我正在寻找以下方法的任何替代方法，以创建包含1到N的JavaScript数组，其中N仅在运行时已知。
var foo = [];
for (var i = 1; i <= N; i++) {
foo.push(i);
}
对我来说，我觉得应该有一种不用循环的方法。
当前回答
以下是摘要（在控制台中运行）：
// setup:
var n = 10000000;
function* rangeIter(a, b) {
for (let i = a; i <= b; ++i) yield i;
}
function range(n) {
let a = []
for (; n--; a[n] = n);
return a;
}
function sequence(max, step = 1) {
return {
[Symbol.iterator]: function* () {
for (let i = 1; i <= max; i += step) yield i
}
}
}
var t0, t1, arr;
// tests
t0 = performance.now();
arr = Array.from({ length: n }, (a, i) => 1)
t1 = performance.now();
console.log("Array.from({ length: n }, (a, i) => 1) Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = range(n);
t1 = performance.now();
console.log("range(n) Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = Array.from(rangeIter(0, n));
t1 = performance.now();
console.log("Array.from(rangeIter(0, n)) Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = [...rangeIter(0, n)];
t1 = performance.now();
console.log("[...rangeIter(0, n)] Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = Array.from(sequence(n));
t1 = performance.now();
console.log("Array.from(sequence(n)) Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = [...sequence(n)];
t1 = performance.now();
console.log("[...sequence(n)] Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = Array(n).fill(0).map(Number.call, Number);
t1 = performance.now();
console.log("Array(n).fill(0).map(Number.call, Number) Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = Array.from(Array(n).keys());
t1 = performance.now();
console.log("Array.from(Array(n).keys()) Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = [...Array(n).keys()];
t1 = performance.now();
console.log("[...Array(n).keys()] Took " + (t1 - t0) + " milliseconds.");
最快的是Array（n）.fill（0）.map（Number.call，Number），第二个是[…Array（n）.keys（）]
但是。。。rangeIter的方式非常方便（可以内联），速度快，功能更强大
其他回答
//不分配N大小数组（ES6，带有一些流注释）的解决方案：函数*zeroToN（N/*：数字*/）/*：生成器＜number，void，empty＞*/{对于（设n＝0；n＜＝n；n＋＝1），得到n；}//通过这一代，您可以拥有您的阵列console.log（[…zeroToN（10-1）]）//但是让我们定义一个助手迭代器函数函数mapIterator（迭代器，映射）{常量arr=[]；for（let result=iterater.next（）！result.done；result=iterator.next（））{arr.push（映射（result.value））；}返回arr；}//现在您有了一个map函数，不需要分配0…N-1数组console.log（mapIterator（zeroToN（10-1），n=>n*n））；
数组（…数组（9））.map（（_，i）=>i）；console.log（数组（…数组（9））.map（（_，i）=>i））
ES5版本，效率很低，但可能是最短的一个，它是一个表达式，而不是一个变量填充有例如for循环的语句：
(Array(N)+'').split(',').map(function(d,i){return i})
在v8中填充数组的最快方法是：
[...Array(5)].map((_,i) => i);
结果将为：[0，1，2，3，4]
对我来说，这是更有用的实用程序：
/**
* create an array filled with integer numbers from base to length
* @param {number} from
* @param {number} to
* @param {number} increment
* @param {Array} exclude
* @return {Array}
*/
export const count = (from = 0, to = 1, increment = 1, exclude = []) => {
const array = [];
for (let i = from; i <= to; i += increment) !exclude.includes(i) && array.push(i);
return array;
};
