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
有一个小功能，它允许使用像[1，2]。范围（3，4）->[1，2，3，4]也适用于负参数。享受
Array.prototype.range = function(from, to)
{
var range = (!to)? from : Math.abs(to - from) + 1, increase = from < to;
var tmp = Array.apply(this, {"length": range}).map(function()
{
return (increase)?from++ : from--;
}, Number);
return this.concat(tmp);
};
对以上内容进行改进：
var range = function (n) {
return Array(n).join().split(',').map(function(e, i) { return i; });
}
可以获得以下选项：
1） Array.init设置为值v
var arrayInitTo = function (n,v) {
return Array(n).join().split(',').map(function() { return v; });
};
2） 获得反向范围：
var rangeRev = function (n) {
return Array(n).join().split(',').map(function() { return n--; });
};
您可以使用此选项：
new Array(/*any number which you want*/)
.join().split(',')
.map(function(item, index){ return ++index;})
例如
new Array(10)
.join().split(',')
.map(function(item, index){ return ++index;})
将创建以下数组：
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
“_”.重复（5）.拆分（“”）.映射（（_，i）=>i+1）将产生[1，2，3，4，5]
您可以使用Es6中的阵列填充和映射；就像一些人在回答这个问题时建议的那样。以下是一些示例：
Example-One: Array(10).fill(0).map((e,i)=>i+1)
Result-One: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Example-Two: Array(100/10).fill(0).map((e,i)=>(i*10)+10)
Result-Two:[10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
我更喜欢这样，因为我觉得这很简单。
