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
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;
}
其他回答
破坏性似乎是最简单的,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;
}
非破坏性的要稍微复杂一点,因为我们要跟踪指标:
/* 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;
}
这是我使用的一个非常简单的实现。它是无损的,也确保不复制整个。
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;
}
我写了一个相交函数,它甚至可以根据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}]
我们想要基于id属性的交集,那么输出应该是:
[{id: 20}]
因此,相同(注:ES6代码)的函数为:
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)));
};
你可以这样调用这个函数:
intersect(arr1, arr2, [elem => elem.id, elem => elem.id])
还要注意:该函数查找交集时考虑到第一个数组是主数组,因此交集结果将是主数组的结果。
对这里最小的一个(filter/indexOf解决方案)稍作调整,即使用JavaScript对象在其中一个数组中创建值的索引,将从O(N*M)减少到“可能”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要多),也不是最快的解决方案(可能比intersect_safe()慢一个常数因子),但似乎是一个很好的平衡。它非常简单,同时提供了良好的性能,并且不需要预先排序的输入。
在coffescript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()