blocks|key|3178391|text|Prototype有一个array.partition函数和一个eachSlice()函数。听起来eachSlice()就是您要找的。如果你正在使用jquery，有一个插件可以使用原型函数。这是它的链接。http://www.learningjquery.com/2009/02/implementing-prototypes-array-methods-in-jquery|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3178392|entityMap|0|LINK|mutability|MUTABLE|url|http://www.learningjquery.com/2009/02/implementing-prototypes-array-methods-in-jquery^0|2U|2D|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Prototype has an array.partition function as well as an eachSlice() function. Sounds like eachSlice() is what you're looking for. If you're using jquery, there's a plug in to be able to use prototype functions. Here's a link to it... <a href="http://www.learningjquery.com/2009/02/implementing-prototypes-array-methods-in-jquery" rel="nofollow">http://www.learningjquery.com/2009/02/implementing-prototypes-array-methods-in-jquery</a>

blocks|key|3178414|text|您可以编写自己的原型方法来完成此操作|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3178415|Array.prototype.partition+=+function(length)+{
++var+result+=+[];
++for(var+i+=+0;+i+<+this.length;+i%2B%2B)+{
++++if(i+%25+length+===+0)+result.push([]);
++++result[result.length+-+1].push(this[i]);
++}
++return+result;
};|code-block|syntax|javascript|3178416|如果不想添加到原生原型中，可以编写一个简单的函数：|3178417|var+partition+=+function(arr,+length)+{
++var+result+=+[];
++for(var+i+=+0;+i+<+arr.length;+i%2B%2B)+{
++++if(i+%25+length+===+0)+result.push([]);
++++result[result.length+-+1].push(arr[i]);
++}
++return+result;
};|3178418|你可以使用see+it+in+action+on+this+jsFiddle+demo。|offset|length|3178419|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/FPwpS/^0|0|0|0|0|5|12|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 write your own prototype method to do this

<pre><code>Array.prototype.partition = function(length) {
 var result = [];
 for(var i = 0; i &lt; this.length; i++) {
 if(i % length === 0) result.push([]);
 result[result.length - 1].push(this[i]);
 }
 return result;
};
</code></pre>

If you prefer not to add to the native prototype, you can write a simple function:

<pre><code>var partition = function(arr, length) {
 var result = [];
 for(var i = 0; i &lt; arr.length; i++) {
 if(i % length === 0) result.push([]);
 result[result.length - 1].push(arr[i]);
 }
 return result;
};
</code></pre>

You can <a href="http://jsfiddle.net/FPwpS/" rel="nofollow">see it in action on this jsFiddle demo</a>.

blocks|key|1541477|text|如果使用Underscore.js，则可以使用groupBy()和values()实现此功能|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1541478|function+partition(items,+size)+{
++++var+result+=+_.groupBy(items,+function(item,+i)+{
++++++++return+Math.floor(i/size);
++++});
++++return+_.values(result);
}|code-block|syntax|javascript|1541479|(这是一个不那么难看的in+CoffeeScript。)|1541480|jsFiddle：http://jsfiddle.net/MW3BS/|1541481|entityMap|0|LINK|mutability|MUTABLE|url|http://underscorejs.org/|1|http://underscorejs.org/#groupBy|2|http://underscorejs.org/#values|3|http://jsfiddle.net/MW3BS/3/|4|http://jsfiddle.net/MW3BS/^0|N|9|X|8|4|D|0|N|9|1|X|8|2|0|0|B|F|3|0|9|Q|4|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@$9|19|A|1A|1|1B]|$9|1C|A|1D|1|1E]|$9|1F|A|1G|1|1H]]|E|$]]|$1|F|3|G|5|H|7|1I|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1J|8|@]|D|@$9|1K|A|1L|1|1M]]|E|$]]|$1|M|3|N|5|6|7|1N|8|@]|D|@$9|1O|A|1P|1|1Q]]|E|$]]|$1|O|3|-4|5|6|7|1R|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]|Y|$5|R|S|T|E|$U|Z]]|10|$5|R|S|T|E|$U|11]]|12|$5|R|S|T|E|$U|13]]]]

If using <a href="http://underscorejs.org/" rel="noreferrer">Underscore.js</a>, you can implement this with <a href="http://underscorejs.org/#groupBy" rel="noreferrer"><code>groupBy()</code></a> and <a href="http://underscorejs.org/#values" rel="noreferrer"><code>values()</code></a>

<pre><code>function partition(items, size) {
 var result = _.groupBy(items, function(item, i) {
 return Math.floor(i/size);
 });
 return _.values(result);
}
</code></pre>

(This is less ugly <a href="http://jsfiddle.net/MW3BS/3/" rel="noreferrer">in CoffeeScript</a>.)

jsFiddle: <a href="http://jsfiddle.net/MW3BS/" rel="noreferrer">http://jsfiddle.net/MW3BS/</a>

blocks|key|3178467|text|我认为你必须使用for循环，不知道有什么内置的函数...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3178468|尝试此函数：|3178469|function+splitarray(input,+spacing)
{
++++var+output+=+[];

++++for+(var+i+=+0;+i+<+input.length;+i+%2B=+spacing)
++++{
++++++++output[output.length]+=+input.slice(i,+i+%2B+spacing);
++++}

++++return+output;
}|code-block|syntax|javascript|3178470|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I think you will have to use a for loop, don't know of any inbuilt functions...

Try this function:

<pre><code>function splitarray(input, spacing)
{
 var output = [];

 for (var i = 0; i &lt; input.length; i += spacing)
 {
 output[output.length] = input.slice(i, i + spacing);
 }

 return output;
}
</code></pre>

blocks|key|1541522|text|另一个解决方案，没有外部库：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1541523|function+partition(items,+size)+{
++++var+p+=+[];
++++for+(var+i=Math.floor(items.length/size);+i-->0;+)+{
++++++++p[i]=items.slice(i*size,+(i%2B1)*size);
++++}
++++return+p;
}|code-block|syntax|javascript|1541524|演示：http://jsfiddle.net/dystroy/xtHXZ/|offset|length|1541525|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/dystroy/xtHXZ/^0|0|0|3|Y|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

One more solution, with no external library :

<pre><code>function partition(items, size) {
 var p = [];
 for (var i=Math.floor(items.length/size); i--&gt;0; ) {
 p[i]=items.slice(i*size, (i+1)*size);
 }
 return p;
}
</code></pre>

Demonstration : <a href="http://jsfiddle.net/dystroy/xtHXZ/" rel="nofollow">http://jsfiddle.net/dystroy/xtHXZ/</a>

blocks|key|1824382|text|下面是一个递归解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1824383|function+partition(array,+n)+{
++return+array.length+?+[array.splice(0,+n)].concat(partition(array,+n))+:+[];
}++++|code-block|syntax|javascript|1824384|这利用了Array#splice破坏性地删除指定项的事实，并将它们作为函数值返回。注意，这将销毁输入数组，使其保留为空的。|offset|length|style|CODE|BOLD|1824385|entityMap^0|0|0|4|C|15|J|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Q|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|R|8|@$I|S|J|T|K|L]|$I|U|J|V|K|M]]|9|@]|A|$]]|$1|N|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|O|$]]

Here's a recursive solution:
<pre><code>function partition(array, n) {
 return array.length ? [array.splice(0, n)].concat(partition(array, n)) : [];
} 
</code></pre>
This takes advantage of the fact that <code>Array#splice</code> destructively remove the specified items, and returns them as the function value. Note that this will destroy the input array, leaving it empty.

blocks|key|3178681|text|我已经将这个解决方案添加到@dystroy's+jspref+here中，它的运行速度似乎是其他解决方案的两倍。编辑:在Safari和Chrome中，但不在Firefox中|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3178682|这里有一个函数式的解决方案，可以添加到这里的混合答案中。|3178683|它是一个名为toPartitions的高阶函数，它返回下划线的reduce方法或本机数组reduce方法的回调。|CODE|3178684|示例用法：|3178685|[1,2,3,4,5,6,7,8,9].reduce(+toPartitions(+3+),+[]+);|code-block|syntax|javascript|3178686|函数：|3178687|function+toPartitions+(+size+)+{
++++var+partition+=+[];
++++return+function+(+acc,+v+)+{
++++++++partition.push(+v+);
++++++++if+(+partition.length+===+size+)+{
++++++++++++acc.push(+partition+);
++++++++++++partition+=+[];
++++++++}
++++++++return+acc;
++++};
}|3178688|与Clojure的partition一样，当没有足够的元素时，它不会包含尾部分区。|3178689|在您的示例中，您可以这样做：|3178690|arrayALLPartionned+=+arrayAll.reduce(+toPartitions(+3+),+[]+)+);|3178691|如果你不想在reduce中使用这个函数，而只想使用一个接受数组和分区大小的函数，你可以这样做：|3178692|function+partition+(+arr,+size+)+{
++++return+arr.reduce(+toPartitions(+size+),+[]+);
}|3178693|因此，解决方案就是：|3178694|arrayALLPartionned+=+partition(+arrayAll,+3+);|3178695|entityMap|0|LINK|mutability|MUTABLE|url|http://jsperf.com/partition/3|1|https://clojuredocs.org/clojure.core/partition^0|1K|T|D|M|0|0|0|6|C|0|0|0|0|0|9|9|1|0|0|0|6|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1J|8|@$9|1K|A|1L|B|C]]|D|@$9|1M|A|1N|1|1O]]|E|$]]|$1|F|3|G|5|6|7|1P|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|1Q|8|@$9|1R|A|1S|B|J]]|D|@]|E|$]]|$1|K|3|L|5|6|7|1T|8|@]|D|@]|E|$]]|$1|M|3|N|5|O|7|1U|8|@]|D|@]|E|$P|Q]]|$1|R|3|S|5|6|7|1V|8|@]|D|@]|E|$]]|$1|T|3|U|5|O|7|1W|8|@]|D|@]|E|$P|Q]]|$1|V|3|W|5|6|7|1X|8|@]|D|@$9|1Y|A|1Z|1|20]]|E|$]]|$1|X|3|Y|5|6|7|21|8|@]|D|@]|E|$]]|$1|Z|3|10|5|O|7|22|8|@]|D|@]|E|$P|Q]]|$1|11|3|12|5|6|7|23|8|@$9|24|A|25|B|J]]|D|@]|E|$]]|$1|13|3|14|5|O|7|26|8|@]|D|@]|E|$P|Q]]|$1|15|3|16|5|6|7|27|8|@]|D|@]|E|$]]|$1|17|3|18|5|O|7|28|8|@]|D|@]|E|$P|Q]]|$1|19|3|-4|5|6|7|29|8|@]|D|@]|E|$]]]|1A|$1B|$5|1C|1D|1E|E|$1F|1G]]|1H|$5|1C|1D|1E|E|$1F|1I]]]]

I've added this solution to <a href="http://jsperf.com/partition/3" rel="nofollow">@dystroy's jspref here</a> and it appears to run twice as fast as the other solutions. Edit: in Safari &amp; Chrome but not Firefox

Here is functional style solution to add to the mix of answers here. 

It is a higher order function called <code>toPartitions</code> which returns a callback for underscore's reduce method or the native array reduce method.

Example usage:

<pre><code>[1,2,3,4,5,6,7,8,9].reduce( toPartitions( 3 ), [] );
</code></pre>

The function:

<pre><code>function toPartitions ( size ) {
 var partition = [];
 return function ( acc, v ) {
 partition.push( v );
 if ( partition.length === size ) {
 acc.push( partition );
 partition = [];
 }
 return acc;
 };
}
</code></pre>

Like Clojure's <a href="https://clojuredocs.org/clojure.core/partition" rel="nofollow">partition</a> it will not include a tail partition when there are not enough elements.

In your example you could do:

<pre><code>arrayALLPartionned = arrayAll.reduce( toPartitions( 3 ), [] ) );
</code></pre>

If you don't want to use this with <code>reduce</code>, but just have a function which takes an array and partition size you could do:

<pre><code>function partition ( arr, size ) {
 return arr.reduce( toPartitions( size ), [] );
}
</code></pre>

Therefore the solution would just be:

<pre><code>arrayALLPartionned = partition( arrayAll, 3 );
</code></pre>

Please consider an array such as :

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

Is there a package that enable to do partitioning to obtain :

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

I can see how to do this with a for loop but would appreciate a "pre-made" function if existing.

Partitioning in JavaScript

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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请考虑如下数组：arrayAll = [1,2,3,4,5,6,7,8,9]有没有包可以进行分区来获取：arrayALLPartionned = [[1,2,3],[4,5,6],[7,8,9]]我可以看到如何使用for循环来做到这一点，但如果存在“预制”函数，则会更好。

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