要紧密复制的类型脚本函数

/**
 * Create a generator from 0 to stop, useful for iteration. Similar to range in Python.
 * See: https://stackoverflow.com/questions/3895478/does-javascript-have-a-method-like-range-to-generate-a-range-within-the-supp
 * See: https://docs.python.org/3/library/stdtypes.html#ranges
 * @param {number | BigNumber} stop
 * @returns {Iterable<number>}
 */
export function range(stop: number | BigNumber): Iterable<number>
/**
 * Create a generator from start to stop, useful for iteration. Similar to range in Python.
 * See: https://stackoverflow.com/questions/3895478/does-javascript-have-a-method-like-range-to-generate-a-range-within-the-supp
 * See: https://docs.python.org/3/library/stdtypes.html#ranges
 * @param {number | BigNumber} start
 * @param {number | BigNumber} stop
 * @returns {Iterable<number>}
 */
export function range(
  start: number | BigNumber,
  stop: number | BigNumber,
): Iterable<number>

/**
 * Create a generator from start to stop while skipping every step, useful for iteration. Similar to range in Python.
 * See: https://stackoverflow.com/questions/3895478/does-javascript-have-a-method-like-range-to-generate-a-range-within-the-supp
 * See: https://docs.python.org/3/library/stdtypes.html#ranges
 * @param {number | BigNumber} start
 * @param {number | BigNumber} stop
 * @param {number | BigNumber} step
 * @returns {Iterable<number>}
 */
export function range(
  start: number | BigNumber,
  stop: number | BigNumber,
  step: number | BigNumber,
): Iterable<number>
export function* range(a: unknown, b?: unknown, c?: unknown): Iterable<number> {
  const getNumber = (val: unknown): number =>
    typeof val === 'number' ? val : (val as BigNumber).toNumber()
  const getStart = () => (b === undefined ? 0 : getNumber(a))
  const getStop = () => (b === undefined ? getNumber(a) : getNumber(b))
  const getStep = () => (c === undefined ? 1 : getNumber(c))

  for (let i = getStart(); i < getStop(); i += getStep()) {
    yield i
  }
}

(from, to) => [...Array(to - from)].map((_,i)=> i + from)

对于具有良好向后兼容性的更像红宝石的方法：

范围（[开始]，结束=0），其中开始和结束是数字

var range = function(begin, end) {
  if (typeof end === "undefined") {
    end = begin; begin = 0;
  }
  var result = [], modifier = end > begin ? 1 : -1;
  for ( var i = 0; i <= Math.abs(end - begin); i++ ) {
    result.push(begin + i * modifier);
  }
  return result;
}

示例：

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

数字

[...Array(5).keys()];
 => [0, 1, 2, 3, 4]

字符迭代

String.fromCharCode(...[...Array('D'.charCodeAt(0) - 'A'.charCodeAt(0) + 1).keys()].map(i => i + 'A'.charCodeAt(0)));
 => "ABCD"

迭代

for (const x of Array(5).keys()) {
  console.log(x, String.fromCharCode('A'.charCodeAt(0) + x));
}
 => 0,"A" 1,"B" 2,"C" 3,"D" 4,"E"

作为函数

function range(size, startAt = 0) {
    return [...Array(size).keys()].map(i => i + startAt);
}

function characterRange(startChar, endChar) {
    return String.fromCharCode(...range(endChar.charCodeAt(0) -
            startChar.charCodeAt(0), startChar.charCodeAt(0)))
}

类型化函数

function range(size:number, startAt:number = 0):ReadonlyArray<number> {
    return [...Array(size).keys()].map(i => i + startAt);
}

function characterRange(startChar:string, endChar:string):ReadonlyArray<string> {
    return String.fromCharCode(...range(endChar.charCodeAt(0) -
            startChar.charCodeAt(0), startChar.charCodeAt(0)))
}

lodash.js_.range（）函数

_.range(10);
 => [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
_.range(1, 11);
 => [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
_.range(0, 30, 5);
 => [0, 5, 10, 15, 20, 25]
_.range(0, -10, -1);
 => [0, -1, -2, -3, -4, -5, -6, -7, -8, -9]
String.fromCharCode(..._.range('A'.charCodeAt(0), 'D'.charCodeAt(0) + 1));
 => "ABCD"

没有库的旧非es6浏览器：

Array.apply(null, Array(5)).map(function (_, i) {return i;});
 => [0, 1, 2, 3, 4]

console.log（[…Array（5）.keys（）]）；

（ES6归功于尼尔斯·彼得索恩和其他评论者）

Array.from(Array((m - n + 1)), (v, i) => n + i); // m > n and both of them are integers.

为了在任何一个给定的数字可能更大的地方工作，我写了这样一个：

function getRange(start, end) {
  return Array.from({
    length: 1 + Math.abs(end - start)
  }, (_, i) => end > start ? start + i : start - i);
}