我正在寻找以下方法的任何替代方法,以创建包含1到N的JavaScript数组,其中N仅在运行时已知。
var foo = [];
for (var i = 1; i <= N; i++) {
foo.push(i);
}
对我来说,我觉得应该有一种不用循环的方法。
我正在寻找以下方法的任何替代方法,以创建包含1到N的JavaScript数组,其中N仅在运行时已知。
var foo = [];
for (var i = 1; i <= N; i++) {
foo.push(i);
}
对我来说,我觉得应该有一种不用循环的方法。
当前回答
以下是摘要(在控制台中运行):
// setup:
var n = 10000000;
function* rangeIter(a, b) {
for (let i = a; i <= b; ++i) yield i;
}
function range(n) {
let a = []
for (; n--; a[n] = n);
return a;
}
function sequence(max, step = 1) {
return {
[Symbol.iterator]: function* () {
for (let i = 1; i <= max; i += step) yield i
}
}
}
var t0, t1, arr;
// tests
t0 = performance.now();
arr = Array.from({ length: n }, (a, i) => 1)
t1 = performance.now();
console.log("Array.from({ length: n }, (a, i) => 1) Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = range(n);
t1 = performance.now();
console.log("range(n) Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = Array.from(rangeIter(0, n));
t1 = performance.now();
console.log("Array.from(rangeIter(0, n)) Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = [...rangeIter(0, n)];
t1 = performance.now();
console.log("[...rangeIter(0, n)] Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = Array.from(sequence(n));
t1 = performance.now();
console.log("Array.from(sequence(n)) Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = [...sequence(n)];
t1 = performance.now();
console.log("[...sequence(n)] Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = Array(n).fill(0).map(Number.call, Number);
t1 = performance.now();
console.log("Array(n).fill(0).map(Number.call, Number) Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = Array.from(Array(n).keys());
t1 = performance.now();
console.log("Array.from(Array(n).keys()) Took " + (t1 - t0) + " milliseconds.");
t0 = performance.now();
arr = [...Array(n).keys()];
t1 = performance.now();
console.log("[...Array(n).keys()] Took " + (t1 - t0) + " milliseconds.");
最快的是Array(n).fill(0).map(Number.call,Number),第二个是[…Array(n).keys()]
但是。。。rangeIter的方式非常方便(可以内联),速度快,功能更强大
其他回答
对于小范围,切片是不错的。N仅在运行时已知,因此:
[0, 1, 2, 3, 4, 5].slice(0, N+1)
在ES6中:
Array.from({length: 1000}, (_, i) => i).slice(1);
或者更好(没有额外的变量_,也没有额外的切片调用):
Array.from({length:1000}, Number.call, i => i + 1)
或者,如果您的列表少于256个结果,您可以使用Uint8Array来获得稍快的结果(或者,您可以根据列表的长度使用其他Uint列表,例如Uint16的最大值为65535,或Uint32的最大值4294967295等。不过,正式地说,这些类型的数组只是在ES6中添加的)。例如:
Uint8Array.from({length:10}, Number.call, i => i + 1)
ES5:
Array.apply(0, {length: 1000}).map(function(){return arguments[1]+1});
或者,在ES5中,对于map函数(类似于上面ES6中Array.from函数的第二个参数),可以使用Number.call
Array.apply(0,{length:1000}).map(Number.call,Number).slice(1)
或者,如果你在这里也反对.sslice,你可以执行上面的ES5等效操作(来自ES6),比如:
Array.apply(0,{length:1000}).map(Number.call, Function("i","return i+1"))
基于高票答案和高票评论。
const range = (from, to) => [...Array(to + 1).keys()].slice(from);
// usage
let test = [];
test = range(5, 10);
console.log(test); // output: [ 5, 6, 7, 8, 9, 10 ]
✅ 简单地说,这对我有用:
[...Array(5)].map(...)
function arrGen(n) {
var a = Array(n)
while (n--) a[n] = n
return a
}
// arrGen(10) => [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]