https://stackoverflow.com/a/49577331/8784402

使用Delta

对于javascript

smallest and one-liner

[...Array(N)].map((v, i) => from + i * step);

示例和其他备选方案

Array.from(Array(10).keys()).map(i => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]

[...Array(10).keys()].map(i => 4 + i * -2);
//=> [4, 2, 0, -2, -4, -6, -8, -10, -12, -14]

Array(10).fill(0).map((v, i) => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]

Array(10).fill().map((v, i) => 4 + i * -2);
//=> [4, 2, 0, -2, -4, -6, -8, -10, -12, -14]

[...Array(10)].map((v, i) => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]

Range Function

const range = (from, to, step) =>
  [...Array(Math.floor((to - from) / step) + 1)].map((_, i) => from + i * step);

range(0, 9, 2);
//=> [0, 2, 4, 6, 8]

// can also assign range function as static method in Array class (but not recommended )
Array.range = (from, to, step) =>
  [...Array(Math.floor((to - from) / step) + 1)].map((_, i) => from + i * step);

Array.range(2, 10, 2);
//=> [2, 4, 6, 8, 10]

Array.range(0, 10, 1);
//=> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Array.range(2, 10, -1);
//=> []

Array.range(3, 0, -1);
//=> [3, 2, 1, 0]

As Iterators

class Range {
  constructor(total = 0, step = 1, from = 0) {
    this[Symbol.iterator] = function* () {
      for (let i = 0; i < total; yield from + i++ * step) {}
    };
  }
}

[...new Range(5)]; // Five Elements
//=> [0, 1, 2, 3, 4]
[...new Range(5, 2)]; // Five Elements With Step 2
//=> [0, 2, 4, 6, 8]
[...new Range(5, -2, 10)]; // Five Elements With Step -2 From 10
//=>[10, 8, 6, 4, 2]
[...new Range(5, -2, -10)]; // Five Elements With Step -2 From -10
//=> [-10, -12, -14, -16, -18]

// Also works with for..of loop
for (i of new Range(5, -2, 10)) console.log(i);
// 10 8 6 4 2

As Generators Only

const Range = function* (total = 0, step = 1, from = 0) {
  for (let i = 0; i < total; yield from + i++ * step) {}
};

Array.from(Range(5, -2, -10));
//=> [-10, -12, -14, -16, -18]

[...Range(5, -2, -10)]; // Five Elements With Step -2 From -10
//=> [-10, -12, -14, -16, -18]

// Also works with for..of loop
for (i of Range(5, -2, 10)) console.log(i);
// 10 8 6 4 2

// Lazy loaded way
const number0toInf = Range(Infinity);
number0toInf.next().value;
//=> 0
number0toInf.next().value;
//=> 1
// ...

带步长/增量的从到

using iterators

class Range2 {
  constructor(to = 0, step = 1, from = 0) {
    this[Symbol.iterator] = function* () {
      let i = 0,
        length = Math.floor((to - from) / step) + 1;
      while (i < length) yield from + i++ * step;
    };
  }
}
[...new Range2(5)]; // First 5 Whole Numbers
//=> [0, 1, 2, 3, 4, 5]

[...new Range2(5, 2)]; // From 0 to 5 with step 2
//=> [0, 2, 4]

[...new Range2(5, -2, 10)]; // From 10 to 5 with step -2
//=> [10, 8, 6]

using Generators

const Range2 = function* (to = 0, step = 1, from = 0) {
  let i = 0,
    length = Math.floor((to - from) / step) + 1;
  while (i < length) yield from + i++ * step;
};

[...Range2(5, -2, 10)]; // From 10 to 5 with step -2
//=> [10, 8, 6]

let even4to10 = Range2(10, 2, 4);
even4to10.next().value;
//=> 4
even4to10.next().value;
//=> 6
even4to10.next().value;
//=> 8
even4to10.next().value;
//=> 10
even4to10.next().value;
//=> undefined

对于字体

class _Array<T> extends Array<T> {
  static range(from: number, to: number, step: number): number[] {
    return Array.from(Array(Math.floor((to - from) / step) + 1)).map(
      (v, k) => from + k * step
    );
  }
}
_Array.range(0, 9, 1);

最终总结报告。。Drrruummm Rolll公司-

这是在不使用ES6的情况下生成大小为N（此处为10）的数组的最短代码。Cocco上面的版本很接近，但不是最短的。

(function(n){for(a=[];n--;a[n]=n+1);return a})(10)

但这场“代码高尔夫”（用最少的源代码字节来解决特定问题的比赛）无可争议的赢家是Niko Ruotsalainen。使用数组构造函数和ES6扩展运算符。（大多数ES6语法都是有效的typeScript，但下面不是。所以在使用它时要谨慎）

[...Array(10).keys()]

你可以这样做：

var arr=数组.from（数组（10）.keys（））arr.shift（）控制台日志（arr）

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--; });
};

如果您碰巧像我一样在应用程序中使用d3.js，d3会提供一个助手函数来为您执行此操作。

因此，要获得从0到4的数组，非常简单：

d3.range(5)
[0, 1, 2, 3, 4]

并获得从1到5的数组，如您所请求的：

d3.range(1, 5+1)
[1, 2, 3, 4, 5]

查看本教程了解更多信息。

尝试在Number的原型中添加迭代器。

Number.prototype[Symbol.iterator] = function *(){
  let i = 0;
  while(i < this) yield i++;
  return;
}

现在数字是可迭代的，只需将一个数字传递给Array.from

Array.from(10);//[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

或者任何其他需要迭代的地方，比如。。。循环。

for(const number of 10) console.log(number);//logs 0 through 9 sequentially

这有点复杂，但也很酷。