Javascript Array.map 1 个元素到多个元素
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Array.map 1 element to multiple element
提问by Boyang
I have [3, 16, 120]. when I do [3, 16, 120].map(mapper), I want to achieve, for example [4,5, 17,18, 121,122]i.e. each element map to n+1 and n+2. This is of course an example - what I want is to simply push multiple values from mapper function
我有[3, 16, 120]。当我这样做时[3, 16, 120].map(mapper)，我想实现，例如，[4,5, 17,18, 121,122]即每个元素映射到 n+1 和 n+2。这当然是一个例子 - 我想要的是简单地从映射器函数中推送多个值
Do I have to use Array.each and push to an array, or is it possible to do it with Array.map (or other built-in api)
我是否必须使用 Array.each 并推送到数组，或者是否可以使用 Array.map （或其他内置 api）
回答by Nenad Vracar
You can use reduce()and add to array e+1, e+2of each element.
您可以使用reduce()并添加到e+1, e+2每个元素的数组中。
var ar = [3, 16, 120];
var result = ar.reduce(function(r, e) {
r.push(e+1, e+2);
return r;
}, []);
console.log(result)This is ES6 version with arrow function
这是带有箭头功能的 ES6 版本
var ar = [3, 16, 120];
var result = ar.reduce((r, e) => r.push(e+1, e+2) && r, []);
console.log(result)PS: Array.push seems to be faster and has no Maximum call stack..error, see below:
PS：Array.push 好像更快，没有Maximum call stack..错误，见下图：
a = Array(1000000).fill(1); st = Date.now(); Array.prototype.concat.apply([], a.map(function (n) { return [n+1, n+2]; })); console.log(`${Date.now() - st}ms `);
> RangeError: Maximum call stack size exceeded
a = Array(1000000).fill(1); st = Date.now(); a.reduce((r, e) => r.push(e+1, e+2) && r, []); console.log(`${Date.now() - st}ms `);
> 180ms
So .push is preferable comparing to accepted solution.
因此，与公认的解决方案相比，.push 更可取。
回答by Boyang
I come up with one myself, using spread operator.
我自己想出了一个，使用扩展运算符。
[].concat(...[3, 16, 120].map(x => [x+1, x+2]))
[].concat(...[3, 16, 120].map(x => [x+1, x+2]))
回答by melpomene
Not particularly nice, but it is a possible solution:
不是特别好，但这是一个可能的解决方案：
var arr = [3, 16, 120];
console.log([].concat.apply([], arr.map(function (n) { return [n+1, n+2]; })));回答by Olivier Boissé
you could produce an array for each items, then concat all these arrays :
您可以为每个项目生成一个数组，然后连接所有这些数组：
[3, 16, 120].map(x => [x+1, x+2] ).reduce( (acc,val) => acc.concat(val), []);
回答by Nina Scholz
You could use Array#reducein combination with Array#concat.
console.log([3, 16, 120].reduce(function (r, a) {
return r.concat(a + 1, a + 2);
}, []));ES6
ES6
console.log([3, 16, 120].reduce((r, a) => r.concat(a + 1, a + 2), []));回答by Freewalker
Immutable solution, with the spread operator:
不可变解决方案，使用扩展运算符：
[3, 16, 120].reduce((a, v) => [...a, v+1, v+2], [])
回答by jabacchetta
2019 Update
2019年更新
Use Array.prototype.flatMap(), introduced in ES10.
使用Array.prototype.flatMap()，在 ES10 中引入。
const oddNumbers = [1, 3, 5, 7, 9];
const allNumbers = oddNumbers.flatMap((number) => [number, number + 1]);
console.log(allNumbers); // [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
回答by eltonkamami
using Array#concatand Array#map
使用Array#concat和Array#map
Array.prototype.concat.apply([], [3, 16, 120].map(x => [x+1, x+2] ));
回答by Me.Name
Just for fun, an ES6 solution with a generator:
只是为了好玩，一个带有生成器的 ES6 解决方案：
var arr = [3, 16, 120];
var [...result] = (function*() { for( i of arr){ yield ++i; yield ++i; }})();
console.log(result);回答by HynekS
Using Array.prototype.flat():
使用 Array.prototype.flat()：
const doubled = [3, 16, 120].map(item => [item + 1, item + 2]).flat();
console.log(doubled)Fair warning – not a standard method to this date (posted 12/2018).
公平警告 – 到目前为止还不是标准方法（发布于 12/2018）。
