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
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' ]
其他回答
使用jQuery:
var a = [1,2,3];
var b = [2,3,4,5];
var c = $(b).not($(b).not(a));
alert(c);
//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。
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