blocks|key|2152465|text|您可以使用给定数组的副本并生成不重复的随机项。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2152466|​|2152467|function+generate(count,+values)+{
++++return+Array.apply(null,+{+length:+count+}).map(function+()+{
++++++++var+r+=+[],
++++++++++++array+=+values.slice();
++++++++while+(array.length)+{
++++++++++++r.push(array.splice(Math.floor(Math.random()+*+array.length),+1)[0]);
++++++++}
++++++++return+r;
++++});
}

console.log(generate(7,+[1,+2,+3,+4,+5,+6,+7]));|code-block|syntax|javascript|2152468|2152469|随机化单个数组|2152470|2152471|function+generate(values)+{
++++var+r+=+[],
++++++++array+=+values.slice();
++++while+(array.length)+{
++++++++r.push(array.splice(Math.floor(Math.random()+*+array.length),+1)[0]);
++++}
++++return+r;
}

console.log(generate([1,+2,+3,+4,+5,+6,+7]));|2152472|2152473|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|V|8|@]|9|@]|A|$]]|$1|L|3|C|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|X|8|@]|9|@]|A|$G|H]]|$1|O|3|C|5|6|7|Y|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|Q|$]]

<p>You could use an copy of the given array and generate random items without repeat.</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 generate(count, values) {
    return Array.apply(null, { length: count }).map(function () {
        var r = [],
            array = values.slice();
        while (array.length) {
            r.push(array.splice(Math.floor(Math.random() * array.length), 1)[0]);
        }
        return r;
    });
}

console.log(generate(7, [1, 2, 3, 4, 5, 6, 7]));</code></pre>
</div>
</div>
</p>

<p>Randomize a single 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>function generate(values) {
    var r = [],
        array = values.slice();
    while (array.length) {
        r.push(array.splice(Math.floor(Math.random() * array.length), 1)[0]);
    }
    return r;
}

console.log(generate([1, 2, 3, 4, 5, 6, 7]));</code></pre>
</div>
</div>
</p>


blocks|key|2152485|text|如果我们想把这个问题推广到一个N维数组的一般情形中，那么一个可重用的方法如下；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2152486|我们将使用两个我非常喜欢使用的泛型函数：Array.prototype.clone()和Array.prototype.shuffle()。我们的arrayND函数有不确定的参数数。最后一个参数将指定要填充的随机整数的最小(基)。以前的参数将给出每个维度的长度。因此，在特殊情况下，由于我们需要一个2D+7x7矩阵来填充随机但唯一的数字，从1开始，我们将调用我们作为arrayND(7,7,1)的函数。|offset|length|style|CODE|2152487|​|2152488|Array.prototype.shuffle+=+function(){
++var+i+=+this.length,
++++++j;
++while+(i+>+1)+{
++++j+=+~~(Math.random()*i--);
++++[this[i],this[j]]+=+[this[j],this[i]];
++}
return+this;
};

Array.prototype.clone+=+function(){
++return+this.map(e+=>+Array.isArray(e)+?+e.clone()+:+e);
};

function+arrayND(...n){
++return+n.reduceRight((p,c)+=>+c+=+Array(...Array(c)).map((e,i)+=>+Array.isArray(p)+?+p.clone().shuffle()+:+i%2Bp));
}

var+arr+=+arrayND(7,7,1);
console.log(arr);|code-block|syntax|javascript|2152489|2152490|entityMap^0|0|K|N|18|P|21|7|53|E|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@$D|T|E|U|F|G]|$D|V|E|W|F|G]|$D|X|E|Y|F|G]|$D|Z|E|10|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|11|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|12|8|@]|9|@]|A|$M|N]]|$1|O|3|I|5|6|7|13|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|Q|$]]

<p>If we would like to extend the question for a general case of an N dimensional array filled with random integers then a reusable approach could be as follows;</p>

<p>We will use two generic functions that i like to use very much; <code>Array.prototype.clone()</code> and <code>Array.prototype.shuffle()</code>. Our <code>arrayND</code> function takes indefinite number of arguments. The last argument will designate the minimum (base) of the random integer to be filled. Previous arguments will give the length of each dimension. So in the particular case as we will need a 2D 7x7 matrice to be filled with random but unique numbers in each starting from 1, we shall invoke our function as <code>arrayND(7,7,1)</code></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>Array.prototype.shuffle = function(){
  var i = this.length,
      j;
  while (i &gt; 1) {
    j = ~~(Math.random()*i--);
    [this[i],this[j]] = [this[j],this[i]];
  }
return this;
};

Array.prototype.clone = function(){
  return this.map(e =&gt; Array.isArray(e) ? e.clone() : e);
};

function arrayND(...n){
  return n.reduceRight((p,c) =&gt; c = Array(...Array(c)).map((e,i) =&gt; Array.isArray(p) ? p.clone().shuffle() : i+p));
}

var arr = arrayND(7,7,1);
console.log(arr);</code></pre>
</div>
</div>
</p>


blocks|key|1137011|text|由于您希望将这些数组放在一起处理，所以应该将它们放在一个数据结构中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1137012|var+spans+=+[+one,+two,+tree,+...,+seven+]|code-block|syntax|javascript|1137013|现在，您想要一个不重复的随机顺序，将可能的索引放入Array并对其进行洗牌：|1137014|var+indices+=+[0,1,2,3,4,5,6];|1137015|还有一些洗牌的人：|1137016|function+shuffle(arr){
++++for(var+i=arr.length,+j,+tmp;+i--+>+1;+){
++++++++tmp+=+arr[j+=+0%7C(Math.random()*i)];
++++++++arr[j]+=+arr[i];
++++++++arr[i]+=+tmp;
++++}
++++return+arr;
}|1137017|现在，如果您想要节点总数：|1137018|var+total+=+shuffle(indices).map((index,+n)+=>+spans[n][index]);|1137019|至少就我理解你的问题/代码/意图而言。|1137020|entityMap^0|0|0|0|0|0|0|0|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]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

<p>since you want these Arrays to be processed together you should put them together into a data-structure:</p>

<pre><code>var spans = [ one, two, tree, ..., seven ]
</code></pre>

<p>now you want a random order without repetition, put the possible indices into an Array and shuffle that:</p>

<pre><code>var indices = [0,1,2,3,4,5,6];
</code></pre>

<p>and some shuffler:</p>

<pre><code>function shuffle(arr){
    for(var i=arr.length, j, tmp; i-- &gt; 1; ){
        tmp = arr[j = 0|(Math.random()*i)];
        arr[j] = arr[i];
        arr[i] = tmp;
    }
    return arr;
}
</code></pre>

<p>now if you want the Nodes for total:</p>

<pre><code>var total = shuffle(indices).map((index, n) =&gt; spans[n][index]);
</code></pre>

<p>at least as far as I understand your question/code/intentions.</p>


<p>I have an array of array with elements for example:</p>

<pre><code>var one = [1span,2span,3span,4span,5span,6span,7span];
var two = [1span,2span,3span,4span,5span,6span,7span];
var three = [1span,2span,3span,4span,5span,6span,7span];
var ...till seven.
var total = [one,two,three ..till 7]
</code></pre>

<p>so basically we have 7 arrays and <code>total variable</code> will display 7 elements for each one of all 7 arrays. </p>

<p>now I have a function that should populate my variables with distinct numbers from 1 to 7 on each section.</p>

<pre><code>   var bar = [1, 2, 3, 4, 5, 6, 7];
    for(var j=0; j&lt;total.length; j++) {
      Array.from(total[j]).forEach(function(e, i, a) {
       e.textContent = bar[Math.round(Math.random()*(bar.length-1))];
      console.log(e,i,k);
});
    }
</code></pre>

<p>all good my function does that but unfortunately is populating span elements with values from <code>bar variable</code> for each variable from <code>total var</code> 7 times for each and should populate just once for each variable.</p>

<p>So my problem is:</p>

<ol>
<li>I want to populate each variable from <code>total var</code> with values from <code>bar array</code> just once.</li>
<li>The values should be randomly and unique for each variable.</li>
</ol>


populating array of arrays with distinct values

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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我有一个包含元素的数组，例如：var one = [1span,2span,3span,4span,5span,6span,7span];var two = [1span,2span,3span,4span,5span,6span,7span];var three = [1span,2span,3span,4span,...

问使用不同的值填充数组数组
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