在PHP中,您可以。。。

range(1, 3); // Array(1, 2, 3)
range("A", "C"); // Array("A", "B", "C")

也就是说,有一个函数可以通过传递上下限来获得一系列数字或字符。

JavaScript本机是否有内置的功能?如果没有,我将如何实施?


当前回答

方便的函数来完成这个技巧,请运行下面的代码片段

功能范围(开始、结束、步长、偏移){var len=(数学.abs(结束-开始)+((偏移量||0)*2))/(步骤||1)+1;var方向=开始<结束?1 : -1;var startingPoint=开始-(方向*(偏移||0));var stepSize=方向*(步骤||1);return Array(len).fill(0).map(函数(_,索引){return startingPoint+(stepSize*索引);});}console.log('范围(1,5)=>'+范围(1、5));console.log('范围(5,1)=>'+范围(5、1));console.log('范围(5,5)=>'+范围(5、5));console.log('范围(-5,5)=>'+范围(-5、5));console.log('范围(-10,5,5)=>'+范围(-10、5,5));console.log('范围(1,5,1,2)=>'+范围(1、5、1,2;

下面是如何使用它

范围(开始,结束,步长=1,偏移=0);

包括-前进档(5,10)//[5,6,7,8,9,10]包括-向后范围(10,5)//[10,9,8,7,6,5]后退范围(10,2,2)//[10,8,6,4,2]排他-前进范围(5,10,0,-1)//[6,7,8,9]而不是5,10本身偏移-扩展范围(5,10,0,1)//[4,5,6,7,8,9,10,11]偏移-收缩范围(5,10,0,-2)//[7,8]步进-扩展范围(10,0,2,2)//[12,10,8,6,4,2,0,-2]

希望你觉得它有用。


这就是它的工作原理。

基本上,我首先计算得到的数组的长度,并创建一个长度为零的填充数组,然后用所需的值填充

(step | |1)=>其他类似的方法使用step的值,如果没有提供,则使用1我们首先使用(Math.abs(end-start)+((offset | |0)*2))/(step | |1)+1)计算结果数组的长度,以使其更简单(两个方向上的差值*offset/step)获得长度后,我们使用newArray(length).fill(0)创建一个带有初始化值的空数组;在此处检查现在我们有一个数组[0,0,0,..],其长度是我们想要的。我们对其进行映射,并使用array.map(function(){})返回一个具有所需值的新数组var方向=开始<结束?1 : 0; 显然,如果起点不小于终点,我们就需要后退。我的意思是从0到5,反之亦然在每次迭代中,startingPoint+stepSize*索引将为我们提供所需的值

其他回答

这个也反过来。

const range = ( a , b ) => Array.from( new Array( b > a ? b - a : a - b ), ( x, i ) => b > a ? i + a : a - i );

range( -3, 2 ); // [ -3, -2, -1, 0, 1 ]
range( 1, -4 ); // [ 1, 0, -1, -2, -3 ]

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

带增量/步长

smallest and one-liner
[...Array(N)].map((_, i) => from + i * step);

示例和其他备选方案

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

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

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

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

https://stackoverflow.com/a/64599169/8784402

用一行代码生成字符列表

constcharList=(a,z,d=1)=>(a=a.charCodeAt(),z=z.charCodeAt(),[…数组(Math.floor((z-a)/d)+1)].map((_,i)=>String.fromCharCode(a+i*d)));console.log(“从A到G”,charList('A','G'));console.log(“从A到Z,步长/增量为2”,charList('A','Z',2));console.log(“从Z到P的反向顺序”,charList('Z','P',-1));console.log(“从0到5”,charList(“0”,“5”,1));console.log(“从9到5”,charList('9','5',-1));console.log(“从0到8,步骤2”,charList('0','8',2));console.log(“从α到ω”,charList(“α”,“ω”));console.log(“印地语字符来自क 到ह“,charList('क', 'ह'));console.log(“从А到Е的俄语字符”,charList(“А”,“Е”));

For TypeScript
const charList = (p: string, q: string, d = 1) => {
  const a = p.charCodeAt(0),
    z = q.charCodeAt(0);
  return [...Array(Math.floor((z - a) / d) + 1)].map((_, i) =>
    String.fromCharCode(a + i * d)
  );
};

这里有一个npm模块bereich(“bereich”是德语中“范围”的意思)。它利用了现代JavaScript的迭代器,因此您可以以各种方式使用它,例如:

console.log(...bereich(1, 10));
// => 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

const numbers = Array.from(bereich(1, 10));
// => [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]

for (const number of bereich(1, 10)) {
  // ...
}

它还支持递减范围(通过简单地交换最小值和最大值),并且还支持除1以外的步骤。

免责声明:我是本模块的作者,所以请对我的答案持保留态度。

使用范围([start,]stop[,step])签名完成ES6实现:

function range(start, stop, step=1){
  if(!stop){stop=start;start=0;}
  return Array.from(new Array(int((stop-start)/step)), (x,i) => start+ i*step)
}

如果要自动负步进,请添加

if(stop<start)step=-Math.abs(step)

或者更简单地说:

range=(b, e, step=1)=>{
  if(!e){e=b;b=0}
  return Array.from(new Array(int((e-b)/step)), (_,i) => b<e? b+i*step : b-i*step)
}

如果你有巨大的射程,看看保罗·莫雷蒂的发电机方法

尚未实施!

使用新的Number.range建议(第1阶段):

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