blocks|key|2088665|text|我想出了一些类似于tymeJV的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2088666|[2,+3,+1,+4].reduce(function+(res,+current,+index,+array)+{
++++return+res.concat([current,+current]);
},+[]);|code-block|syntax|javascript|2088667|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/1686485/tymejv^0|9|6|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

I came up with something similar to <a href="https://stackoverflow.com/users/1686485/tymejv">tymeJV</a>'s answer

<pre><code>[2, 3, 1, 4].reduce(function (res, current, index, array) {
 return res.concat([current, current]);
}, []);
</code></pre>

blocks|key|2088669|text|基本上：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2088670|a+=+[2,+3,+1,+4];
b=[];

for(var+i+=+0;+i<+a.length;%2B%2Bi){
++b.push(a[i]);
++b.push(a[i]);
}

a=b;|code-block|syntax|javascript|2088671|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Basically:

<pre><code>a = [2, 3, 1, 4];
b=[];

for(var i = 0; i&lt; a.length;++i){
 b.push(a[i]);
 b.push(a[i]);
}

a=b;
</code></pre>

blocks|key|2088750|text|基本上，您可以在ES19中使用flatMap|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2088751|a+=+[1,+2,+3,+4];
a.flatMap(i+=>+[i,i]);+//+[1,+1,+2,+2,+3,+3,+4,+4]|code-block|syntax|javascript|2088752|您还可以像这样自定义重复次数：|2088753|a+=+[1,+2,+3,+4];
const+dublicateItems+=+(arr,+numberOfRepetitions)+=>+
++++arr.flatMap(i+=>+Array.from({+length:+numberOfRepetitions+}).fill(i));

dublicateItems(a,+3);|2088754|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/flatMap^0|F|7|F|7|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

Basically you can use <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/flatMap" rel="noreferrer"><code>flatMap</code></a> in ES19 

<pre class="lang-js prettyprint-override"><code>a = [1, 2, 3, 4];
a.flatMap(i =&gt; [i,i]); // [1, 1, 2, 2, 3, 3, 4, 4]
</code></pre>

Also you can customize the repetitions number like this:

<pre class="lang-js prettyprint-override"><code>a = [1, 2, 3, 4];
const dublicateItems = (arr, numberOfRepetitions) =&gt; 
 arr.flatMap(i =&gt; Array.from({ length: numberOfRepetitions }).fill(i));

dublicateItems(a, 3);
</code></pre>

blocks|key|1874527|text|我想你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1874528|var+duplicated+=+a.map(function(item)+{
++++return+[item,+item];
}).reduce(function(a,+b)+{+return+a.concat(b)+});

//duplicated:+[2,+2,+3,+3,+1,+1,+4,+4]|code-block|syntax|javascript|1874529|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I suppose you could do:

<pre><code>var duplicated = a.map(function(item) {
 return [item, item];
}).reduce(function(a, b) { return a.concat(b) });

//duplicated: [2, 2, 3, 3, 1, 1, 4, 4]
</code></pre>

blocks|key|1874569|text|我在我的应用程序中遇到了这个问题，所以我创建了这个函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1874570|function+duplicateElements(array,+times)+{
++return+array.reduce((res,+current)+=>+{
++++++return+res.concat(Array(times).fill(current));
++},+[]);
}|code-block|syntax|javascript|1874571|要使用它，只需传递数组和希望复制元素的次数：|1874572|duplicateElements([2,+3,+1,+4],+2);
//+returns:+[2,+2,+3,+3,+1,+1,+4,+4]|1874573|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I came across this issue in my application, so I created this function:

<pre><code>function duplicateElements(array, times) {
 return array.reduce((res, current) =&gt; {
 return res.concat(Array(times).fill(current));
 }, []);
}
</code></pre>

To use it, simple pass the array, and the number of times you want the elements to be duplicated:

<pre><code>duplicateElements([2, 3, 1, 4], 2);
// returns: [2, 2, 3, 3, 1, 1, 4, 4]
</code></pre>

blocks|key|1874554|text|只要拼接一下就行了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1874555|​|1874556|var+a+=+[2,+3,+1,+4],
++++i+=+a.length;
while+(i--)+{
++++a.splice(i,+0,+a[i]);
}
document.write('<pre>'+%2B+JSON.stringify(a,+0,+4)+%2B+'</pre>');|code-block|syntax|javascript|1874557|1874558|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

Just splice a little bit.

<div class="snippet" data-lang="js" data-hide="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var a = [2, 3, 1, 4],
 i = a.length;
while (i--) {
 a.splice(i, 0, a[i]);
}
document.write('&lt;pre&gt;' + JSON.stringify(a, 0, 4) + '&lt;/pre&gt;');</code></pre>
</div>
</div>

blocks|key|2088697|text|这些函数可能有助于查看.sort()、.concat()|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2088698|function+duplicate(arr)+{
++++return+arr.concat(arr).sort()
}+
console.log(duplicate([1,2,3,4,5]))|code-block|syntax|javascript|2088699|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

These functions may help see .sort(), .concat()

<pre><code>function duplicate(arr) {
 return arr.concat(arr).sort()
} 
console.log(duplicate([1,2,3,4,5]))
</code></pre>

blocks|key|1874607|text|ES6生活方式(基于axelduch's+answer)|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1874608|​|1874609|const+arr+=+[2,+3,+1,+4].reduce((res,+current)+=>+[...res,+current,+current],+[]);

console.log(arr);|code-block|syntax|javascript|1874610|1874611|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/2389720/axelduch^0|A|A|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|D|3|E|5|6|7|X|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|Y|8|@]|9|@]|C|$I|J]]|$1|K|3|E|5|6|7|Z|8|@]|9|@]|C|$]]|$1|L|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|M|$N|$5|O|P|Q|C|$R|S]]]]

ES6 way of life (based on <a href="https://stackoverflow.com/users/2389720/axelduch">axelduch's</a> answer)

<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const arr = [2, 3, 1, 4].reduce((res, current) =&gt; [...res, current, current], []);

console.log(arr);</code></pre>
</div>
</div>

blocks|key|1874613|text|这个怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1874614|for+(i=a.length-1;i>=0;i--)a.splice(i,0,a[i]);|code-block|syntax|javascript|1874615|向后迭代被低估了，在这种情况下，它保持了索引的完整性;)|1874616|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

how 'bout that?

<pre><code>for (i=a.length-1;i&gt;=0;i--)a.splice(i,0,a[i]);
</code></pre>

iterating backwards is undervalued, in this case it keeps the index intact ;)

blocks|key|2088711|text|0/2++=++0++++=++0++%7C0++=++0
1/2++=++0.5++=++0.5%7C0++=++0
2/2++=++1++++=++1++%7C0++=++1
3/2++=++1.5++=++1.5%7C0++=++1
4/2++=++2++++=++2++%7C0++=++2
5/2++=++2.5++=++2.5%7C0++=++2
6/2++=++3++++=++3++%7C0++=++3
7/2++=++3.5++=++3.5%7C0++=++3|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2088712|将%7C0视为Math.floor|unstyled|offset|length|style|CODE|2088713|2088714|在代码中，这可能如下所示：|2088715|for+(let+i+=+0;+i+<+a.length+*+2;+i%2B%2B)+{
++a[i]+=+a[i+/+2+%7C 0]
}|2088716|因为不变性更可取，所以可以这样做：|2088717|function+repeatItems(a,+n)+{
++const+b+=+new+Array(a.length+*+n)
++for+(let+i+=+0;+i+<+b.length;+i%2B%2B)+{
++++b[i]+=+a[i+/+n+%7C+0]
++}
++return+b
}|2088718|不可读的ES6意大利面条代码：|2088719|const+repeatItems+=+(a,+n)+=>+Array.from(Array(a.length+*+n),+(_,+i)+=>+a[i+/+n+%7C+0])|2088720|entityMap^0|0|1|2|5|A|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|10|8|@$G|11|H|12|I|J]|$G|13|H|14|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|15|8|@]|9|@]|A|$]]|$1|L|3|M|5|F|7|16|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|17|8|@]|9|@]|A|$B|C]]|$1|P|3|Q|5|F|7|18|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|19|8|@]|9|@]|A|$B|C]]|$1|T|3|U|5|F|7|1A|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1B|8|@]|9|@]|A|$B|C]]|$1|X|3|-4|5|F|7|1C|8|@]|9|@]|A|$]]]|Y|$]]

<pre><code>0/2 = 0 = 0 |0 = 0
1/2 = 0.5 = 0.5|0 = 0
2/2 = 1 = 1 |0 = 1
3/2 = 1.5 = 1.5|0 = 1
4/2 = 2 = 2 |0 = 2
5/2 = 2.5 = 2.5|0 = 2
6/2 = 3 = 3 |0 = 3
7/2 = 3.5 = 3.5|0 = 3
</code></pre>

Treat <code>|0</code> as <code>Math.floor</code>

<hr>

In code this could look like this:

<pre><code>for (let i = 0; i &lt; a.length * 2; i++) {
 a[i] = a[i / 2 | 0]
}
</code></pre>

Because immutability is preferable, one could do something like this:

<pre><code>function repeatItems(a, n) {
 const b = new Array(a.length * n)
 for (let i = 0; i &lt; b.length; i++) {
 b[i] = a[i / n | 0]
 }
 return b
}
</code></pre>

Unreadable ES6 spaghetti code:

<pre><code>const repeatItems = (a, n) =&gt; Array.from(Array(a.length * n), (_, i) =&gt; a[i / n | 0])
</code></pre>

blocks|key|1874618|text|有很多方法可以将项添加到数组中，如上图所示。我已经比较过了，您可以在控制台中自己查看性能。无论计时是什么，都在两个"console.time“之间进行。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1874619|console.clear();

function+loopMyArray(){
+var+loopArray+=+new+Array(10000);
+for(var+i+=+0;+i+<+loopArray.length;+i%2B%2B){
+++loopArray[i]+=+i%2B1;
+}
+console.log(loopArray);
}
console.time('loopMyArray');
loopMyArray();
console.timeEnd('loopMyArray');

function+fillArray(){
+let+x+=+0;
+let+filledArray+=+new+Array(10000).fill(null).map(()=>+%2B%2Bx);
+console.log(filledArray);
}

console.time('fillArray');
fillArray();
console.timeEnd('fillArray');

function+keyMyArray(){
+let+fromKeyArray+=+Array.from(Array(10000).keys());
+console.log(fromKeyArray);
}
console.time('keyMyArray');
keyMyArray();
console.timeEnd('keyMyArray');

function+spreadKeysArray(){
+let+spreadArray+=+[...Array(10000).keys()];
+console.log(spreadArray);
}
console.time('spreadKeysArray');
spreadKeysArray();
console.timeEnd('spreadKeysArray');

console.log('+Start+from+1');

function+mapKeyArray(){
+//let+mapArray+=+([...Array(1000).keys()].map(x+=>+x%2B%2B));+//increment+after+return
+let+mapArray+=+[...Array(10000).keys()].map(x+=>+%2B%2Bx);
+console.log(mapArray);
}

console.time('mapKeyArray');
mapKeyArray();
console.timeEnd('mapKeyArray');

function+sliceKeyArray(){
+let+sliceFirstElementArray+=+[...Array(10000%2B1).keys()].slice(1);
+console.log(sliceFirstElementArray);
}
console.time('sliceKeyArray');
sliceKeyArray();
console.timeEnd('sliceKeyArray');

function+calcFromLength(){
+let+fromLengthArray+=+Array.from({length:+10000},+(v,+k)+=>+k%2B1);
+console.log(fromLengthArray);
}
console.time('calcFromLength');
calcFromLength();
console.timeEnd('calcFromLength');

console.log('========+add+a+double+for+every+item+========');

function+loopDoubleArray(){
+var+first5000Array+=+[...Array(5000%2B1).keys()].slice(1);
+var+double5000Array+=[];

+for(var+i+=+0;+i<+first500Array.length;%2B%2Bi){
+++double5000Array.push(first5000Array[i]);
+++double5000Array.push(first5000Array[i]);
+}
+console.log(double5000Array);
}
console.time('loopDoubleArray');
loopDoubleArray();+//+Whatever+is+timed+goes+between+the+two+"console.time"
console.timeEnd('loopDoubleArray');

function+mapDoubleArray(){
+//+adding+1,1,2,2,3,3+etc
+let+doubleArray+=+[...Array(10000).keys()].map(x+=>+Math.floor(%2B%2Bx/2)+%2B+x%252);
+console.log(doubleArray);
}
console.time('mapDoubleArray');
mapDoubleArray();
console.timeEnd('mapDoubleArray');

function+fromDoubleArray(){
+let+fromDoubleArray+=+Array.from({length:+10000},+(v,+x)+=>+Math.floor(%2B%2Bx/2)+%2B+x%252);
+console.log(fromDoubleArray);
}
console.time('fromDoubleArray');
fromDoubleArray();+//+Whatever+is+timed+goes+between+the+two+"console.time"
console.timeEnd('fromDoubleArray');

function+doubleSpreadArray(){
+let+keyArray+=+[...Array(500%2B1).keys()].slice(1);
+let+doubleSpreadArray+=+[...keyArray,...keyArray];
+console.log(doubleSpreadArray);
}
console.time('doubleSpreadArray');
doubleSpreadArray();+//+Whatever+is+timed+goes+between+the+two+"console.time"
console.timeEnd('doubleSpreadArray');

function+reduceDoubleArray(){
+let+reduceDoubleArray+=+Array.from({length:+5000},+(v,+k)+=>+k%2B1).reduce((m,i)+=>+m.concat([i,i]),+[]);
+console.log(reduceDoubleArray);
}
console.time('reduceDoubleArray');
reduceDoubleArray();+//+Whatever+is+timed+goes+between+the+two+"console.time"
console.timeEnd('reduceDoubleArray');|code-block|syntax|javascript|1874620|我比较了一些速度，看起来似乎没有太多差异。map、slice和from之间的计算速度(对于10000个项目，以毫秒为单位)差别不大(切片1似乎更好一点)。在双项目数组的计算中使用reduce比使用其他方法要慢得多|offset|length|style|CODE|1874621|%7C+mapMyArray+++++%7C+5,342041016+%7C+5,21484375++%7C+8,424804688+%7C+5,516113281+%7C
%7C+sliceKeyArray++%7C+5,221191406+%7C+4,854248047+%7C+6,069091797+%7C+4,940185547+%7C
%7C+calcFromLength+%7C+6,156005859+%7C+5,988037109+%7C+6,031982422+%7C+6,739990234+%7C|1874622|entityMap^0|0|0|L|3|P|5|V|4|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

Lot of ways to add items to an array as seen above. I have compared it and you can view the performance by yourself in the console.
Whatever is timed goes between the two "console.time"

<pre><code>console.clear();

function loopMyArray(){
 var loopArray = new Array(10000);
 for(var i = 0; i &lt; loopArray.length; i++){
 loopArray[i] = i+1;
 }
 console.log(loopArray);
}
console.time('loopMyArray');
loopMyArray();
console.timeEnd('loopMyArray');

function fillArray(){
 let x = 0;
 let filledArray = new Array(10000).fill(null).map(()=&gt; ++x);
 console.log(filledArray);
}

console.time('fillArray');
fillArray();
console.timeEnd('fillArray');

function keyMyArray(){
 let fromKeyArray = Array.from(Array(10000).keys());
 console.log(fromKeyArray);
}
console.time('keyMyArray');
keyMyArray();
console.timeEnd('keyMyArray');

function spreadKeysArray(){
 let spreadArray = [...Array(10000).keys()];
 console.log(spreadArray);
}
console.time('spreadKeysArray');
spreadKeysArray();
console.timeEnd('spreadKeysArray');

console.log(' Start from 1');

function mapKeyArray(){
 //let mapArray = ([...Array(1000).keys()].map(x =&gt; x++)); //increment after return
 let mapArray = [...Array(10000).keys()].map(x =&gt; ++x);
 console.log(mapArray);
}

console.time('mapKeyArray');
mapKeyArray();
console.timeEnd('mapKeyArray');

function sliceKeyArray(){
 let sliceFirstElementArray = [...Array(10000+1).keys()].slice(1);
 console.log(sliceFirstElementArray);
}
console.time('sliceKeyArray');
sliceKeyArray();
console.timeEnd('sliceKeyArray');

function calcFromLength(){
 let fromLengthArray = Array.from({length: 10000}, (v, k) =&gt; k+1);
 console.log(fromLengthArray);
}
console.time('calcFromLength');
calcFromLength();
console.timeEnd('calcFromLength');

console.log('======== add a double for every item ========');

function loopDoubleArray(){
 var first5000Array = [...Array(5000+1).keys()].slice(1);
 var double5000Array =[];

 for(var i = 0; i&lt; first500Array.length;++i){
 double5000Array.push(first5000Array[i]);
 double5000Array.push(first5000Array[i]);
 }
 console.log(double5000Array);
}
console.time('loopDoubleArray');
loopDoubleArray(); // Whatever is timed goes between the two "console.time"
console.timeEnd('loopDoubleArray');

function mapDoubleArray(){
 // adding 1,1,2,2,3,3 etc
 let doubleArray = [...Array(10000).keys()].map(x =&gt; Math.floor(++x/2) + x%2);
 console.log(doubleArray);
}
console.time('mapDoubleArray');
mapDoubleArray();
console.timeEnd('mapDoubleArray');

function fromDoubleArray(){
 let fromDoubleArray = Array.from({length: 10000}, (v, x) =&gt; Math.floor(++x/2) + x%2);
 console.log(fromDoubleArray);
}
console.time('fromDoubleArray');
fromDoubleArray(); // Whatever is timed goes between the two "console.time"
console.timeEnd('fromDoubleArray');

function doubleSpreadArray(){
 let keyArray = [...Array(500+1).keys()].slice(1);
 let doubleSpreadArray = [...keyArray,...keyArray];
 console.log(doubleSpreadArray);
}
console.time('doubleSpreadArray');
doubleSpreadArray(); // Whatever is timed goes between the two "console.time"
console.timeEnd('doubleSpreadArray');

function reduceDoubleArray(){
 let reduceDoubleArray = Array.from({length: 5000}, (v, k) =&gt; k+1).reduce((m,i) =&gt; m.concat([i,i]), []);
 console.log(reduceDoubleArray);
}
console.time('reduceDoubleArray');
reduceDoubleArray(); // Whatever is timed goes between the two "console.time"
console.timeEnd('reduceDoubleArray');

</code></pre>

I compared some speeds and it looks as if there are not so many differences. The speed(in ms for 10000 items) of calculation between <code>map</code>, <code>slice</code> and <code>from</code> are not so different (slice one seems a bit better). Using reduce in the calculation for double items array is way more slower then the rest of methods

<pre><code>| mapMyArray | 5,342041016 | 5,21484375 | 8,424804688 | 5,516113281 |
| sliceKeyArray | 5,221191406 | 4,854248047 | 6,069091797 | 4,940185547 |
| calcFromLength | 6,156005859 | 5,988037109 | 6,031982422 | 6,739990234 |
</code></pre>

Whats the easiest way (with "native" javascript) to duplicate every element in a javascript array?

The order matters.

For example:

<pre><code>a = [2, 3, 1, 4]
// do something with a
a
// a is now [2, 2, 3, 3, 1, 1, 4, 4]
</code></pre>

How to duplicate elements in a js array?

复制javascript数组中的每个元素的最简单方法(使用“原生”javascript)是什么？顺序很重要。例如：a = [2, 3, 1, 4]// do something with aa// a is now [2, 2, 3, 3, 1, 1, 4, 4]

问如何在js数组中复制元素？
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