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?
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
函数范围(firstNum,lastNum){let rangeList=[];如果(firstNum>lastNum){return console.error(“第一个数字不能大于最后一个数字”);}让counter=firstNum;while(计数器<=lastNum){rangeList.push(计数器);计数器++;}返回范围列表;}
其他回答
根据我的理解:
JS的运行时环境不支持尾部调用优化。编写任何递归函数来生成一个大范围的函数都会给你带来麻烦。如果我们要处理大量数据,那么创建循环数组可能不是最好的做法。写入大循环会导致事件队列变慢。
function range(start, end, step = 1) {
const _range = _start => f => {
if (_start < end) {
f(_start);
setTimeout(() => _range(_start + step)(f), 0);
}
}
return {
map: _range(start),
};
}
range(0, 50000).map(console.log);
此函数不会引发上述问题。
我想补充一点,我认为这是一个非常可调的版本,速度非常快。
const range = (start, end) => {
let all = [];
if (typeof start === "string" && typeof end === "string") {
// Return the range of characters using utf-8 least to greatest
const s = start.charCodeAt(0);
const e = end.charCodeAt(0);
for (let i = s; i <= e; i++) {
all.push(String.fromCharCode(i));
}
} else if (typeof start === "number" && typeof end === "number") {
// Return the range of numbers from least to greatest
for(let i = end; i >= start; i--) {
all.push(i);
}
} else {
throw new Error("Did not supply matching types number or string.");
}
return all;
}
// usage
const aTod = range("a", "d");
如果您愿意,也可以使用打字机
const range = (start: string | number, end: string | number): string[] | number[] => {
const all: string[] | number[] = [];
if (typeof start === "string" && typeof end === "string") {
const s: number = start.charCodeAt(0);
const e: number = end.charCodeAt(0);
for (let i = s; i <= e; i++) {
all.push(String.fromCharCode(i));
}
} else if (typeof start === "number" && typeof end === "number") {
for (let i = end; i >= start; i--) {
all.push(i);
}
} else {
throw new Error("Did not supply matching types number or string.");
}
return all;
}
// Usage
const negTenToten: number[] = range(-10, 10) as number[];
受到其他答案的影响。用户已离开。
您可以使用以下一行代码使事情简短明了
var启动=4;var端=20;console.log(数组(end-start+1).fill(start).map((x,y)=>x+y));
/**
* @param {!number|[!number,!number]} sizeOrRange Can be the `size` of the range (1st signature) or a
* `[from, to]`-shape array (2nd signature) that represents a pair of the *starting point (inclusive)* and the
* *ending point (exclusive)* of the range (*mathematically, a left-closed/right-open interval: `[from, to)`*).
* @param {!number} [fromOrStep] 1st signature: `[from=0]`. 2nd signature: `[step=1]`
* @param {!number} [stepOrNothing] 1st signature: `[step=1]`. 2nd signature: NOT-BEING-USED
* @example
* range(5) ==> [0, 1, 2, 3, 4] // size: 5
* range(4, 5) ==> [5, 6, 7, 8] // size: 4, starting from: 5
* range(4, 5, 2) ==> [5, 7, 9, 11] // size: 4, starting from: 5, step: 2
* range([2, 5]) ==> [2, 3, 4] // [2, 5) // from: 2 (inclusive), to: 5 (exclusive)
* range([1, 6], 2) ==> [1, 3, 5] // from: 1, to: 6, step: 2
* range([1, 7], 2) ==> [1, 3, 5] // from: 1, to: 7 (exclusive), step: 2
* @see {@link https://stackoverflow.com/a/72388871/5318303}
*/
export function range (sizeOrRange, fromOrStep, stepOrNothing) {
let from, to, step, size
if (sizeOrRange instanceof Array) { // 2nd signature: `range([from, to], step)`
[from, to] = sizeOrRange
step = fromOrStep ?? 1
size = Math.ceil((to - from) / step)
} else { // 1st signature: `range(size, from, step)`
size = sizeOrRange
from = fromOrStep ?? 0
step = stepOrNothing ?? 1
}
return Array.from({length: size}, (_, i) => from + i * step)
}
示例:
控制台日志(范围(5),//[0,1,2,3,4]//size:5范围([2,5]),//[2,3,4]//[2、5)//从:2(含)到:5(不含)范围(4,2),//[2,3,4,5]//大小:4,从:2开始范围([1,6],2),//[1,3,5]//从:1到:6,步骤:2范围([1,7],2),//[1,3,5]//从:1到:7(不含),步骤:2)<脚本>函数范围(sizeOrRange、fromOrStep、stepOrNothing){让从、到、步长、大小if(sizeOrRange instanceof Array){//第二个签名:`range([from,to],step)`[from,to]=sizeOrRange步骤=来自或步骤??1.size=数学ceil((to-from)/步长)}else{//第一个签名:`range(大小,从,步)`size=sizeOrRangefrom=来自或步骤??0step=stepOrNothing??1.}return Array.from({length:size},(_,i)=>from+i*step)}</script>
我的实施
export function stringRange(a: string, b: string) {
let arr = [a + ''];
const startPrefix = a.match(/([\D])+/g);
const endPrefix = b.match(/([\D])+/g);
if ((startPrefix || endPrefix) && (Array.isArray(startPrefix) && startPrefix[0]) !== (Array.isArray(endPrefix) && endPrefix[0])) {
throw new Error('Series number does not match');
}
const startNum = a.match(/([\d])+/g);
const endNum = b.match(/([\d])+/g);
if (!startNum || !endNum) {
throw new Error('Range is not valid');
}
let start = parseInt(startNum[0], 10);
let end = parseInt(endNum[0], 10);
if (start > end) {
throw new Error('Ending value should be lessesr that starting value');
}
while (start !== end) {
start++;
arr.push(startPrefix ? startPrefix[0] + (start + '').padStart(startNum[0].length, '0') : start + '');
}
return arr;
}
样本结果
// console.log(range('0', '10'));
// console.log(range('10', '10'));
// console.log(range('10', '20'));
// console.log(range('10', '20000'));
// console.log(range('ABC10', 'ABC23'));
// console.log(range('ABC10', 'ABC2300'));
// console.log(range('ABC10', 'ABC09')); --> Failure case
// console.log(range('10', 'ABC23')); --> Failure case
// console.log(range('ABC10', '23')); --> Failure case