blocks|key|283067|text|我建议使用Fisher-Yates+shuffle对数组的副本进行混洗，然后取一片：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|283068|function+getRandomSubarray(arr,+size)+{
++++var+shuffled+=+arr.slice(0),+i+=+arr.length,+temp,+index;
++++while+(i--)+{
++++++++index+=+Math.floor((i+%2B+1)+*+Math.random());
++++++++temp+=+shuffled[index];
++++++++shuffled[index]+=+shuffled[i];
++++++++shuffled[i]+=+temp;
++++}
++++return+shuffled.slice(0,+size);
}

var+x+=+[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];
var+fiveRandomMembers+=+getRandomSubarray(x,+5);|code-block|syntax|javascript|283069|请注意，对于获取大型数组的小随机子集而言，这不是最有效的方法，因为它会对整个数组进行不必要的打乱。为了获得更好的性能，您可以改为执行部分混洗：|283070|function+getRandomSubarray(arr,+size)+{
++++var+shuffled+=+arr.slice(0),+i+=+arr.length,+min+=+i+-+size,+temp,+index;
++++while+(i--+>+min)+{
++++++++index+=+Math.floor((i+%2B+1)+*+Math.random());
++++++++temp+=+shuffled[index];
++++++++shuffled[index]+=+shuffled[i];
++++++++shuffled[i]+=+temp;
++++}
++++return+shuffled.slice(min);
}|283071|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Fisher%25E2%2580%2593Yates_shuffle^0|5|K|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

<p>I suggest shuffling a copy of the array using the <a href="http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle" rel="noreferrer">Fisher-Yates shuffle</a> and taking a slice:</p>

<pre><code>function getRandomSubarray(arr, size) {
    var shuffled = arr.slice(0), i = arr.length, temp, index;
    while (i--) {
        index = Math.floor((i + 1) * Math.random());
        temp = shuffled[index];
        shuffled[index] = shuffled[i];
        shuffled[i] = temp;
    }
    return shuffled.slice(0, size);
}

var x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];
var fiveRandomMembers = getRandomSubarray(x, 5);
</code></pre>

<p>Note that this will not be the most efficient method for getting a small random subset of a large array because it shuffles the whole array unnecessarily. For better performance you could do a partial shuffle instead:</p>

<pre><code>function getRandomSubarray(arr, size) {
    var shuffled = arr.slice(0), i = arr.length, min = i - size, temp, index;
    while (i-- &gt; min) {
        index = Math.floor((i + 1) * Math.random());
        temp = shuffled[index];
        shuffled[index] = shuffled[i];
        shuffled[i] = temp;
    }
    return shuffled.slice(min);
}
</code></pre>


blocks|key|283137|text|可以在选择图元时将其从阵列副本中删除。性能可能不是很理想，但对于您所需要的可能是可以的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|283138|function+getRandom(arr,+size)+{
++var+copy+=+arr.slice(0),+rand+=+[];
++for+(var+i+=+0;+i+<+size+&&+i+<+copy.length;+i%2B%2B)+{
++++var+index+=+Math.floor(Math.random()+*+copy.length);
++++rand.push(copy.splice(index,+1)[0]);
++}
++return+rand;
}|code-block|syntax|javascript|283139|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|$]]

<p>You can remove the elements from a copy of the array as you select them. Performance is probably not ideal, but it might be OK for what you need:</p>

<pre><code>function getRandom(arr, size) {
  var copy = arr.slice(0), rand = [];
  for (var i = 0; i &lt; size &amp;&amp; i &lt; copy.length; i++) {
    var index = Math.floor(Math.random() * copy.length);
    rand.push(copy.splice(index, 1)[0]);
  }
  return rand;
}
</code></pre>


blocks|key|283185|text|或者..。如果您使用underscore.js...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|283186|_und+=+require('underscore');

...

function+sample(a,+n)+{
++++return+_und.take(_und.shuffle(a),+n);
}|code-block|syntax|javascript|283187|很简单。|283188|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|$]]

<p>Or... if you use underscore.js...</p>

<pre><code>_und = require('underscore');

...

function sample(a, n) {
    return _und.take(_und.shuffle(a), n);
}
</code></pre>

<p>Simple enough.</p>


blocks|key|283239|text|参加聚会有点晚了，但这可以用下划线的新sample方法解决(下划线1.5.2+-2013年9月)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|283240|var+x+=+[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];

var+randomFiveNumbers+=+_.sample(x,+5);|code-block|syntax|javascript|283241|entityMap|0|LINK|mutability|MUTABLE|url|http://underscorejs.org/#sample^0|J|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]]]]

<p>A little late to the party but this could be solved with underscore's new <a href="http://underscorejs.org/#sample">sample</a>  method (underscore 1.5.2 - Sept 2013):</p>

<pre><code>var x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];

var randomFiveNumbers = _.sample(x, 5);
</code></pre>


blocks|key|283323|text|虽然我强烈支持使用Fisher-Yates+Shuffle，作为suggested+by+Tim+Down，但这里有一个非常简短的方法，用于根据请求实现随机子集，数学上正确，包括空集和给定集本身。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|283324|注意:解决方案依赖于lodash+/+underscore|283325|Lodash+v4|283326|const+_+=+require('loadsh')

function+subset(arr)+{
++++return+_.sampleSize(arr,+_.random(arr.length))
}|code-block|syntax|javascript|283327|Lodash+v3|283328|const+_+=+require('loadsh')

function+subset(arr)+{
++++return+_.sample(arr,+_.random(arr.length));
}|283329|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/11935263/3873012|1|http://devdocs.io/lodash/index#sample|2|http://underscorejs.org/#sample^0|W|L|0|0|A|6|1|J|A|2|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|6|7|16|8|@]|9|@$A|17|B|18|1|19]|$A|1A|B|1B|1|1C]]|C|$]]|$1|F|3|G|5|6|7|1D|8|@]|9|@]|C|$]]|$1|H|3|I|5|J|7|1E|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|1F|8|@]|9|@]|C|$]]|$1|O|3|P|5|J|7|1G|8|@]|9|@]|C|$K|L]]|$1|Q|3|-4|5|6|7|1H|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]|Y|$5|T|U|V|C|$W|Z]]|10|$5|T|U|V|C|$W|11]]]]

<p>While I strongly support using the Fisher-Yates Shuffle, as <a href="https://stackoverflow.com/a/11935263/3873012">suggested by Tim Down</a>, here's a very short method for achieving a random subset as requested, mathematically correct, including the empty set, and the given set itself.</p>

<p>Note solution depends on <a href="http://devdocs.io/lodash/index#sample" rel="nofollow noreferrer">lodash</a> / <a href="http://underscorejs.org/#sample" rel="nofollow noreferrer">underscore</a>:</p>

<h3>Lodash v4</h3>

<pre><code>const _ = require('loadsh')

function subset(arr) {
    return _.sampleSize(arr, _.random(arr.length))
}
</code></pre>

<h3>Lodash v3</h3>

<pre><code>const _ = require('loadsh')

function subset(arr) {
    return _.sample(arr, _.random(arr.length));
}
</code></pre>


blocks|key|283415|text|这是另一个基于Fisher-Yates+Shuffle的实现。但这种方法针对样本大小明显小于数组长度的情况进行了优化。这个实现不会扫描整个数组，也不会分配与原始数组一样大的数组。它使用稀疏数组来减少内存分配。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|283416|function+getRandomSample(array,+count)+{
++++var+indices+=+[];
++++var+result+=+new+Array(count);
++++for+(let+i+=+0;+i+<+count;+i%2B%2B+)+{
++++++++let+j+=+Math.floor(Math.random()+*+(array.length+-+i)+%2B+i);
++++++++result[i]+=+array[indices[j]+===+undefined+?+j+:+indices[j]];
++++++++indices[j]+=+indices[i]+===+undefined+?+i+:+indices[i];
++++}
++++return+result;
}|code-block|syntax|javascript|283417|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|$]]

<p>Here is another implementation based on Fisher-Yates Shuffle. But this one is optimized for the case where the sample size is significantly smaller than the array length. This implementation doesn't scan the entire array nor allocates arrays as large as the original array. It uses sparse arrays to reduce memory allocation.</p>

<pre><code>function getRandomSample(array, count) {
    var indices = [];
    var result = new Array(count);
    for (let i = 0; i &lt; count; i++ ) {
        let j = Math.floor(Math.random() * (array.length - i) + i);
        result[i] = array[indices[j] === undefined ? j : indices[j]];
        indices[j] = indices[i] === undefined ? i : indices[i];
    }
    return result;
}
</code></pre>


blocks|key|500154|text|如果你使用的是lodash，API在4.x中发生了变化：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|500155|const+oneItem+=+_.sample(arr);
const+nItems+=+_.sampleSize(arr,+n);|code-block|syntax|javascript|500156|https://lodash.com/docs#sampleSize|offset|length|500157|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|Y|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|T|8|@]|9|@$I|U|J|V|1|W]]|A|$]]|$1|K|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|H]]]]

<p>If you're using lodash the API changed in 4.x:</p>

<pre><code>const oneItem = _.sample(arr);
const nItems = _.sampleSize(arr, n);
</code></pre>

<p><a href="https://lodash.com/docs#sampleSize" rel="nofollow">https://lodash.com/docs#sampleSize</a></p>


blocks|key|500195|text|您可以通过以下方式获取5个元素的示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|500196|var+sample+=+[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
.map(a+=>+[a,Math.random()])
.sort((a,b)+=>+{return+a[1]+<+b[1]+?+-1+:+1;})
.slice(0,5)
.map(a+=>+a[0]);|code-block|syntax|javascript|500197|您可以将其定义为在代码中使用的函数：|500198|var+randomSample+=+function(arr,num){+return+arr.map(a+=>+[a,Math.random()]).sort((a,b)+=>+{return+a[1]+<+b[1]+?+-1+:+1;}).slice(0,num).map(a+=>+a[0]);+}|500199|或者将其添加到Array对象本身：|500200|++++Array.prototype.sample+=+function(num){+return+this.map(a+=>+[a,Math.random()]).sort((a,b)+=>+{return+a[1]+<+b[1]+?+-1+:+1;}).slice(0,num).map(a+=>+a[0]);+};|500201|如果您愿意，您可以将的代码分开，以拥有两个功能(Shuffle和Sample)：|500202|++++Array.prototype.shuffle+=+function(){+return+this.map(a+=>+[a,Math.random()]).sort((a,b)+=>+{return+a[1]+<+b[1]+?+-1+:+1;}).map(a+=>+a[0]);+};
++++Array.prototype.sample+=+function(num){+return+this.shuffle().slice(0,num);+};|500203|entityMap^0|0|0|0|0|0|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]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

<p>You can get a 5 elements sample by this way:</p>

<pre><code>var sample = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
.map(a =&gt; [a,Math.random()])
.sort((a,b) =&gt; {return a[1] &lt; b[1] ? -1 : 1;})
.slice(0,5)
.map(a =&gt; a[0]);
</code></pre>

<p>You can define it as a function to use in your code:</p>

<pre><code>var randomSample = function(arr,num){ return arr.map(a =&gt; [a,Math.random()]).sort((a,b) =&gt; {return a[1] &lt; b[1] ? -1 : 1;}).slice(0,num).map(a =&gt; a[0]); }
</code></pre>

<p>Or add it to the Array object itself:</p>

<pre><code>    Array.prototype.sample = function(num){ return this.map(a =&gt; [a,Math.random()]).sort((a,b) =&gt; {return a[1] &lt; b[1] ? -1 : 1;}).slice(0,num).map(a =&gt; a[0]); };
</code></pre>

<p>if you want, you can separate the code for to have 2 functionalities (Shuffle and Sample):</p>

<pre><code>    Array.prototype.shuffle = function(){ return this.map(a =&gt; [a,Math.random()]).sort((a,b) =&gt; {return a[1] &lt; b[1] ? -1 : 1;}).map(a =&gt; a[0]); };
    Array.prototype.sample = function(num){ return this.shuffle().slice(0,num); };
</code></pre>


blocks|key|500218|text|也许我遗漏了一些东西，但似乎有一种解决方案不需要复杂性或潜在的洗牌开销：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|500219|function+sample(array,size)+{
++const+results+=+[],
++++sampled+=+{};
++while(results.length<size+&&+results.length<array.length)+{
++++const+index+=+Math.trunc(Math.random()+*+array.length);
++++if(!sampled[index])+{
++++++results.push(array[index]);
++++++sampled[index]+=+true;
++++}
++}
++return+results;
}|code-block|syntax|javascript|500220|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|$]]

<p>Perhaps I am missing something, but it seems there is a solution that does not require the complexity or potential overhead of a shuffle:</p>

<pre><code>function sample(array,size) {
  const results = [],
    sampled = {};
  while(results.length&lt;size &amp;&amp; results.length&lt;array.length) {
    const index = Math.trunc(Math.random() * array.length);
    if(!sampled[index]) {
      results.push(array[index]);
      sampled[index] = true;
    }
  }
  return results;
}
</code></pre>


blocks|key|283577|text|这些答案中有很多都是关于克隆、洗牌、切片原始数组的。我很好奇为什么从熵/分布的角度来看这是有帮助的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|283578|我不是专家，但我确实写了一个使用索引的示例函数，以避免任何数组突变-它确实会添加到集合中。我也不知道这上面的随机分布是怎么回事，但代码很简单，我认为这里应该有一个答案。|283579|​|283580|function+sample(array,+size+=+1)+{
++const+{+floor,+random+}+=+Math;
++let+sampleSet+=+new+Set();
++for+(let+i+=+0;+i+<+size;+i%2B%2B)+{
++++let+index;
++++do+{+index+=+floor(random()+*+array.length);+}
++++while+(sampleSet.has(index));
++++sampleSet.add(index);
++}
++return+[...sampleSet].map(i+=>+array[i]);
}

const+words+=+[
++'confused',+'astonishing',+'mint',+'engine',+'team',+'cowardly',+'cooperative',
++'repair',+'unwritten',+'detailed',+'fortunate',+'value',+'dogs',+'air',+'found',
++'crooked',+'useless',+'treatment',+'surprise',+'hill',+'finger',+'pet',
++'adjustment',+'alleged',+'income'
];

console.log(sample(words,+4));|code-block|syntax|javascript|283581|283582|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|P|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|Q|8|@]|9|@]|A|$I|J]]|$1|K|3|E|5|6|7|R|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|M|$]]

<p>A lot of these answers talk about cloning, shuffling, slicing the original array. I was curious why this helps from a entropy/distribution perspective.</p>
<p>I'm no expert but I did write a sample function using the indexes to avoid any array mutations — it does add to a Set though. I also don't know how the random distribution on this but the code was simple enough to I think warrant an answer here.</p>
<p><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 sample(array, size = 1) {
  const { floor, random } = Math;
  let sampleSet = new Set();
  for (let i = 0; i &lt; size; i++) {
    let index;
    do { index = floor(random() * array.length); }
    while (sampleSet.has(index));
    sampleSet.add(index);
  }
  return [...sampleSet].map(i =&gt; array[i]);
}

const words = [
  'confused', 'astonishing', 'mint', 'engine', 'team', 'cowardly', 'cooperative',
  'repair', 'unwritten', 'detailed', 'fortunate', 'value', 'dogs', 'air', 'found',
  'crooked', 'useless', 'treatment', 'surprise', 'hill', 'finger', 'pet',
  'adjustment', 'alleged', 'income'
];

console.log(sample(words, 4));</code></pre>
</div>
</div>
</p>


blocks|key|283636|text|对于非常大的数组，使用索引比使用数组的成员更有效。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|283637|这就是我在这个页面上找不到我喜欢的东西后得到的结果。|283638|/**
+*+Get+a+random+subset+of+an+array
+*+@param+{Array}+arr+-+Array+to+take+a+smaple+of.
+*+@param+{Number}+sample_size+-+Size+of+sample+to+pull.
+*+@param+{Boolean}+return_indexes+-+If+true,+return+indexes+rather+than+members
+*+@returns+{Array%7CBoolean}+-+An+array+containing+random+a+subset+of+the+members+or+indexes.
+*/
function+getArraySample(arr,+sample_size,+return_indexes+=+false)+{
++++if(sample_size+>+arr.length)+return+false;
++++const+sample_idxs+=+[];
++++const+randomIndex+=+()+=>+Math.floor(Math.random()+*+arr.length);
++++while(sample_size+>+sample_idxs.length){
++++++++let+idx+=+randomIndex();
++++++++while(sample_idxs.includes(idx))+idx+=+randomIndex();
++++++++sample_idxs.push(idx);
++++}
++++sample_idxs.sort((a,+b)+=>+a+>+b+?+1+:+-1);
++++if(return_indexes)+return+sample_idxs;
++++return+sample_idxs.map(i+=>+arr[i]);
}|code-block|syntax|javascript|283639|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|$]]

<p>For very large arrays, it's more efficient to work with indexes rather than the members of the array.</p>
<p>This is what I ended up with after not finding anything I liked on this page.</p>
<pre><code>/**
 * Get a random subset of an array
 * @param {Array} arr - Array to take a smaple of.
 * @param {Number} sample_size - Size of sample to pull.
 * @param {Boolean} return_indexes - If true, return indexes rather than members
 * @returns {Array|Boolean} - An array containing random a subset of the members or indexes.
 */
function getArraySample(arr, sample_size, return_indexes = false) {
    if(sample_size &gt; arr.length) return false;
    const sample_idxs = [];
    const randomIndex = () =&gt; Math.floor(Math.random() * arr.length);
    while(sample_size &gt; sample_idxs.length){
        let idx = randomIndex();
        while(sample_idxs.includes(idx)) idx = randomIndex();
        sample_idxs.push(idx);
    }
    sample_idxs.sort((a, b) =&gt; a &gt; b ? 1 : -1);
    if(return_indexes) return sample_idxs;
    return sample_idxs.map(i =&gt; arr[i]);
}
</code></pre>


blocks|key|500370|text|我的方法是创建一个getRandomIndexes方法，您可以使用该方法创建一个索引数组，该数组将从主数组中提取。在本例中，我添加了一个简单的逻辑，以避免在示例中使用相同的索引。这就是它的工作原理|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|500371|const+getRandomIndexes+=+(length,+size)+=>+{
++const+indexes+=+[];
++const+created+=+{};

++while+(indexes.length+<+size)+{
++++const+random+=+Math.floor(Math.random()+*+length);
++++if+(!created[random])+{
++++++indexes.push(random);
++++++created[random]+=+true;
++++}
++}
++return+indexes;
};|code-block|syntax|javascript|500372|这个函数独立于您所拥有的内容，它将为您提供一个索引数组，您可以使用该数组从长度为length的数组中提取值，因此可以通过以下方式进行采样|500373|const+myArray+=+['a',+'b',+'c',+'d',+'e',+'f',+'g',+'h',+'i',+'j']

getRandomIndexes(myArray.length,+3).map(i+=>+myArray[i])|500374|每次调用该方法时，您都会得到一个不同的myArray示例。在这一点上，这个解决方案很酷，但对于不同大小的样本可能会更好。如果你想这样做，你可以使用|500375|getRandomIndexes(myArray.length,+Math.ceil(Math.random()+*+6)).map(i+=>+myArray[i])|500376|每次调用它时，都会给你一个从1到6的不同样本大小。|500377|我希望这能有所帮助:D|500378|entityMap^0|9|G|0|0|14|6|0|0|J|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|16|8|@$9|17|A|18|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1A|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|1B|8|@]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|X|$]]

<p>My approach on this is to create a <code>getRandomIndexes</code> method that you can use to create an array of the indexes that you will pull from the main array. In this case, I added a simple logic to avoid the same index in the sample. this is how it works</p>
<pre class="lang-js prettyprint-override"><code>const getRandomIndexes = (length, size) =&gt; {
  const indexes = [];
  const created = {};

  while (indexes.length &lt; size) {
    const random = Math.floor(Math.random() * length);
    if (!created[random]) {
      indexes.push(random);
      created[random] = true;
    }
  }
  return indexes;
};
</code></pre>
<p>This function independently of whatever you have is going to give you an array of indexes that you can use to pull the values from your array of length <code>length</code>, so could be sampled by</p>
<pre class="lang-js prettyprint-override"><code>const myArray = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']

getRandomIndexes(myArray.length, 3).map(i =&gt; myArray[i])
</code></pre>
<p>Every time you call the method you are going to get a different sample of <code>myArray</code>. at this point, this solution is cool but could be even better to sample different sizes. if you want to do that you can use</p>
<pre class="lang-js prettyprint-override"><code>getRandomIndexes(myArray.length, Math.ceil(Math.random() * 6)).map(i =&gt; myArray[i])
</code></pre>
<p>will give you a different sample size from 1-6 every time you call it.</p>
<p>I hope this has helped :D</p>


blocks|key|283745|text|Underscore.js大小约为70kb。如果你不需要所有额外的东西，rando.js只有2kb+(97%25小)，它是这样工作的：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|283746|​|283747|console.log(randoSequence([8,+6,+7,+5,+3,+0,+9]).slice(-5));|code-block|syntax|javascript|283748|<script+src="https://randojs.com/2.0.0.js"></script>|283749|283750|您可以看到，默认情况下，它会跟踪原始索引，以防两个值相同，但您仍然关心选择的是哪一个。如果您不需要这些，您可以只添加一个地图，如下所示：|283751|283752|console.log(randoSequence([8,+6,+7,+5,+3,+0,+9]).slice(-5).map((i)+=>+i.value));|283753|<script+src="https://randojs.com/2.0.0.js"></script>|283754|283755|entityMap|0|LINK|mutability|MUTABLE|url|https://underscorejs.org/|1|https://github.com/nastyox/Rando.js^0|0|D|0|10|8|1|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@$A|16|B|17|1|18]|$A|19|B|1A|1|1B]]|C|$]]|$1|D|3|E|5|6|7|1C|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|1D|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|H|7|1E|8|@]|9|@]|C|$I|J]]|$1|M|3|E|5|6|7|1F|8|@]|9|@]|C|$]]|$1|N|3|O|5|6|7|1G|8|@]|9|@]|C|$]]|$1|P|3|E|5|6|7|1H|8|@]|9|@]|C|$]]|$1|Q|3|R|5|H|7|1I|8|@]|9|@]|C|$I|J]]|$1|S|3|T|5|H|7|1J|8|@]|9|@]|C|$I|J]]|$1|U|3|E|5|6|7|1K|8|@]|9|@]|C|$]]|$1|V|3|-4|5|6|7|1L|8|@]|9|@]|C|$]]]|W|$X|$5|Y|Z|10|C|$11|12]]|13|$5|Y|Z|10|C|$11|14]]]]

<p><a href="https://underscorejs.org/" rel="nofollow noreferrer">Underscore.js</a> is about 70kb. if you don't need all the extra crap, <a href="https://github.com/nastyox/Rando.js" rel="nofollow noreferrer">rando.js</a> is only about 2kb (97% smaller), and it works like this:</p>
<p><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>console.log(randoSequence([8, 6, 7, 5, 3, 0, 9]).slice(-5));</code></pre>
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;script src="https://randojs.com/2.0.0.js"&gt;&lt;/script&gt;</code></pre>
</div>
</div>
</p>
<p>You can see that it keeps track of the original indices by default in case two values are the same but you still care about which one was picked. If you don't need those, you can just add a map, like this:</p>
<p><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>console.log(randoSequence([8, 6, 7, 5, 3, 0, 9]).slice(-5).map((i) =&gt; i.value));</code></pre>
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;script src="https://randojs.com/2.0.0.js"&gt;&lt;/script&gt;</code></pre>
</div>
</div>
</p>


blocks|key|500500|text|D3-array的shuffle使用费舍尔-叶芝打乱算法对数组进行随机重新排序，这是一个变异函数--这意味着原始数组在原地重新排序，这对性能很好。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|500501|D3是针对浏览器的--与node一起使用会更加复杂。|500502|https://github.com/d3/d3-array#shuffle|500503|npm+install+d3-array|code-block|syntax|javascript|500504|​|500505|++++//import+{shuffle}+from+"d3-array"+
++++
++++let+x+=+[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];

++++d3.shuffle(x)

++++console.log(x)+//+it+is+shuffled|500506|<script+src="https://cdnjs.cloudflare.com/ajax/libs/d3/5.0.0/d3.min.js"></script>|500507|500508|如果您不想改变原始数组|500509|500510|++++let+x+=+[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];

++++let+shuffled_x+=+d3.shuffle(x.slice())+//calling+slice+with+no+parameters+returns+a+copy+of+the+original+array

++++console.log(x)+//+not+shuffled
++++console.log(shuffled_x)+|500511|<script+src="https://cdnjs.cloudflare.com/ajax/libs/d3/5.0.0/d3.min.js"></script>|500512|500513|entityMap|0|LINK|mutability|MUTABLE|url|https://observablehq.com/@d3/d3-shuffle|1^0|0|8|9|7|0|0|0|0|12|1|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|F|3|G|5|6|7|1I|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|1J|8|@]|D|@$9|1K|A|1L|1|1M]]|E|$]]|$1|J|3|K|5|L|7|1N|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1O|8|@]|D|@]|E|$]]|$1|Q|3|R|5|L|7|1P|8|@]|D|@]|E|$M|N]]|$1|S|3|T|5|L|7|1Q|8|@]|D|@]|E|$M|N]]|$1|U|3|P|5|6|7|1R|8|@]|D|@]|E|$]]|$1|V|3|W|5|6|7|1S|8|@]|D|@]|E|$]]|$1|X|3|P|5|6|7|1T|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|L|7|1U|8|@]|D|@]|E|$M|N]]|$1|10|3|11|5|L|7|1V|8|@]|D|@]|E|$M|N]]|$1|12|3|P|5|6|7|1W|8|@]|D|@]|E|$]]|$1|13|3|-4|5|6|7|1X|8|@]|D|@]|E|$]]]|14|$15|$5|16|17|18|E|$19|1A]]|1B|$5|16|17|18|E|$19|I]]]]

<p><code>D3-array</code>'s <a href="https://observablehq.com/@d3/d3-shuffle" rel="nofollow noreferrer">shuffle</a> uses the Fisher-Yeates shuffle algorithm to randomly re-order arrays. It is a mutating function - meaning that the original array is re-ordered in place, which is good for performance.</p>
<p>D3 is for the browser - it is more complicated to use with node.</p>
<p><a href="https://github.com/d3/d3-array#shuffle" rel="nofollow noreferrer">https://github.com/d3/d3-array#shuffle</a></p>
<pre class="lang-sh prettyprint-override"><code>npm install d3-array

</code></pre>
<p><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>    //import {shuffle} from "d3-array" 
    
    let x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];

    d3.shuffle(x)

    console.log(x) // it is shuffled</code></pre>
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;script src="https://cdnjs.cloudflare.com/ajax/libs/d3/5.0.0/d3.min.js"&gt;&lt;/script&gt;</code></pre>
</div>
</div>
</p>
<p>If you don't want to mutate the original array</p>
<p><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>    let x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];

    let shuffled_x = d3.shuffle(x.slice()) //calling slice with no parameters returns a copy of the original array

    console.log(x) // not shuffled
    console.log(shuffled_x) </code></pre>
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;script src="https://cdnjs.cloudflare.com/ajax/libs/d3/5.0.0/d3.min.js"&gt;&lt;/script&gt;</code></pre>
</div>
</div>
</p>


<p>What is a clean way of taking a random sample, without replacement from an array in javascript? So suppose there is an array</p>

<pre><code>x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
</code></pre>

<p>and I want to randomly sample 5 unique values; i.e. generate a random subset of length 5. To generate one random sample one could do something like:</p>

<pre><code>x[Math.floor(Math.random()*x.length)];
</code></pre>

<p>But if this is done multiple times, there is a risk of a grabbing the same entry multiple times.</p>


Sampling a random subset from an array

在javascript中不替换数组的情况下，随机抽样的干净方法是什么？假设有一个数组x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]我想随机采样5个唯一的值；即生成一个长度为5的随机子集。要生成一个随机样本，可以这样做：x[Math.floor(Math.random()*x.leng...

问从数组中采样随机子集
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问从数组中采样随机子集EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问从数组中采样随机子集
EN