javascript中数组交集的最简单代码

结合使用 Array.prototype.filter 和 Array.prototype.indexOf :

array1.filter(value => -1 !== array2.indexOf(value))

或作为 vrugtehagel 在评论中建议,您可以使用最近的 Array.prototype.includes 对于更简单的代码:

array1.filter(value => array2.includes(value))

破坏性看起来最简单,特别是如果我们可以假设输入是排序的:

/* destructively finds the intersection of 
 * two arrays in a simple fashion.  
 *
 * PARAMS
 *  a - first array, must already be sorted
 *  b - second array, must already be sorted
 *
 * NOTES
 *  State of input arrays is undefined when
 *  the function returns.  They should be 
 *  (prolly) be dumped.
 *
 *  Should have O(n) operations, where n is 
 *    n = MIN(a.length, b.length)
 */
function intersection_destructive(a, b)
{
  var result = [];
  while( a.length > 0 && b.length > 0 )
  {  
     if      (a[0] < b[0] ){ a.shift(); }
     else if (a[0] > b[0] ){ b.shift(); }
     else /* they're equal */
     {
       result.push(a.shift());
       b.shift();
     }
  }

  return result;
}

非破坏性必须更加复杂,因为我们必须跟踪指数:

/* finds the intersection of 
 * two arrays in a simple fashion.  
 *
 * PARAMS
 *  a - first array, must already be sorted
 *  b - second array, must already be sorted
 *
 * NOTES
 *
 *  Should have O(n) operations, where n is 
 *    n = MIN(a.length(), b.length())
 */
function intersect_safe(a, b)
{
  var ai=0, bi=0;
  var result = [];

  while( ai < a.length && bi < b.length )
  {
     if      (a[ai] < b[bi] ){ ai++; }
     else if (a[ai] > b[bi] ){ bi++; }
     else /* they're equal */
     {
       result.push(a[ai]);
       ai++;
       bi++;
     }
  }

  return result;
}

如果您的环境支持 ECMAScript 6 Set ,一种简单而有效的方法(见规范链接):

function intersect(a, b) {
  var setA = new Set(a);
  var setB = new Set(b);
  var intersection = new Set([...setA].filter(x => setB.has(x)));
  return Array.from(intersection);
}

较短,但可读性较低(也不创建附加交叉口 Set ):

function intersect(a, b) {
      return [...new Set(a)].filter(x => new Set(b).has(x));
}

避新 集合 从 b 每一次:

function intersect(a, b) {
      var setB = new Set(b);
      return [...new Set(a)].filter(x => setB.has(x));
}

注意,当使用集合时,您将只获得不同的值,因此 new Set[1,2,3,3].size 评估为 3 .

使用 不足之处 或 J.S.

_.intersection( [0,345,324] , [1,0,324] )  // gives [0,324]

我在ES6学期的贡献。一般来说,它会发现数组与作为参数提供的不定数量数组的交集。

Array.prototype.intersect = function(...a) {
  return [this,...a].reduce((p,c) => p.filter(e => c.includes(e)));
}
var arrs = [[0,2,4,6,8],[4,5,6,7],[4,6]],
     arr = [0,1,2,3,4,5,6,7,8,9];

document.write("<pre>" + JSON.stringify(arr.intersect(...arrs)) + "</pre>");

只使用关联数组怎么样?

function intersect(a, b) {
    var d1 = {};
    var d2 = {};
    var results = [];
    for (var i = 0; i < a.length; i++) {
        d1[a[i]] = true;
    }
    for (var j = 0; j < b.length; j++) {
        d2[b[j]] = true;
    }
    for (var k in d1) {
        if (d2[k]) 
            results.push(k);
    }
    return results;
}

编辑:

// new version
function intersect(a, b) {
    var d = {};
    var results = [];
    for (var i = 0; i < b.length; i++) {
        d[b[i]] = true;
    }
    for (var j = 0; j < a.length; j++) {
        if (d[a[j]]) 
            results.push(a[j]);
    }
    return results;
}

@atk的原语排序数组实现的性能可以通过使用.pop而不是.shift来提高。

function intersect(array1, array2) {
   var result = [];
   // Don't destroy the original arrays
   var a = array1.slice(0);
   var b = array2.slice(0);
   var aLast = a.length - 1;
   var bLast = b.length - 1;
   while (aLast >= 0 && bLast >= 0) {
      if (a[aLast] > b[bLast] ) {
         a.pop();
         aLast--;
      } else if (a[aLast] < b[bLast] ){
         b.pop();
         bLast--;
      } else /* they're equal */ {
         result.push(a.pop());
         b.pop();
         aLast--;
         bLast--;
      }
   }
   return result;
}

我使用jsper创建了一个基准: http://bit.ly/P9FrZK . 用起来快三倍。砰。

// Return elements of array a that are also in b in linear time:
function intersect(a, b) {
  return a.filter(Set.prototype.has, new Set(b));
}

// Example:
console.log(intersect([1,2,3], [2,3,4,5]));

我推荐上述简洁的解决方案,它在大型输入上优于其他实现。如果小输入的性能很重要,请检查下面的备选方案。

备选方案和性能比较:

有关替代实现和检查,请参见以下代码段 https://jsperf.com/array-intersection-comparison 用于性能比较。

function intersect_for(a, b) {
  const result = [];
  const alen = a.length;
  const blen = b.length;
  for (let i = 0; i < alen; ++i) {
    const ai = a[i];
    for (let j = 0; j < blen; ++j) {
      if (ai === b[j]) {
        result.push(ai);
        break;
      }
    }
  } 
  return result;
}

function intersect_filter_indexOf(a, b) {
  return a.filter(el => b.indexOf(el) !== -1);
}

function intersect_filter_in(a, b) {
  const map = b.reduce((map, el) => {map[el] = true; return map}, {});
  return a.filter(el => el in map);
}

function intersect_for_in(a, b) {
  const result = [];
  const map = {};
  for (let i = 0, length = b.length; i < length; ++i) {
    map[b[i]] = true;
  }
  for (let i = 0, length = a.length; i < length; ++i) {
    if (a[i] in map) result.push(a[i]);
  }
  return result;
}

function intersect_filter_includes(a, b) {
  return a.filter(el => b.includes(el));
}

function intersect_filter_has_this(a, b) {
  return a.filter(Set.prototype.has, new Set(b));
}

function intersect_filter_has_arrow(a, b) {
  const set = new Set(b);
  return a.filter(el => set.has(el));
}

function intersect_for_has(a, b) {
  const result = [];
  const set = new Set(b);
  for (let i = 0, length = a.length; i < length; ++i) {
    if (set.has(a[i])) result.push(a[i]);
  }
  return result;
}

火狐53的结果:

• 大型阵列上的每秒操作数(10000个元素):

  filter + has (this)               523 (this answer)
  for + has                         482
  for-loop + in                     279
  filter + in                       242
  for-loops                          24
  filter + includes                  14
  filter + indexOf                   10
  

• 小阵列上的每秒操作数(100个元素):

  for-loop + in                 384,426
  filter + in                   192,066
  for-loops                     159,137
  filter + includes             104,068
  filter + indexOf               71,598
  filter + has (this)            43,531 (this answer)
  filter + has (arrow function)  35,588
  

使用 JQuery 以下内容:

var a = [1,2,3];
var b = [2,3,4,5];
var c = $(b).not($(b).not(a));
alert(c);

1. 排序它
2. 从索引0中逐个检查,然后从中创建新数组。

像这样的东西,但测试不好。

function intersection(x,y){
 x.sort();y.sort();
 var i=j=0;ret=[];
 while(i<x.length && j<y.length){
  if(x[i]<y[j])i++;
  else if(y[j]<x[i])j++;
  else {
   ret.push(x[i]);
   i++,j++;
  }
 }
 return ret;
}

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

PS:该算法只适用于数字和普通字符串,任意对象数组的交集可能不起作用。

对于只包含字符串或数字的数组,可以按照其他一些答案进行排序。对于任意对象数组的一般情况,我认为您不能避免这么做。下面将给出作为参数提供的任意数量数组的交集 arrayIntersection :

var arrayContains = Array.prototype.indexOf ?
    function(arr, val) {
        return arr.indexOf(val) > -1;
    } :
    function(arr, val) {
        var i = arr.length;
        while (i--) {
            if (arr[i] === val) {
                return true;
            }
        }
        return false;
    };

function arrayIntersection() {
    var val, arrayCount, firstArray, i, j, intersection = [], missing;
    var arrays = Array.prototype.slice.call(arguments); // Convert arguments into a real array

    // Search for common values
    firstArray = arrays.pop();
    if (firstArray) {
        j = firstArray.length;
        arrayCount = arrays.length;
        while (j--) {
            val = firstArray[j];
            missing = false;

            // Check val is present in each remaining array 
            i = arrayCount;
            while (!missing && i--) {
                if ( !arrayContains(arrays[i], val) ) {
                    missing = true;
                }
            }
            if (!missing) {
                intersection.push(val);
            }
        }
    }
    return intersection;
}

arrayIntersection( [1, 2, 3, "a"], [1, "a", 2], ["a", 1] ); // Gives [1, "a"]; 

使用ES2015和SETS非常短。接受类似数组的值(如字符串)并删除重复项。

let intersection = function(a, b) {
  a = new Set(a), b = new Set(b);
  return [...a].filter(v => b.has(v));
};

console.log(intersection([1,2,1,2,3], [2,3,5,4,5,3]));

console.log(intersection('ccaabbab', 'addb').join(''));

对这里最小的一个 filter/indexOf solution ,也就是说,使用一个javascript对象在一个数组中创建一个值的索引,将把它从O(n*m)减少到“可能”的线性时间。 source1 source2

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

这不是最简单的解决方案(它的代码比 filter+indexOf 它也不是最快的(可能比 intersect_safe() 但似乎是一个很好的平衡。这是关于 非常 简单的一面,同时提供良好的性能,而且不需要预先排序的输入。

另一种索引方法可以同时处理任意数量的数组:

// Calculate intersection of multiple array or object values.
function intersect (arrList) {
    var arrLength = Object.keys(arrList).length;
        // (Also accepts regular objects as input)
    var index = {};
    for (var i in arrList) {
        for (var j in arrList[i]) {
            var v = arrList[i][j];
            if (index[v] === undefined) index[v] = 0;
            index[v]++;
        };
    };
    var retv = [];
    for (var i in index) {
        if (index[i] == arrLength) retv.push(i);
    };
    return retv;
};

它只适用于可以作为字符串计算的值,您应该将它们作为数组传递,如下所示:

intersect ([arr1, arr2, arr3...]);

…但它透明地接受对象作为参数或作为要相交的任何元素(始终返回公共值数组)。示例:

intersect ({foo: [1, 2, 3, 4], bar: {a: 2, j:4}}); // [2, 4]
intersect ([{x: "hello", y: "world"}, ["hello", "user"]]); // ["hello"]

编辑: 我只是注意到,从某种程度上说,这辆车有点小。

也就是说:我编写代码时认为输入数组本身不能包含重复(如示例所示)。

但如果输入数组恰好包含重复,则会产生错误的结果。示例(使用以下实现):

intersect ([[1, 3, 4, 6, 3], [1, 8, 99]]);
// Expected: [ '1' ]
// Actual: [ '1', '3' ]

幸运的是,通过简单地添加二级索引,这很容易解决。即:

变化:

        if (index[v] === undefined) index[v] = 0;
        index[v]++;

通过:

        if (index[v] === undefined) index[v] = {};
        index[v][i] = true; // Mark as present in i input.

……和:

         if (index[i] == arrLength) retv.push(i);

签署人:

         if (Object.keys(index[i]).length == arrLength) retv.push(i);

完整示例:

// Calculate intersection of multiple array or object values.
function intersect (arrList) {
    var arrLength = Object.keys(arrList).length;
        // (Also accepts regular objects as input)
    var index = {};
    for (var i in arrList) {
        for (var j in arrList[i]) {
            var v = arrList[i][j];
            if (index[v] === undefined) index[v] = {};
            index[v][i] = true; // Mark as present in i input.
        };
    };
    var retv = [];
    for (var i in index) {
        if (Object.keys(index[i]).length == arrLength) retv.push(i);
    };
    return retv;
};

intersect ([[1, 3, 4, 6, 3], [1, 8, 99]]); // [ '1' ]

function intersection(A,B){
var result = new Array();
for (i=0; i<A.length; i++) {
    for (j=0; j<B.length; j++) {
        if (A[i] == B[j] && $.inArray(A[i],result) == -1) {
            result.push(A[i]);
        }
    }
}
return result;
}

由于对数据有一些限制,您可以在 线性的 时间!

为了 正整数 :使用数组将值映射到“已看到/未看到”布尔值。

function intersectIntegers(array1,array2) { 
   var seen=[],
       result=[];
   for (var i = 0; i < array1.length; i++) {
     seen[array1[i]] = true;
   }
   for (var i = 0; i < array2.length; i++) {
     if ( seen[array2[i]])
        result.push(array2[i]);
   }
   return result;
}

有一种类似的技术 物体 :取一个伪键,将array1中的每个元素设置为“true”,然后在array2的元素中查找该键。完成后清理干净。

function intersectObjects(array1,array2) { 
   var result=[];
   var key="tmpKey_intersect"
   for (var i = 0; i < array1.length; i++) {
     array1[i][key] = true;
   }
   for (var i = 0; i < array2.length; i++) {
     if (array2[i][key])
        result.push(array2[i]);
   }
   for (var i = 0; i < array1.length; i++) {
     delete array1[i][key];
   }
   return result;
}

当然,您需要确保密钥以前没有出现,否则您将破坏您的数据…

我将为我的工作做出贡献:

if (!Array.prototype.intersect){
Array.prototype.intersect = function (arr1) {

    var r = [], o = {}, l = this.length, i, v;
    for (i = 0; i < l; i++) {
        o[this[i]] = true;
    }
    l = arr1.length;
    for (i = 0; i < l; i++) {
        v = arr1[i];
        if (v in o) {
            r.push(v);
        }
    }
    return r;
};
}

IE 9.0、Chrome、Firefox、Opera的“indexof”

    function intersection(a,b){
     var rs = [], x = a.length;
     while (x--) b.indexOf(a[x])!=-1 && rs.push(a[x]);
     return rs.sort();
    }

intersection([1,2,3], [2,3,4,5]);
//Result:  [2,3]

这可能是最简单的一个, 此外 list1.filter(n=>list2.includes(n))。

var list1 = ['bread', 'ice cream', 'cereals', 'strawberry', 'chocolate']
var list2 = ['bread', 'cherry', 'ice cream', 'oats']

function check_common(list1, list2){
	
	list3 = []
	for (let i=0; i<list1.length; i++){
		
		for (let j=0; j<list2.length; j++){	
			if (list1[i] === list2[j]){
				list3.push(list1[i]);				
			}		
		}
		
	}
	return list3
	
}

check_common(list1, list2) // ["bread", "ice cream"]

ES2015的功能方法

函数方法必须考虑只使用没有副作用的纯函数,每个函数只涉及单个作业。

这些限制增强了相关函数的可组合性和可重用性。

// small, reusable auxiliary functions

const createSet = xs => new Set(xs);
const filter = f => xs => xs.filter(apply(f));
const apply = f => x => f(x);


// intersection

const intersect = xs => ys => {
  const zs = createSet(ys);
  return filter(x => zs.has(x)
     ? true
     : false
  ) (xs);
};


// mock data

const xs = [1,2,2,3,4,5];
const ys = [0,1,2,3,3,3,6,7,8,9];


// run it

console.log( intersect(xs) (ys) );

请注意,当地人 Set 使用类型,这有利于 查找性能。

避免重复

显然从一开始就反复出现 Array 保存,而第二个 数组 重复数据消除。这可能是或可能不是所需的行为。如果您需要一个独特的结果,只需应用 dedupe 第一个论点:

// auxiliary functions

const apply = f => x => f(x);
const comp = f => g => x => f(g(x));
const afrom = apply(Array.from);
const createSet = xs => new Set(xs);
const filter = f => xs => xs.filter(apply(f));


// intersection

const intersect = xs => ys => {
  const zs = createSet(ys);
  return filter(x => zs.has(x)
     ? true
     : false
  ) (xs);
};


// de-duplication

const dedupe = comp(afrom) (createSet);


// mock data

const xs = [1,2,2,3,4,5];
const ys = [0,1,2,3,3,3,6,7,8,9];


// unique result

console.log( intersect(dedupe(xs)) (ys) );

计算任意数的交集 数组 S

如果要计算任意数 数组 只是组成 intersect 具有 foldl . 这是一个便利功能:

// auxiliary functions

const apply = f => x => f(x);
const uncurry = f => (x, y) => f(x) (y);
const createSet = xs => new Set(xs);
const filter = f => xs => xs.filter(apply(f));
const foldl = f => acc => xs => xs.reduce(uncurry(f), acc);


// intersection

const intersect = xs => ys => {
  const zs = createSet(ys);
  return filter(x => zs.has(x)
     ? true
     : false
  ) (xs);
};


// intersection of an arbitrarily number of Arrays

const intersectn = (head, ...tail) => foldl(intersect) (head) (tail);


// mock data

const xs = [1,2,2,3,4,5];
const ys = [0,1,2,3,3,3,6,7,8,9];
const zs = [0,1,2,3,4,5,6];


// run

console.log( intersectn(xs, ys, zs) );

为简单起见:

// Usage
const intersection = allLists
  .reduce(intersect, allValues)
  .reduce(removeDuplicates, []);


// Implementation
const intersect = (intersection, list) =>
  intersection.filter(item =>
    list.some(x => x === item));

const removeDuplicates = (uniques, item) =>
  uniques.includes(item) ? uniques : uniques.concat(item);


// Example Data
const somePeople = [bob, doug, jill];
const otherPeople = [sarah, bob, jill];
const morePeople = [jack, jill];

const allPeople = [...somePeople, ...otherPeople, ...morePeople];
const allGroups = [somePeople, otherPeople, morePeople];

// Example Usage
const intersection = allGroups
  .reduce(intersect, allPeople)
  .reduce(removeDuplicates, []);

intersection; // [jill]

效益:

• 简单污垢
• 以数据为中心
• 适用于任意数量的列表
• 适用于任意长度的列表
• 适用于任意类型的值
• 适用于任意排序顺序
• 保留形状(任何数组中第一个外观的顺序)
• 尽可能提前退出
• 内存安全,不会篡改函数/数组原型

缺点:

• 更高的内存使用率
• 更高的CPU使用率
• 需要了解reduce
• 需要了解数据流

您不想将其用于3D引擎或内核工作,但是如果在基于事件的应用程序中运行它时遇到问题,那么您的设计就有更大的问题。

.reduce 绘制地图,以及 .filter 找到十字路口。 delete 内 滤波器 允许我们将第二个数组视为一个唯一的集合。

function intersection (a, b) {
  var seen = a.reduce(function (h, k) {
    h[k] = true;
    return h;
  }, {});

  return b.filter(function (k) {
    var exists = seen[k];
    delete seen[k];
    return exists;
  });
}

我觉得这种方法很容易理解。它在恒定时间内执行。

这里是 underscore.js 实施:

_.intersection = function(array) {
  if (array == null) return [];
  var result = [];
  var argsLength = arguments.length;
  for (var i = 0, length = array.length; i < length; i++) {
    var item = array[i];
    if (_.contains(result, item)) continue;
    for (var j = 1; j < argsLength; j++) {
      if (!_.contains(arguments[j], item)) break;
    }
    if (j === argsLength) result.push(item);
  }
  return result;
};

来源: http://underscorejs.org/docs/underscore.html#section-62

var listA = [1,2,3,4,5,6,7];
var listB = [2,4,6,8];

var result = listA.filter(itemA=> {
    return listB.some(itemB => itemB === itemA);
});

function getIntersection(arr1, arr2){
    var result = [];
    arr1.forEach(function(elem){
        arr2.forEach(function(elem2){
            if(elem === elem2){
                result.push(elem);
            }
        });
    });
    return result;
}

getIntersection([1,2,3], [2,3,4,5]); // [ 2, 3 ]

如果需要让它处理多个数组的相交:

const intersect = (a, b, ...rest) => {
  if (rest.length === 0) return [...new Set(a)].filter(x => new Set(b).has(x));
  return intersect(a, intersect(b, ...rest));
};

console.log(intersect([1,2,3,4,5], [1,2], [1, 2, 3,4,5], [2, 10, 1])) // [1,2]

相反,使用indexof也可以使用 Array.protype.includes .

function intersection(arr1, arr2) {
  return arr1.filter((ele => {
    return arr2.includes(ele);
  }));
}

console.log(intersection([1,2,3], [2,3,4,5]));

自ECMA 2016以来,您可以使用:

const intersection = (arr1, arr2) => arr1.filter(el => arr2.includes(el));