如果只有两个数组要连接，并且实际上需要追加其中一个数组，而不是创建一个新数组，则应该使用push或循环。

基准:https://jsperf.com/concat-small-arrays-vs-push-vs-loop/

使用Array.prototype.concat.apply来处理多个数组的连接:

var resultArray = Array.prototype.concat.apply([], arrayOfArraysToConcat);

例子:

var a1 = [1, 2, 3],
    a2 = [4, 5],
    a3 = [6, 7, 8, 9];
Array.prototype.concat.apply([], [a1, a2, a3]); // [1, 2, 3, 4, 5, 6, 7, 8, 9]

[].concat.apply([], [array1, array2, ...])

效率证明:http://jsperf.com/multi-array-concat/7

Tim Supinie mentions in the comments that this may cause the interpreter to exceed the call stack size. This is perhaps dependent on the js engine, but I've also gotten "Maximum call stack size exceeded" on Chrome at least. Test case: [].concat.apply([], Array(300000).fill().map(_=>[1,2,3])). (I've also gotten the same error using the currently accepted answer, so one is anticipating such use cases or building a library for others, special testing may be necessary no matter which solution you choose.)

最快的10倍是迭代数组，就像它们是一个数组一样，而不实际连接它们(如果可以的话)。

我很惊讶concat比push稍微快一点，除非测试不公平。

const arr1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']; const arr2 = ['j', 'k', 'l', 'i', 'm', 'n', 'o', 'p', 'q', 'r', 's']; const arr3 = ['t', 'u', 'v', 'w']; const arr4 = ['x', 'y', 'z']; let start; // Not joining but iterating over all arrays - fastest // at about 0.06ms start = performance.now() const joined = [arr1, arr2, arr3, arr4]; for (let j = 0; j < 1000; j++) { let i = 0; while (joined.length) { // console.log(joined[0][i]); if (i < joined[0].length - 1) i++; else { joined.shift() i = 0; } } } console.log(performance.now() - start); // Concating (0.51ms). start = performance.now() for (let j = 0; j < 1000; j++) { const a = [].concat(arr1, arr2, arr3, arr4); } console.log(performance.now() - start); // Pushing on to an array (mutating). Slowest (0.77ms) start = performance.now() const joined2 = [arr1, arr2, arr3, arr4]; for (let j = 0; j < 1000; j++) { const arr = []; for (let i = 0; i < joined2.length; i++) { Array.prototype.push.apply(arr, joined2[i]) } } console.log(performance.now() - start);

如果你抽象它，你可以让不加入的迭代更干净，它仍然是原来的两倍:

const arr1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']; const arr2 = ['j', 'k', 'l', 'i', 'm', 'n', 'o', 'p', 'q', 'r', 's']; const arr3 = ['t', 'u', 'v', 'w']; const arr4 = ['x', 'y', 'z']; function iterateArrays(arrays, onEach) { let i = 0; while (joined.length) { onEach(joined[0][i]); if (i < joined[0].length - 1) i++; else { joined.shift(); i = 0; } } } // About 0.23ms. let start = performance.now() const joined = [arr1, arr2, arr3, arr4]; for (let j = 0; j < 1000; j++) { iterateArrays(joined, item => { //console.log(item); }); } console.log(performance.now() - start);

轻松使用concat函数:

var a = [1,2,3];
var b = [2,3,4];
a = a.concat(b);
>> [1,2,3,2,3,4]

concat()方法用于连接两个或多个数组。它不会更改现有数组，只返回已连接数组的副本。

array1 = array1.concat(array2, array3, array4, ..., arrayN);