blocks|key|2802750|text|你可以在没有第三方库的情况下做到这一点，这样会更有效率。(这只会删除找到的值的第一个实例，而不是多个)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2802751|Javascript|2802752|var+a+=+[0,+1,+2,+3,+4,+6,+7,+8,+9],
++++b+=+5,
++++c+=+6;

function+addOrRemove(array,+value)+{
++++var+index+=+array.indexOf(value);

++++if+(index+===+-1)+{
++++++++array.push(value);
++++}+else+{
++++++++array.splice(index,+1);
++++}
}

console.log(a);

addOrRemove(a,+b);
console.log(a);

addOrRemove(a,+c);
console.log(a);|code-block|syntax|javascript|2802753|输出|2802754|[0,+1,+2,+3,+4,+6,+7,+8,+9]
[0,+1,+2,+3,+4,+6,+7,+8,+9,+5]
[0,+1,+2,+3,+4,+7,+8,+9,+5]+|2802755|在jsfiddle上|offset|length|2802756|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/xwVFy/^0|0|0|0|0|0|1|8|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|11|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|13|8|@]|9|@$O|14|P|15|1|16]]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

You could do it without a 3rd party library, this would be more efficient, like this. (this only removes the first instance of a value if found, not multiple)

Javascript

<pre><code>var a = [0, 1, 2, 3, 4, 6, 7, 8, 9],
 b = 5,
 c = 6;

function addOrRemove(array, value) {
 var index = array.indexOf(value);

 if (index === -1) {
 array.push(value);
 } else {
 array.splice(index, 1);
 }
}

console.log(a);

addOrRemove(a, b);
console.log(a);

addOrRemove(a, c);
console.log(a);
</code></pre>

Output

<pre><code>[0, 1, 2, 3, 4, 6, 7, 8, 9]
[0, 1, 2, 3, 4, 6, 7, 8, 9, 5]
[0, 1, 2, 3, 4, 7, 8, 9, 5] 
</code></pre>

On <a href="http://jsfiddle.net/xwVFy/">jsfiddle</a>

blocks|key|183872|text|如果你关心效率，那么使用一个数组来实现一个集合可能不是一个好主意。例如，使用一个你可以做的对象：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|183873|function+toggle(S,+x)+{
++++S[x]+=+1+-+(S[x]%7C0);
}|code-block|syntax|javascript|183874|然后，在多次添加/删除操作之后，您只能保留值为1的键|183875|这样，每次添加/删除都是O(1)的，您只需执行一次O(n)操作即可获得最终结果。|offset|length|style|CODE|183876|如果密钥都是“小”数字，则位掩码甚至值得努力(未测试)|183877|function+toggle(S,+x)+{
++++var+i+=+x+>>+4;
++++S[i]+=+(S[i]%7C0)+%5E+(1<<(x&15));
}|183878|entityMap^0|0|0|0|C|4|P|4|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@$K|Y|L|Z|M|N]|$K|10|L|11|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

If you care about efficiency then may be using an array to implement a set is a bad idea. For example using an object you could do:

<pre><code>function toggle(S, x) {
 S[x] = 1 - (S[x]|0);
}
</code></pre>

then after many add/remove operations you can keep only keys where the value is 1

This way every addition/removal is <code>O(1)</code> and you need only one <code>O(n)</code> operation to get the final result.

If keys are all "small" numbers may be a bitmask is even worth the effort (not tested)

<pre><code>function toggle(S, x) {
 var i = x &gt;&gt; 4;
 S[i] = (S[i]|0) ^ (1&lt;&lt;(x&amp;15));
}
</code></pre>

blocks|key|2802810|text|使用underscorejs|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2802811|function+toggle(a,b)
{
return+_.indexOf(a,b)==-1?_.union(a,[b]):_.without(a,b);
}|code-block|syntax|javascript|2802812|用法：|2802813|var+a+=+[1,2,3];
var+b+=+[4];
a+=+toggle(a,b);+//+[1,2,3,4]
a+=+toggle(a,b);+//+[1,2,3]|2802814|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|$]]

Using underscorejs

<pre><code>function toggle(a,b)
{
return _.indexOf(a,b)==-1?_.union(a,[b]):_.without(a,b);
}
</code></pre>

Usage:

<pre><code>var a = [1,2,3];
var b = [4];
a = toggle(a,b); // [1,2,3,4]
a = toggle(a,b); // [1,2,3]
</code></pre>

blocks|key|186164|text|看看类似问题的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|186165|Lodash+issue|offset|length|186166|Lodash+gist|186167|代码：|style|BOLD|186168|function+toggle(collection,+item)+{
++var+idx+=+collection.indexOf(item);
++if+(idx+!==+-1)+{
++++collection.splice(idx,+1);
++}+else+{
++++collection.push(item);
++}
}|code-block|syntax|javascript|186169|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/lodash/lodash/issues/1677|1|https://gist.github.com/chodorowicz/1a77655681a3f67ea36de974a1357882/^0|0|0|C|0|0|0|B|1|0|0|3|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@$D|12|E|13|1|14]]|A|$]]|$1|F|3|G|5|6|7|15|8|@]|9|@$D|16|E|17|1|18]]|A|$]]|$1|H|3|I|5|6|7|19|8|@$D|1A|E|1B|J|K]]|9|@]|A|$]]|$1|L|3|M|5|N|7|1C|8|@]|9|@]|A|$O|P]]|$1|Q|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]|Y|$5|T|U|V|A|$W|Z]]]]

Look at <a href="https://stackoverflow.com/a/18765477/2179748">this answer</a> of similar question.
<a href="https://github.com/lodash/lodash/issues/1677" rel="nofollow noreferrer">Lodash issue</a>
<a href="https://gist.github.com/chodorowicz/1a77655681a3f67ea36de974a1357882/" rel="nofollow noreferrer">Lodash gist</a>
Code:
<pre><code>function toggle(collection, item) {
 var idx = collection.indexOf(item);
 if (idx !== -1) {
 collection.splice(idx, 1);
 } else {
 collection.push(item);
 }
}
</code></pre>

blocks|key|183952|text|您可以使用lodash函数"xor"：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|183953|_.xor([2,+1],+[2,+3]);
//+=>+[1,+3]|code-block|syntax|javascript|183954|如果没有数组作为第二个参数，可以简单地将变量包装到一个数组中|183955|var+variableToInsertOrRemove+=+2;
_.xor([2,+1],+[variableToInsertOrRemove]);
//+=>+[1]
_.xor([1,+3],+[variableToInsertOrRemove]);
//+=>+[1,+2,+3]|183956|这里是文档：https://lodash.com/docs/4.16.4#xor|offset|length|183957|entityMap|0|LINK|mutability|MUTABLE|url|https://lodash.com/docs/4.16.4#xor^0|0|0|0|0|6|Y|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@$M|11|N|12|1|13]]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

You can use the lodash function "xor":

<pre><code>_.xor([2, 1], [2, 3]);
// =&gt; [1, 3]
</code></pre>

If you don't have an array as 2nd parameter you can simpy wrap the variable into an array

<pre><code>var variableToInsertOrRemove = 2;
_.xor([2, 1], [variableToInsertOrRemove]);
// =&gt; [1]
_.xor([1, 3], [variableToInsertOrRemove]);
// =&gt; [1, 2, 3]
</code></pre>

Here's the doc: <a href="https://lodash.com/docs/4.16.4#xor">https://lodash.com/docs/4.16.4#xor</a>

blocks|key|2802966|text|对于不可变克隆状态(克隆阵列)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2802967|const+addOrRemove+=+(arr,+item)+=>+arr.includes(item)+?+arr.filter(i+=>+i+!==+item)+:+[+...arr,+item+];|code-block|syntax|javascript|2802968|另请参阅：Remove+properties+from+objects+(JavaScript)|2802969|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/208105/remove-properties-from-objects-javascript/56030135#56030135^0|2|E|0|0|0|1C|5|17|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|M|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

For immutable state ( clone array ):
<pre><code>const addOrRemove = (arr, item) =&gt; arr.includes(item) ? arr.filter(i =&gt; i !== item) : [ ...arr, item ];
</code></pre>
see also:
<a href="https://stackoverflow.com/questions/208105/remove-properties-from-objects-javascript/56030135#56030135">Remove properties from objects (JavaScript)</a>

blocks|key|2802977|text|如果“类型”可以是一个集合，那么|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2802978|let+toggle+=+type_id+=>+this.types.delete(type_id)+%7C%7C+this.types.add(type_id);|code-block|syntax|javascript|2802979|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|$]]

If "types" can be a Set then

<pre><code>let toggle = type_id =&gt; this.types.delete(type_id) || this.types.add(type_id);
</code></pre>

blocks|key|2802994|text|扩展Xotic750's+answer这将始终确保被切换的元素在数组中只出现一次。如果你的数组像用户输入一样有点随机，你就会这样做。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2802995|​|2802996|function+toggleValueInArray(array,+value)+{
++var+index+=+array.indexOf(value);

++if+(index+==+-1)+{
++++array.push(value);
++}+else+{
++++do+{
++++++array.splice(index,+1);
++++++index+=+array.indexOf(value);
++++}+while+(index+!=+-1);
++}
}


var+a+=+[0,+1,+2,+3,+4,+5,+5,+6,+7,+8,+9],
++b+=+5,
++c+=+10;

//+toString+just+for+good+output
console.log(a.toString());

toggleValueInArray(a,+b);
console.log(a.toString());

toggleValueInArray(a,+c);
console.log(a.toString());|code-block|syntax|javascript|2802997|2802998|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/17153457/9921110^0|14|Q|2|H|0|0|0|1|0|0|0|1|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@$9|Y|A|Z|1|10]]|E|$]]|$1|F|3|G|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]]]

Extending <a href="https://stackoverflow.com/a/17153457/9921110">Xotic750's answer</a> this will always ensure that toggled elements occur only once in array. You this if your arrays are bit random like user inputs.

<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>function toggleValueInArray(array, value) {
 var index = array.indexOf(value);

 if (index == -1) {
 array.push(value);
 } else {
 do {
 array.splice(index, 1);
 index = array.indexOf(value);
 } while (index != -1);
 }
}


var a = [0, 1, 2, 3, 4, 5, 5, 6, 7, 8, 9],
 b = 5,
 c = 10;

// toString just for good output
console.log(a.toString());

toggleValueInArray(a, b);
console.log(a.toString());

toggleValueInArray(a, c);
console.log(a.toString());</code></pre>
</div>
</div>

Here is my code for javascript for this simple task:

<ol>
<li>Remove the element if it exists in an array. </li>
<li>Add the element if it is not in an array.

<pre><code>if(_.contains(this.types,type_id)){
 var index = this.types.indexOf(type_id);
 this.types.splice(index,1);
}
else{
 this.types.push(type_id);
}
</code></pre></li>
</ol>

Is there a more efficient way to do this?

How to toggle an element in array using JavaScript?

下面是我为这个简单任务编写的javascript代码：如果元素存在于数组中，则对其执行Remove操作。如果元素不在数组中，则使用Add。if(_.contains(this.types，type_id)){ var index = this.types.indexOf(type_id)；this.types.spli...

问如何使用JavaScript切换数组中的元素？
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