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

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

为了好玩,我想借鉴伊恩·亨利的答案。

当然,var array=new array(N);将为您提供大小为N的数组,但键和值将相同。。。。然后,要将数组缩短为M大小,请使用array.length=M…但对于一些添加的功能,请尝试:

function range()
{
    // This function takes optional arguments:
    // start, end, increment
    //    start may be larger or smaller than end
    // Example:  range(null, null, 2);

    var array = []; // Create empty array

      // Get arguments or set default values:
    var start = (arguments[0] ? arguments[0] : 0);
    var end   = (arguments[1] ? arguments[1] : 9);
      // If start == end return array of size 1
    if (start == end) { array.push(start); return array; }
    var inc   = (arguments[2] ? Math.abs(arguments[2]) : 1);

    inc *= (start > end ? -1 : 1); // Figure out which direction to increment.

      // Loop ending condition depends on relative sizes of start and end
    for (var i = start; (start < end ? i <= end : i >= end) ; i += inc)
        array.push(i);

    return array;
}

var foo = range(1, -100, 8.5)

for(var i=0;i<foo.length;i++){
  document.write(foo[i] + ' is item: ' + (i+1) + ' of ' + foo.length + '<br/>'); 
}​

上述输出:

1是第1项,共12项-7.5为第2项,共12项-16是第3项,共12项-24.5为第4项,共12项-33是第5项,共12项-41.5是第6项,共12项-50是第7项,共12项-58.5为第8项,共12项-67是第9项,共12项-75.5为第10项,共12项-84是第11项,共12项-92.5是第12项,共12项

jsFiddle示例

此函数使用自动生成的参数数组。

该函数创建一个数组,该数组中的值以大小增量开始,以大小增量结束,其中

range(start, end, increment);

每个值都有一个默认值,增量的符号无关紧要,因为增量的方向取决于开始和结束的相对大小。

function range(start, end) {
    var foo = [];
    for (var i = start; i <= end; i++) {
        foo.push(i);
    }
    return foo;
}

然后由调用

var foo = range(1, 5);

在Javascript中没有内置的方法来实现这一点,但如果您需要多次执行,则可以创建一个非常有效的实用程序函数。

编辑:在我看来,以下是一个更好的范围函数。也许只是因为我对LINQ有偏见,但我认为它在更多情况下更有用。您的里程数可能有所不同。

function range(start, count) {
    if(arguments.length == 1) {
        count = start;
        start = 0;
    }

    var foo = [];
    for (var i = 0; i < count; i++) {
        foo.push(start + i);
    }
    return foo;
}

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);
};

对于小范围,切片是不错的。N仅在运行时已知,因此:

[0, 1, 2, 3, 4, 5].slice(0, N+1)