JavaScript中数组交集的最简单代码

2020/09/23 00:21 · javascript ·  · 0评论

在javascript中实现数组交集的最简单,无库代码是什么?我想写

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

并得到

[2, 3]

使用的组合Array.prototype.filterArray.prototype.includes

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

对于较旧的浏览器,带有Array.prototype.indexOf和不带有箭头功能:

array1.filter(function(n) {
    return array2.indexOf(n) !== -1;
})

破坏性似乎最简单,特别是如果我们可以假设输入已排序:

/* 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;
}

由于我们必须跟踪索引,因此非破坏性的头发必须更加复杂。

/* 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,那么这是一种简单有效的方法(请参阅规范链接):

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,但可读性较差(也没有创建其他交集Set):

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

请注意,使用集时,您将仅获得不同的值,因此new Set([1, 2, 3, 3]).size计算为3

使用 Underscore.jslodash.js

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

我推荐以上简洁的解决方案,该解决方案在大输入量方面胜过其他实现。如果小投入的性能很重要,请检查以下替代方案。

替代方案和性能比较:

请参阅以下代码段以了解替代实现,并检查https://jsperf.com/array-intersection-comparison以进行性能比较。

Firefox 53中的结果:

  • 大型阵列(10,000个元素)上的运算/秒:

    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
  • 小型阵列(100个元素)上的运算/秒:

    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

我对ES6的贡献。通常,它查找具有作为参数提供的不确定数目的数组的数组的交集。

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;
}
  1. 解决
  2. 从索引0逐一检查,从中创建新数组。

像这样的东西,虽然没有很好的测试。

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:仅适用于数字和普通字符串的算法,任意对象数组的交集可能不起作用。

通过使用.pop而不是.shift可以改善@atk对基元排序数组的实现的性能。

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jsPerf创建了一个基准测试:http ://bit.ly/P9FrZK 使用.pop的速度大约快三倍。

使用jQuery

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

对于仅包含字符串或数字的数组,您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样做。下面将为您提供作为参数提供的任意数量的数组的交集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"]; 

对此处的最小调整进行细微调整(filter / indexOf解决方案),即使用JavaScript对象在数组之一中创建值的索引,会将其从O(N * M)减少为“大概”线性时间。源1 源2

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(它比filter + indexOf的代码更多),也不是最快的(可能比intersect_safe()慢一个常数),但似乎是一个很好的平衡。它在非常简单的方面,同时提供了良好的性能,并且不需要预先排序的输入。

另一种索引方法能够一次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;
};

它仅适用于可以评估为字符串的值,您应该将它们作为数组传递,例如:

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

...但是它透明地接受对象作为参数或要相交的任何元素(总是返回公共值数组)。例子:

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,从某种意义上来说,这是有问题的。

那就是:我对它进行了编码,以为输入数组本身不能包含重复(如提供的示例所没有)。

但是,如果输入数组碰巧包含重复,那将产生错误的结果。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,只需添加二级索引即可轻松解决此问题。那是:

更改:

        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);

通过:

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

完整的例子:

// 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' ]

With some restrictions on your data, you can do it in linear time!

For positive integers: use an array mapping the values to a "seen/not seen" boolean.

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;
}

There is a similar technique for objects: take a dummy key, set it to "true" for each element in array1, then look for this key in elements of array2. Clean up when you're done.

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;
}

Of course you need to be sure the key didn't appear before, otherwise you'll be destroying your data...

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;
}

我将以最适合我的方式做出贡献:

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;
};
}

.reduce绘制地图并.filter找到相交点。delete中的.filter允许我们将第二个数组视为唯一的数组。

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;
  });
}

我发现这种方法很容易推论。它以恒定的时间执行。

我编写了一个积分函数,该函数甚至可以根据那些对象的特定属性来检测对象数组的交集。

例如,

if arr1 = [{id: 10}, {id: 20}]
and arr2 =  [{id: 20}, {id: 25}]

并且我们希望基于该id属性的交集,则输出应为:

[{id: 20}]

因此,相同的功能(请注意:ES6代码)为:

const intersect = (arr1, arr2, accessors = [v => v, v => v]) => {
    const [fn1, fn2] = accessors;
    const set = new Set(arr2.map(v => fn2(v)));
    return arr1.filter(value => set.has(fn1(value)));
};

您可以将函数调用为:

intersect(arr1, arr2, [elem => elem.id, elem => elem.id])

还要注意:此函数在考虑第一个数组是主数组的情况下找到交集,因此交集结果将是主数组的交集。

这是underscore.js的实现:

_.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;
};

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

ES2015的功能性方法

功能方法必须考虑仅使用没有副作用的纯函数,而每个副作用仅与一项工作有关。

这些限制增强了所涉及功能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,使用了本机Set类型,它具有优越的查找性能。

避免重复

显然,从第一个重复出现的项Array被保留,而第二Array个重复项被去重复。这可能是或可能不是所需的行为。如果您需要唯一的结果,只需将其应用于dedupe第一个参数:

// 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) );

计算任意数量的Arrays 的交集

如果你要计算的任意数目的交集Array的只有我撰写intersectfoldl这是一个便捷功能:

// 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]

好处:

  • 简单的污垢
  • 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
  • 保持形状(任何阵列中的首次出现顺序)
  • 尽可能提前退出
  • 内存安全,不影响功能/阵列原型

缺点:

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

您不想将其用于3D引擎或内核工作,但是如果您无法在基于事件的应用程序中运行它,则会遇到较大的问题。

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

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"]

如果需要让它处理相交的多个数组:

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]

使用一个数组创建一个对象,并遍历第二个数组以检查该值是否作为键存在。

function intersection(arr1, arr2) {
  var myObj = {};
  var myArr = [];
  for (var i = 0, len = arr1.length; i < len; i += 1) {
    if(myObj[arr1[i]]) {
      myObj[arr1[i]] += 1; 
    } else {
      myObj[arr1[i]] = 1;
    }
  }
  for (var j = 0, len = arr2.length; j < len; j += 1) {
    if(myObj[arr2[j]] && myArr.indexOf(arr2[j]) === -1) {
      myArr.push(arr2[j]);
    }
  }
  return myArr;
}

这是我正在使用的非常幼稚的实现。它是非破坏性的,并且还确保不重复整个过程。

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;
}

我扩展了tarulen的答案以使用任何数量的数组。它也应该使用非整数值。

function intersect() { 
    const last = arguments.length - 1;
    var seen={};
    var result=[];
    for (var i = 0; i < last; i++)   {
        for (var j = 0; j < arguments[i].length; j++)  {
            if (seen[arguments[i][j]])  {
                seen[arguments[i][j]] += 1;
            }
            else if (!i)    {
                seen[arguments[i][j]] = 1;
            }
        }
    }
    for (var i = 0; i < arguments[last].length; i++) {
        if ( seen[arguments[last][i]] === last)
            result.push(arguments[last][i]);
        }
    return result;
}
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 ]

如果您的数组已排序,则应在O(n)中运行,其中n为min(a.length,b.length)

function intersect_1d( a, b ){
    var out=[], ai=0, bi=0, acurr, bcurr, last=Number.MIN_SAFE_INTEGER;
    while( ( acurr=a[ai] )!==undefined && ( bcurr=b[bi] )!==undefined ){
        if( acurr < bcurr){
            if( last===acurr ){
                out.push( acurr );
            }
            last=acurr;
            ai++;
        }
        else if( acurr > bcurr){
            if( last===bcurr ){
                out.push( bcurr );
            }
            last=bcurr;
            bi++;
        }
        else {
            out.push( acurr );
            last=acurr;
            ai++;
            bi++;
        }
    }
    return out;
}
var arrays = [
    [1, 2, 3],
    [2, 3, 4, 5]
]
function commonValue (...arr) {
    let res = arr[0].filter(function (x) {
        return arr.every((y) => y.includes(x))
    })
    return res;
}
commonValue(...arrays);
function intersectionOfArrays(arr1, arr2) {
    return arr1.filter((element) => arr2.indexOf(element) !== -1).filter((element, pos, self) => self.indexOf(element) == pos);
}
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