javascript中数组交集的最简单代码

Question

在 javascript 中实现数组交集的最简单、无库的代码是什么？我要写

intersection([1,2,3], [2,3,4,5])

得到

[2, 3]

Answer 1

我正在使用地图，甚至可以使用对象。

//find intersection of 2 arrs
const intersections = (arr1,arr2) => {
  let arrf = arr1.concat(arr2)
  let map = new Map();
  let union = [];
  for(let i=0; i<arrf.length; i++){
    if(map.get(arrf[i])){
      map.set(arrf[i],false);
    }else{
      map.set(arrf[i],true);
    }
  }
 map.forEach((v,k)=>{if(!v){union.push(k);}})
 return union;
}

Answer 2

这是我正在使用的一个非常幼稚的实现。它是非破坏性的，并且确保不复制整体。

Array.prototype.contains = function(elem) {
    return(this.indexOf(elem) > -1);
};

Array.prototype.intersect = function( array ) {
    // this is naive--could use some optimization
    var result = [];
    for ( var i = 0; i < this.length; i++ ) {
        if ( array.contains(this[i]) && !result.contains(this[i]) )
            result.push( this[i] );
    }
    return result;
}

Answer 3

coffeescript 中 N 个数组的交集

getIntersection: (arrays) ->
    if not arrays.length
        return []
    a1 = arrays[0]
    for a2 in arrays.slice(1)
        a = (val for val in a1 when val in a2)
        a1 = a
    return a1.unique()

Answer 4

基于 Anon 的出色回答，这个返回两个或多个数组的交集。

function arrayIntersect(arrayOfArrays)
{        
    var arrayCopy = arrayOfArrays.slice(),
        baseArray = arrayCopy.pop();

    return baseArray.filter(function(item) {
        return arrayCopy.every(function(itemList) {
            return itemList.indexOf(item) !== -1;
        });
    });
}

Answer 5

希望这对所有版本都有帮助。

function diffArray(arr1, arr2) {
  var newArr = [];

  var large = arr1.length>=arr2.length?arr1:arr2;
  var small = JSON.stringify(large) == JSON.stringify(arr1)?arr2:arr1;
  for(var i=0;i<large.length;i++){
    var copyExists = false; 
    for(var j =0;j<small.length;j++){
      if(large[i]==small[j]){
        copyExists= true;
        break;
      }
    }
    if(!copyExists)
      {
        newArr.push(large[i]);
      }
  }

  for(var i=0;i<small.length;i++){
    var copyExists = false; 
    for(var j =0;j<large.length;j++){
      if(large[j]==small[i]){
        copyExists= true;
        break;
      }
    }
    if(!copyExists)
      {
        newArr.push(small[i]);
      }
  }


  return newArr;
}

Answer 6

不是关于效率，而是容易理解，这里是集合的并集和交集的例子，它处理集合的数组和集合的集合。

http://jsfiddle.net/zhulien/NF68T/

// process array [element, element...], if allow abort ignore the result
function processArray(arr_a, cb_a, blnAllowAbort_a)
{
    var arrResult = [];
    var blnAborted = false;
    var intI = 0;

    while ((intI < arr_a.length) && (blnAborted === false))
    {
        if (blnAllowAbort_a)
        {
            blnAborted = cb_a(arr_a[intI]);
        }
        else
        {
            arrResult[intI] = cb_a(arr_a[intI]);
        }
        intI++;
    }

    return arrResult;
}

// process array of operations [operation,arguments...]
function processOperations(arrOperations_a)
{
    var arrResult = [];
    var fnOperationE;

    for(var intI = 0, intR = 0; intI < arrOperations_a.length; intI+=2, intR++) 
    {
        var fnOperation = arrOperations_a[intI+0];
        var fnArgs = arrOperations_a[intI+1];
        if (fnArgs === undefined)
        {
            arrResult[intR] = fnOperation();
        }
        else
        {
            arrResult[intR] = fnOperation(fnArgs);
        }
    }

    return arrResult;
}

// return whether an element exists in an array
function find(arr_a, varElement_a)
{
    var blnResult = false;

    processArray(arr_a, function(varToMatch_a)
    {
        var blnAbort = false;

        if (varToMatch_a === varElement_a)
        {
            blnResult = true;
            blnAbort = true;
        }

        return blnAbort;
    }, true);

    return blnResult;
}

// return the union of all sets
function union(arr_a)
{
    var arrResult = [];
    var intI = 0;

    processArray(arr_a, function(arrSet_a)
    {
        processArray(arrSet_a, function(varElement_a)
        {
            // if the element doesn't exist in our result
            if (find(arrResult, varElement_a) === false)
            {
                // add it
                arrResult[intI] = varElement_a;
                intI++;
            }
        });
    });

    return arrResult;
}

// return the intersection of all sets
function intersection(arr_a)
{
    var arrResult = [];
    var intI = 0;

    // for each set
    processArray(arr_a, function(arrSet_a)
    {
        // every number is a candidate
        processArray(arrSet_a, function(varCandidate_a)
        {
            var blnCandidate = true;

            // for each set
            processArray(arr_a, function(arrSet_a)
            {
                // check that the candidate exists
                var blnFoundPart = find(arrSet_a, varCandidate_a);

                // if the candidate does not exist
                if (blnFoundPart === false)
                {
                    // no longer a candidate
                    blnCandidate = false;
                }
            });

            if (blnCandidate)
            {
                // if the candidate doesn't exist in our result
                if (find(arrResult, varCandidate_a) === false)
                {
                    // add it
                    arrResult[intI] = varCandidate_a;
                    intI++;
                }
            }
        });
    });

    return arrResult;
}

var strOutput = ''

var arrSet1 = [1,2,3];
var arrSet2 = [2,5,6];
var arrSet3 = [7,8,9,2];

// return the union of the sets
strOutput = union([arrSet1, arrSet2, arrSet3]);
alert(strOutput);

// return the intersection of 3 sets
strOutput = intersection([arrSet1, arrSet2, arrSet3]);
alert(strOutput);

// of 3 sets of sets, which set is the intersecting set
strOutput = processOperations([intersection,[[arrSet1, arrSet2], [arrSet2], [arrSet2, arrSet3]]]);
alert(strOutput);

Answer 7

IE 中的 Array 不支持“filter”和“indexOf”。这个怎么样：

var array1 = [1, 2, 3];
var array2 = [2, 3, 4, 5];

var intersection = [];
for (i in array1) {
    for (j in array2) {
        if (array1[i] == array2[j]) intersection.push(array1[i]);
    }
}