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
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;
}
其他回答
另一种可以同时处理任意数量数组的索引方法:
// 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,你应该将它们作为一个数组传递:
intersect ([arr1, arr2, arr3...]);
...但它透明地接受对象作为参数或任何要交叉的元素(总是返回公共值的数组)。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,这是,在某种程度上,有点bug。
也就是说:我在编码时认为输入数组本身不能包含重复(正如所提供的示例那样)。
但如果输入数组恰好包含重复,就会产生错误的结果。示例(使用下面的实现):
intersect ([[1, 3, 4, 6, 3], [1, 8, 99]]);
// Expected: [ '1' ]
// Actual: [ '1', '3' ]
幸运的是,这很容易通过添加二级索引来解决。那就是:
变化:
if (index[v] === undefined) index[v] = 0;
index[v]++;
by:
if (index[v] === undefined) index[v] = {};
index[v][i] = true; // Mark as present in i input.
,:
if (index[i] == arrLength) retv.push(i);
by:
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' ]
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;
}
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()
//在线性时间内返回数组a中也在b中的元素: 函数相交(a, b) { 返回a.filter (Set.prototype。new Set(b)); } / /例如: console.log(相交([1,2,3],[2、3、4、5]));
我推荐上述简洁的解决方案,它在大输入上优于其他实现。如果在小输入上的性能很重要,请检查下面的替代方案。
备选方案和性能比较:
有关替代实现,请参阅下面的代码片段,并检查https://jsperf.com/array-intersection-comparison以进行性能比较。
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; }
Firefox 53的结果:
Ops/sec on large arrays (10,000 elements): 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 Ops/sec on small arrays (100 elements): 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
我会用对我来说最有效的方法来贡献:
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;
};
}