用javascript实现数组交叉的最简单、无库代码是什么?我想写
intersection([1,2,3], [2,3,4,5])
并获得
[2, 3]
用javascript实现数组交叉的最简单、无库代码是什么?我想写
intersection([1,2,3], [2,3,4,5])
并获得
[2, 3]
当前回答
如果你需要让它处理交叉多个数组:
Const intersect = (a1, a2,…rest) => { Const a12 = a1。过滤器(value => a2.includes(value)) 如果休息。长度=== 0){返回a12;} 回归相交(a12,…rest); }; console.log(相交([1、2、3、4、5],[1,2],[1、2、3、4、5],[2 10 1]))
其他回答
对于只包含字符串或数字的数组,你可以做一些排序,就像其他答案一样。对于任意对象数组的一般情况,我不认为你可以避免这样做。下面将为您提供任意数量的数组的交集作为arrayIntersection的参数:
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"];
与效率无关,但很容易理解,这里有一个集合的并和交的例子,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);
使用Underscore.js或lodash.js
_.intersection( [0,345,324] , [1,0,324] ) // gives [0,324]
该函数利用字典的强大功能,避免了N^2问题。每个数组只循环一次,第三次更短的循环返回最终结果。 它还支持数字、字符串和对象。
function array_intersect(array1, array2)
{
var mergedElems = {},
result = [];
// Returns a unique reference string for the type and value of the element
function generateStrKey(elem) {
var typeOfElem = typeof elem;
if (typeOfElem === 'object') {
typeOfElem += Object.prototype.toString.call(elem);
}
return [typeOfElem, elem.toString(), JSON.stringify(elem)].join('__');
}
array1.forEach(function(elem) {
var key = generateStrKey(elem);
if (!(key in mergedElems)) {
mergedElems[key] = {elem: elem, inArray2: false};
}
});
array2.forEach(function(elem) {
var key = generateStrKey(elem);
if (key in mergedElems) {
mergedElems[key].inArray2 = true;
}
});
Object.values(mergedElems).forEach(function(elem) {
if (elem.inArray2) {
result.push(elem.elem);
}
});
return result;
}
如果有无法解决的特殊情况,仅通过修改generateStrKey函数就可以解决。这个函数的诀窍在于,它根据类型和值唯一地表示每个不同的数据。
这个变体有一些性能改进。在任何数组为空的情况下避免循环。它还首先遍历较短的数组,因此如果它在第二个数组中找到了第一个数组的所有值,则退出循环。
function array_intersect(array1, array2)
{
var mergedElems = {},
result = [],
firstArray, secondArray,
firstN = 0,
secondN = 0;
function generateStrKey(elem) {
var typeOfElem = typeof elem;
if (typeOfElem === 'object') {
typeOfElem += Object.prototype.toString.call(elem);
}
return [typeOfElem, elem.toString(), JSON.stringify(elem)].join('__');
}
// Executes the loops only if both arrays have values
if (array1.length && array2.length)
{
// Begins with the shortest array to optimize the algorithm
if (array1.length < array2.length) {
firstArray = array1;
secondArray = array2;
} else {
firstArray = array2;
secondArray = array1;
}
firstArray.forEach(function(elem) {
var key = generateStrKey(elem);
if (!(key in mergedElems)) {
mergedElems[key] = {elem: elem, inArray2: false};
// Increases the counter of unique values in the first array
firstN++;
}
});
secondArray.some(function(elem) {
var key = generateStrKey(elem);
if (key in mergedElems) {
if (!mergedElems[key].inArray2) {
mergedElems[key].inArray2 = true;
// Increases the counter of matches
secondN++;
// If all elements of first array have coincidence, then exits the loop
return (secondN === firstN);
}
}
});
Object.values(mergedElems).forEach(function(elem) {
if (elem.inArray2) {
result.push(elem.elem);
}
});
}
return result;
}
下面是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;
};
来源:http://underscorejs.org/docs/underscore.html部分- 62