blocks|key|3934321|text|更好的做法是使用内置方法，提供splice()此方法用于向数组添加项/从数组中删除项，并将从数组中返回已删除的项。你可以在这里了解更多关于Splice的信息。Splice|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3934322|entityMap|0|LINK|mutability|MUTABLE|url|https://www.w3schools.com/jsref/jsref_splice.asp^0|F|8|27|6|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

Better to do with inbuilt methods provided <code>splice()</code> 
This method used to adds/removes items to/from an array and will return the removed item(s) from an array.
you can learn more about Splice here.
<a href="https://www.w3schools.com/jsref/jsref_splice.asp" rel="nofollow noreferrer">Splice</a>

blocks|key|3934401|text|我会说var+result+=+arrayRemove(array,+'b');更好，因为这是：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3934402|if+(array.includes('b'))|code-block|syntax|javascript|3934403|将遍历数组，如果找到b，则如下所示：|3934404|var+result+=+arrayRemove(array,+'b');|3934405|将再次迭代数组以移除它，因此如果元素存在，则有两次迭代，如果元素不存在，则有一次迭代。直接使用arrayRemove(array,+'b')在两种情况下都只迭代一次。|3934406|entityMap^0|3|11|0|0|A|1|0|0|1B|N|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|R|$]]

I'd say <code>var result = arrayRemove(array, 'b');</code> is better, because this:

<pre><code>if (array.includes('b'))
</code></pre>

will iterate over the array, and if it finds <code>b</code>, then this:

<pre><code>var result = arrayRemove(array, 'b');
</code></pre>

will iterate over the array again to remove it, so you have two iterations if the element exists, and one if it doesn't. Using <code>arrayRemove(array, 'b')</code> directly will iterate only once in both cases.

blocks|key|3938357|text|Array.filter函数会创建一个新的数组-所以它总是会比if消耗更多的资源。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3938358|一个简单的测试：|3938359|const+arr+=+Array(10000000).fill(Math.random());
arr[10000]+=+'e';

let+start+=+Date.now();
const+arr2+=+arr.filter(v+=>+v+!==+'d');
console.log('filter+time',+Date.now()+-+start);+//+---+filter+time+1168ms

start+=+Date.now();
if+(arr.includes('e'))+{
++console.log('includes+time',+Date.now()+-+start);+//+---+includes+time+1ms
}|code-block|syntax|javascript|3938360|所以你可以看到，如果数组真的很大，mutch速度会更快|3938361|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Array.filter function creates a new array - so it's always will consume more resources then if.

A simple test:

<pre><code>const arr = Array(10000000).fill(Math.random());
arr[10000] = 'e';

let start = Date.now();
const arr2 = arr.filter(v =&gt; v !== 'd');
console.log('filter time', Date.now() - start); // --- filter time 1168ms

start = Date.now();
if (arr.includes('e')) {
 console.log('includes time', Date.now() - start); // --- includes time 1ms
}
</code></pre>

So as you can see if the array is realy big if would be mutch faster

blocks|key|3934624|text|为此，我创建了一个JS+perf。看起来在检查值是否存在之前直接调用arrayRemove函数会更好。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3934625|一个简单的测试：|3934626|const+arr+=+Array(10000000).fill(Math.random());
arr[10000]+=+'e';

let+start+=+Date.now();
const+arr2+=+arr.filter(v+=>+v+!==+'d');
console.log('filter+time',+Date.now()+-+start);+//+---+filter+time+1168ms

start+=+Date.now();
if+(arr.includes('e'))+{
++console.log('includes+time',+Date.now()+-+start);+//+---+includes+time+1ms
}|code-block|syntax|javascript|3934627|所以你可以看到，如果数组真的很大，速度会快得多|3934628|https://jsperf.com/js-array-filter/1|3934629|entityMap|0|LINK|mutability|MUTABLE|url^0|Y|B|0|0|0|0|0|10|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|10|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|11|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|12|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|13|8|@]|D|@$9|14|A|15|1|16]]|E|$]]|$1|Q|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|P]]]]

I created a JS perf for this. Looks like just calling the <code>arrayRemove</code> function before checking if the value exists, is better.

A simple test:

<pre><code>const arr = Array(10000000).fill(Math.random());
arr[10000] = 'e';

let start = Date.now();
const arr2 = arr.filter(v =&gt; v !== 'd');
console.log('filter time', Date.now() - start); // --- filter time 1168ms

start = Date.now();
if (arr.includes('e')) {
 console.log('includes time', Date.now() - start); // --- includes time 1ms
}
</code></pre>

So as you can see if the array is really big if would be much faster

<a href="https://jsperf.com/js-array-filter/1" rel="nofollow noreferrer">https://jsperf.com/js-array-filter/1</a>

Conceptual question here, when using Javascript to remove an item from a collection by value, what is the more performant way to achieve this? 

Assuming: 

<pre><code>function arrayRemove(arr, value) { 
 return arr.filter(function(ele){ 
 return ele != value;
});

var array = ['a', 'b', 'c', 'd'];
</code></pre>

is simply using 

<pre><code>var result = arrayRemove(array, 'b');
</code></pre>

More appropriate than checking for its existence beforehand

<pre><code>if (array.includes('b')) 
 var result = arrayRemove(array, 'b');
</code></pre>

The first is cleaner and less operations, but the later will not always trigger the filter.

In a Javascript Array, should a value be checked for before being removed?

概念性问题这里，当使用Javascript通过值从集合中删除项时，实现这一点的更好的方法是什么？假设：function arrayRemove(arr, value) {      return arr.filter(function(ele){             return ele != value;});v...

问在Javascript数组中，是否应该在删除值之前检查其值？
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问在Javascript数组中，是否应该在删除值之前检查其值？
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