== -1;
    });
  });
}

Answer 3

Try using Array.prototype.filter() , Array.prototype.indexOf()

var res = sampleOne.filter(function(val) {return sampleTwo.indexOf(val) !== -1})

---

 var sampleOne = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]; var sampleTwo = [5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]; var arr = ["a", "b", "c"]; var arr1 = ["c", "d", "e"]; var arr2 = [2, 7]; function samples() { var args = Array.prototype.slice.call(arguments); var res = []; for (var i = 0, curr, next; i < args.length; i++) { if (args[i + 1]) { // set `curr` to array `i` curr = args[i]; // set `next` to array `i + 1` if it exists next = args[i + 1] } else { // if at last index, set `curr` to `args` : input arrays // flattened to single array , with element at `i` removed curr = [].concat.apply([], args.slice(0, args.length - 1)); console.log(curr) // set next to current index next = args[i]; }; next = next.filter(function(val) { return curr.indexOf(val) !== -1 // filter duplicate entries at `res` && res.indexOf(val) === -1 }); res = res.concat.apply(res, next); }; return res } var sample = samples(sampleOne, sampleTwo, arr, arr1, arr2); console.log(sample); // [5, 6, 7, 8, 9, 10, 11, 12, "c", 2]