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

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--; });
};

其他回答

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

使用不修改Number.prototype的生成器函数的可移植版本。

函数序列(最大值,步长=1){返回{[Symbol.iiterat]:函数*(){对于(设i=1;i<=max;i+=步长),得出i}}}console.log([…序列(10)])

由于有很多好的答案,这可能也是一个选项,您也可以使用下面的创建一个函数,它将适用于任何数字组合

const start = 10;
const end = 30;    
const difference = Math.abs(start-end);
const rangeArray = new Array(difference + 1).fill(undefined).map((val, key) => {
    return start > end ? start - key : start + key;
})

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})

var foo=Array.from(数组(N),(v,i)=>i+1);