// range()              0..10, step=1
// range(max)           0..max, step=1
// range(min,max)       min..max, step=1
// range(min,step,max)  min..max, step=step
// Use:
// console.log(...range(3));
// Array.from(range(5))
// [...range(100)]
// for (const v of range(1,10)) { ... 

function* range(...args) {
    let [min, step, max] = {
        0: [0, 1, 10],
        1: [0, args[0] >= 0 ? 1 : -1, args[0]],
        2: [args[0], args[1] >= args[0] ? 1 : -1, args[1]],
        3: args,
    }[args.length] || [];
    if (min === undefined) throw new SyntaxError("Too many arguments");
    let x = min;
    while (step >= 0 ? x < max : x > max) {
        yield x;
        x += step
    }
}
console.log(...range());      // 0 1 2 3 4 5 6 7 8 9
console.log(...range(3));     // 0 1 2
console.log(...range(2, 5));  // 2 3 4
console.log(...range(5, 2));  // 5 4 3
console.log(...range(3, -3)); // 3 2 1 0 -1 -2
console.log(...range(-3, 3)); // -3 -2 -1 0 1 2
console.log(...range(-5, -2));// -5 -4 -3
console.log(...range(-2, -5));// -2 -3 -4

这里是一个范围函数的定义，它的行为与Python的范围类型完全相同，只是这个函数不是懒惰的。把它变成发电机应该很容易。

范围构造函数的参数必须是数字。如果省略step参数，则默认为1。如果省略了start参数，则默认为0。如果步骤为零，则会引发错误。

range = (start, stop, step=1) => {
    if(step === 0) throw new Error("range() arg 3 must not be zero");

    const noStart = stop == null;
    stop = noStart ? start : stop;
    start = noStart ? 0 : start;
    const length = Math.ceil(((stop - start) / step));

    return Array.from({length}, (_, i) => (i * step) + start);
}

console.log(range(-10, 10, 2));
//output [Array] [-10,-8,-6,-4,-2,0,2,4,6,8]
console.log(range(10));
// [Array] [0,1,2,3,4,5,6,7,8,9]
console.log(3, 12);
// [Array] [3,4,5,6,7,8,9,10,11]

ES6

使用Array.from（此处为文档）：

const range = (start, stop, step) => Array.from({ length: (stop - start) / step + 1}, (_, i) => start + (i * step));

如果我们输入类似[4,2]的内容，我们将得到[2,3,4]作为输出，我们可以使用它。

function createRange(array) {
  var range = [];
  var highest = array.reduce(function(a, b) {
    return Math.max(a, b);
  });
  var lowest = array.reduce(function(a, b) {
    return Math.min(a, b);
  });
  for (var i = lowest; i <= highest; i++) {
    range.push(i);
  }
  return range;
}

它适用于字符和数字，通过可选步骤向前或向后移动。

var range = function(start, end, step) {
    var range = [];
    var typeofStart = typeof start;
    var typeofEnd = typeof end;

    if (step === 0) {
        throw TypeError("Step cannot be zero.");
    }

    if (typeofStart == "undefined" || typeofEnd == "undefined") {
        throw TypeError("Must pass start and end arguments.");
    } else if (typeofStart != typeofEnd) {
        throw TypeError("Start and end arguments must be of same type.");
    }

    typeof step == "undefined" && (step = 1);

    if (end < start) {
        step = -step;
    }

    if (typeofStart == "number") {

        while (step > 0 ? end >= start : end <= start) {
            range.push(start);
            start += step;
        }

    } else if (typeofStart == "string") {

        if (start.length != 1 || end.length != 1) {
            throw TypeError("Only strings with one character are supported.");
        }

        start = start.charCodeAt(0);
        end = end.charCodeAt(0);

        while (step > 0 ? end >= start : end <= start) {
            range.push(String.fromCharCode(start));
            start += step;
        }

    } else {
        throw TypeError("Only string and number types are supported");
    }

    return range;

}

jsFiddle。

如果扩充本机类型是您的事情，那么将其分配给Array.range。

var范围=函数（开始、结束、步骤）{var范围=[]；var typeofStart=启动类型；var typeofEnd=结束类型；如果（步骤==0）{throw TypeError（“步长不能为零。”）；}if（类型开始==“undefined”| |类型结束==“未定义”）{throw TypeError（“必须传递开始和结束参数。”）；}否则如果（typeofStart！=typeofEnd）{throw TypeError（“开始和结束参数必须是相同的类型。”）；}步骤类型==“未定义”&&（步骤=1）；if（结束<开始）{step=-步骤；}if（开始类型==“number”）{while（步骤>0？结束>=开始：结束<=开始）{范围.推（启动）；开始+=步骤；}}否则if（typeofStart==“string”）{如果（start.length！=1 | | end.length；=1）{throw TypeError（“仅支持带有一个字符的字符串。”）；}start=start.charCodeAt（0）；end=end.charCodeAt（0）；while（步骤>0？结束>=开始：结束<=开始）{range.push（String.fromCharCode（开始））；开始+=步骤；}}其他{throw TypeError（“仅支持字符串和数字类型”）；}返回范围；}console.log（范围（“A”，“Z”，1））；console.log（范围（“Z”，“A”，1））；console.log（范围（“A”，“Z”，3））；console.log（范围（0，25，1））；console.log（范围（0，25，5））；console.log（范围（20，5，5））；

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
  }
}