blocks|key|979406|text|如果您不需要保持顺序，并认为45和"45"是相同的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|979407|​|979408|function+union_arrays+(x,+y)+{
++var+obj+=+{};
++for+(var+i+=+x.length-1;+i+>=+0;+--+i)
+++++obj[x[i]]+=+x[i];
++for+(var+i+=+y.length-1;+i+>=+0;+--+i)
+++++obj[y[i]]+=+y[i];
++var+res+=+[]
++for+(var+k+in+obj)+{
++++if+(obj.hasOwnProperty(k))++//+<--+optional
++++++res.push(obj[k]);
++}
++return+res;
}

console.log(union_arrays([34,35,45,48,49],+[44,55]));|code-block|syntax|javascript|979409|979410|entityMap^0|E|2|H|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|V|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|W|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|O|$]]

<p>If you don't need to keep the order, and consider <code>45</code> and <code>"45"</code> to be the same:</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 union_arrays (x, y) {
  var obj = {};
  for (var i = x.length-1; i &gt;= 0; -- i)
     obj[x[i]] = x[i];
  for (var i = y.length-1; i &gt;= 0; -- i)
     obj[y[i]] = y[i];
  var res = []
  for (var k in obj) {
    if (obj.hasOwnProperty(k))  // &lt;-- optional
      res.push(obj[k]);
  }
  return res;
}

console.log(union_arrays([34,35,45,48,49], [44,55]));</code></pre>
</div>
</div>
</p>


blocks|key|982026|text|随着带有sets和splat操作符的ES6的到来(当时它只在火狐中工作，请检查兼容性表)，你可以编写以下神秘的一行代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|982027|​|982028|var+a+=+[34,+35,+45,+48,+49];
var+b+=+[48,+55];
var+union+=+[...new+Set([...a,+...b])];
console.log(union);|code-block|syntax|javascript|982029|982030|关于这一行的简单解释：[...a,+...b]连接两个数组，你也可以使用a.concat(b)。new+Set()为它创建一个集合，从而创建您的联盟。最后一个[...x]将其转换回数组。|offset|length|style|CODE|982031|entityMap^0|0|0|0|0|B|C|10|B|1C|9|27|6|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]]|$1|I|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|V|8|@$L|W|M|X|N|O]|$L|Y|M|Z|N|O]|$L|10|M|11|N|O]|$L|12|M|13|N|O]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|Q|$]]

<p>With the arrival of ES6 with sets and splat operator (at the time of being works only in Firefox, check compatibility table), you can write the following cryptic one liner:</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>var a = [34, 35, 45, 48, 49];
var b = [48, 55];
var union = [...new Set([...a, ...b])];
console.log(union);</code></pre>
</div>
</div>
</p>

<p>Little explanation about this line: <code>[...a, ...b]</code> concatenates two arrays, you can use <code>a.concat(b)</code> as well. <code>new Set()</code> create a set out of it and thus your union. And the last <code>[...x]</code> converts it back to an array. </p>


blocks|key|4334198|text|如果使用库下划线，则可以这样写|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|4334199|​|4334200|var+unionArr+=+_.union([34,35,45,48,49],+[48,55]);
console.log(unionArr);|code-block|syntax|javascript|4334201|<script+src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.9.1/underscore-min.js"></script>|4334202|4334203|参考：http://underscorejs.org/#union|4334204|entityMap|0|LINK|mutability|MUTABLE|url|http://underscorejs.org/#union^0|5|3|0|0|0|0|0|0|2|3|U|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|12|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|13|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|O|3|G|5|6|7|15|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|16|8|@$9|17|A|18|B|C]]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|R|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|S|$T|$5|U|V|W|E|$X|Y]]]]

<p>If you use the library <strong>underscore</strong> you can write 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>var unionArr = _.union([34,35,45,48,49], [48,55]);
console.log(unionArr);</code></pre>
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.9.1/underscore-min.js"&gt;&lt;/script&gt;</code></pre>
</div>
</div>
</p>

<p><strong>Ref</strong>: <a href="http://underscorejs.org/#union" rel="noreferrer">http://underscorejs.org/#union</a></p>


blocks|key|981895|text|我可能是在一条死线上浪费时间。我只是不得不实现它，然后去看看我是不是在浪费时间。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|981896|我很喜欢KennyTM的回答。这就是我解决问题的方式。将这些键合并到一个散列中，以自然地消除重复项，然后提取这些键。如果你真的有jQuery，你可以利用它的优点把它变成一个2行代码的问题，然后把它变成一个扩展。jQuery中的each()将负责不迭代hasOwnProperty()为false的项。|981897|jQuery.fn.extend({
++++union:+function(array1,+array2)+{
++++++++var+hash+=+{},+union+=+[];
++++++++$.each($.merge($.merge([],+array1),+array2),+function+(index,+value)+{+hash[value]+=+value;+});
++++++++$.each(hash,+function+(key,+value)+{+union.push(key);+}+);
++++++++return+union;
++++}
});|code-block|syntax|javascript|981898|请注意，两个原始数组都保持不变。然后你这样叫它：|981899|var+union+=+$.union(array1,+array2);|981900|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

<p>I'm probably wasting time on a dead thread here. I just had to implement this and went looking to see if I was wasting my time. </p>

<p>I really like KennyTM's answer. That's just how I would attack the problem. Merge the keys into a hash to naturally eliminate duplicates and then extract the keys. If you actually have jQuery you can leverage its goodies to make this a 2 line problem and then roll it into an extension. The each() in jQuery will take care of not iterating over items where hasOwnProperty() is false.</p>

<pre><code>jQuery.fn.extend({
    union: function(array1, array2) {
        var hash = {}, union = [];
        $.each($.merge($.merge([], array1), array2), function (index, value) { hash[value] = value; });
        $.each(hash, function (key, value) { union.push(key); } );
        return union;
    }
});
</code></pre>

<p>Note that both of the original arrays are left intact. Then you call it like this:</p>

<pre><code>var union = $.union(array1, array2);
</code></pre>


blocks|key|2924442|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2924443|function+unique(arrayName)
{
++var+newArray=new+Array();
++label:+for(var+i=0;+i<arrayName.length;i%2B%2B+)
++{++
++++for(var+j=0;+j<newArray.length;j%2B%2B+)
++++{
++++++if(newArray[j]==arrayName[i])+
++++++++continue+label;
++++}
++++newArray[newArray.length]+=+arrayName[i];
++}
++return+newArray;
}

var+arr1+=+new+Array(0,2,4,4,4,4,4,5,5,6,6,6,7,7,8,9,5,1,2,3,0);
var+arr2=+new+Array(3,5,8,1,2,32,1,2,1,2,4,7,8,9,1,2,1,2,3,4,5);
var+union+=+unique(arr1.concat(arr2));
console.log(union);|code-block|syntax|javascript|2924444|2924445|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|K|8|@]|9|@]|A|$E|F]]|$1|G|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|M|8|@]|9|@]|A|$]]]|I|$]]

<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 unique(arrayName)
{
  var newArray=new Array();
  label: for(var i=0; i&lt;arrayName.length;i++ )
  {  
    for(var j=0; j&lt;newArray.length;j++ )
    {
      if(newArray[j]==arrayName[i]) 
        continue label;
    }
    newArray[newArray.length] = arrayName[i];
  }
  return newArray;
}

var arr1 = new Array(0,2,4,4,4,4,4,5,5,6,6,6,7,7,8,9,5,1,2,3,0);
var arr2= new Array(3,5,8,1,2,32,1,2,1,2,4,7,8,9,1,2,1,2,3,4,5);
var union = unique(arr1.concat(arr2));
console.log(union);</code></pre>
</div>
</div>
</p>


blocks|key|4334667|text|如果你想连接两个没有重复值的数组，可以试试这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4334668|var+a=[34,+35,+45,+48,+49];
var+b=[48,+55];
var+c=a.concat(b).sort();
var+res=c.filter((value,pos)+=>+{return+c.indexOf(value)+==+pos;}+);|code-block|syntax|javascript|4334669|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>If you wants to concatenate two arrays without any duplicate value,Just try this</p>

<pre><code>var a=[34, 35, 45, 48, 49];
var b=[48, 55];
var c=a.concat(b).sort();
var res=c.filter((value,pos) =&gt; {return c.indexOf(value) == pos;} );
</code></pre>


blocks|key|4334386|text|改编自https://stackoverflow.com/a/4026828/1830259|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4334387|Array.prototype.union+=+function(a)+
{
++++var+r+=+this.slice(0);
++++a.forEach(function(i)+{+if+(r.indexOf(i)+<+0)+r.push(i);+});
++++return+r;
};

Array.prototype.diff+=+function(a)
{
++++return+this.filter(function(i)+{return+a.indexOf(i)+<+0;});
};

var+s1+=+[1,+2,+3,+4];
var+s2+=+[3,+4,+5,+6];

console.log("s1:+"+%2B+s1);
console.log("s2:+"+%2B+s2);
console.log("s1.union(s2):+"+%2B+s1.union(s2));
console.log("s2.union(s1):+"+%2B+s2.union(s1));
console.log("s1.diff(s2):+"+%2B+s1.diff(s2));
console.log("s2.diff(s1):+"+%2B+s2.diff(s1));

//+Output:
//+s1:+1,2,3,4
//+s2:+3,4,5,6
//+s1.union(s2):+1,2,3,4,5,6
//+s2.union(s1):+3,4,5,6,1,2
//+s1.diff(s2):+1,2
//+s2.diff(s1):+5,6+|code-block|syntax|javascript|4334388|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/4026828/1830259^0|3|17|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>Adapted from: <a href="https://stackoverflow.com/a/4026828/1830259">https://stackoverflow.com/a/4026828/1830259</a></p>

<pre><code>Array.prototype.union = function(a) 
{
    var r = this.slice(0);
    a.forEach(function(i) { if (r.indexOf(i) &lt; 0) r.push(i); });
    return r;
};

Array.prototype.diff = function(a)
{
    return this.filter(function(i) {return a.indexOf(i) &lt; 0;});
};

var s1 = [1, 2, 3, 4];
var s2 = [3, 4, 5, 6];

console.log("s1: " + s1);
console.log("s2: " + s2);
console.log("s1.union(s2): " + s1.union(s2));
console.log("s2.union(s1): " + s2.union(s1));
console.log("s1.diff(s2): " + s1.diff(s2));
console.log("s2.diff(s1): " + s2.diff(s1));

// Output:
// s1: 1,2,3,4
// s2: 3,4,5,6
// s1.union(s2): 1,2,3,4,5,6
// s2.union(s1): 3,4,5,6,1,2
// s1.diff(s2): 1,2
// s2.diff(s1): 5,6 
</code></pre>


blocks|key|981962|text|我喜欢Peter+Ajtai的concat-then-unique解决方案，但是代码不是很清楚。这里有一个更好的选择：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|981963|function+unique(x)+{
++return+x.filter(function(elem,+index)+{+return+x.indexOf(elem)+===+index;+});
};
function+union(x,+y)+{
++return+unique(x.concat(y));
};|code-block|syntax|javascript|981964|由于indexOf返回第一次出现的索引，因此我们将其与当前元素的索引(过滤器谓词的第二个参数)进行比较。|981965|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>I like Peter Ajtai's concat-then-unique solution, but the code's not very clear. Here's a nicer alternative:</p>

<pre><code>function unique(x) {
  return x.filter(function(elem, index) { return x.indexOf(elem) === index; });
};
function union(x, y) {
  return unique(x.concat(y));
};
</code></pre>

<p>Since indexOf returns the index of the <em>first</em> occurence, we check this against the current element's index (the second parameter to the filter predicate).</p>


blocks|key|2924708|text|您可以使用jQuery插件：jQuery+Array+Utilities|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2924709|例如，下面的代码|2924710|$.union([1,+2,+2,+3],+[2,+3,+4,+5,+5])|code-block|syntax|javascript|2924711|将返回1,2,3,4,5|2924712|entityMap|0|LINK|mutability|MUTABLE|url|http://plugins.jquery.com/array-utilities/^0|E|M|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|Z|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

<p>You can use a jQuery plugin: <a href="http://plugins.jquery.com/array-utilities/" rel="nofollow">jQuery Array Utilities</a></p>

<p>For example the code below </p>

<pre><code>$.union([1, 2, 2, 3], [2, 3, 4, 5, 5])
</code></pre>

<p>will return [1,2,3,4,5]</p>


blocks|key|4334626|text|kennytm答案的简短版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4334627|function+unionArrays(a,+b)+{
++++const+cache+=+{};

++++a.forEach(item+=>+cache[item]+=+item);
++++b.forEach(item+=>+cache[item]+=+item);

++++return+Object.keys(cache).map(key+=>+cache[key]);
};|code-block|syntax|javascript|4334628|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>Shorter version of kennytm's answer:
</p>

<pre><code>function unionArrays(a, b) {
    const cache = {};

    a.forEach(item =&gt; cache[item] = item);
    b.forEach(item =&gt; cache[item] = item);

    return Object.keys(cache).map(key =&gt; cache[key]);
};
</code></pre>


blocks|key|1728173|text|我将首先连接数组，然后只返回唯一的值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1728174|您必须创建自己的函数来返回唯一值。既然它是一个有用的函数，你也可以使用add+it+in+as+a+functionality+of+the+Array。|offset|length|style|CODE|1728175|在您使用数组array1和array2的情况下，它将如下所示：|1728176|1728177|array1.concat(array2)+-连接两个arrays|ordered-list-item|1728178|array1.concat(array2).unique()+-仅返回唯一值。这里的unique()是您添加到Array.|1728179|原型中的方法|1728180|整个过程看起来是这样的：|1728181|​|1728182|Array.prototype.unique+=+function+()+{
++++var+r+=+new+Array();
++++o:+for(var+i+=+0,+n+=+this.length;+i+<+n;+i%2B%2B)
++++{
++++++++for(var+x+=+0,+y+=+r.length;+x+<+y;+x%2B%2B)
++++++++{
++++++++++++if(r[x]==this[i])
++++++++++++{
++++++++++++++++continue+o;
++++++++++++}
++++++++}
++++++++r[r.length]+=+this[i];
++++}
++++return+r;
}
var+array1+=+[34,35,45,48,49];
var+array2+=+[34,35,45,48,49,55];

//+concatenate+the+arrays+then+return+only+the+unique+values
console.log(array1.concat(array2).unique());|code-block|syntax|javascript|1728183|1728184|entityMap|0|LINK|mutability|MUTABLE|url|http://www.martienus.com/code/javascript-remove-duplicates-from-array.html^0|0|1Z|5|Z|15|0|0|6|6|D|6|0|0|0|L|0|0|U|16|8|1J|5|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|19|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1A|8|@$D|1B|E|1C|F|G]]|9|@$D|1D|E|1E|1|1F]]|A|$]]|$1|H|3|I|5|6|7|1G|8|@$D|1H|E|1I|F|G]|$D|1J|E|1K|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]|$1|K|3|L|5|M|7|1M|8|@$D|1N|E|1O|F|G]]|9|@]|A|$]]|$1|N|3|O|5|M|7|1P|8|@$D|1Q|E|1R|F|G]|$D|1S|E|1T|F|G]|$D|1U|E|1V|F|G]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1W|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|1X|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|V|3|W|5|X|7|1Z|8|@]|9|@]|A|$Y|Z]]|$1|10|3|U|5|6|7|20|8|@]|9|@]|A|$]]|$1|11|3|-4|5|6|7|21|8|@]|9|@]|A|$]]]|12|$13|$5|14|15|16|A|$17|18]]]]

<p>I would first concatenate the arrays, then I would return only the unique value.</p>

<p>You have to create your own function to return unique values. Since it is a useful function, you might as well <a href="http://www.martienus.com/code/javascript-remove-duplicates-from-array.html" rel="nofollow noreferrer">add it in as a functionality of the <code>Array</code></a>.</p>

<p>In your case with arrays <code>array1</code> and <code>array2</code> it would look like this:</p>

<ol>
<li><code>array1.concat(array2)</code> - concatenate the two arrays</li>
<li><code>array1.concat(array2).unique()</code> - return only the unique values. Here <code>unique()</code> is a method you added to the prototype for <code>Array</code>.</li>
</ol>

<p>The whole thing would look 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>Array.prototype.unique = function () {
    var r = new Array();
    o: for(var i = 0, n = this.length; i &lt; n; i++)
    {
        for(var x = 0, y = r.length; x &lt; y; x++)
        {
            if(r[x]==this[i])
            {
                continue o;
            }
        }
        r[r.length] = this[i];
    }
    return r;
}
var array1 = [34,35,45,48,49];
var array2 = [34,35,45,48,49,55];

// concatenate the arrays then return only the unique values
console.log(array1.concat(array2).unique());</code></pre>
</div>
</div>
</p>


blocks|key|979490|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|979491|function+unionArray(arrayA,+arrayB)+{
++var+obj+=+{},
++++++i+=+arrayA.length,
++++++j+=+arrayB.length,
++++++newArray+=+[];
++while+(i--)+{
++++if+(!(arrayA[i]+in+obj))+{
++++++obj[arrayA[i]]+=+true;
++++++newArray.push(arrayA[i]);
++++}
++}
++while+(j--)+{
++++if+(!(arrayB[j]+in+obj))+{
++++++obj[arrayB[j]]+=+true;
++++++newArray.push(arrayB[j]);
++++}
++}
++return+newArray;
}
var+unionArr+=+unionArray([34,+35,+45,+48,+49],+[44,+55]);
console.log(unionArr);|code-block|syntax|javascript|979492|979493|更快的http://jsperf.com/union-array-faster|offset|length|979494|entityMap|0|LINK|mutability|MUTABLE|url|http://jsperf.com/union-array-faster^0|0|0|0|3|10|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|G|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|W|8|@]|9|@$J|X|K|Y|1|Z]]|A|$]]|$1|L|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|S]]]]

<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 unionArray(arrayA, arrayB) {
  var obj = {},
      i = arrayA.length,
      j = arrayB.length,
      newArray = [];
  while (i--) {
    if (!(arrayA[i] in obj)) {
      obj[arrayA[i]] = true;
      newArray.push(arrayA[i]);
    }
  }
  while (j--) {
    if (!(arrayB[j] in obj)) {
      obj[arrayB[j]] = true;
      newArray.push(arrayB[j]);
    }
  }
  return newArray;
}
var unionArr = unionArray([34, 35, 45, 48, 49], [44, 55]);
console.log(unionArr);</code></pre>
</div>
</div>
</p>

<p>Faster 
<a href="http://jsperf.com/union-array-faster" rel="nofollow noreferrer">http://jsperf.com/union-array-faster</a> </p>


blocks|key|4334289|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4334290|function+unionArrays()+{
++++var+args+=+arguments,
++++l+=+args.length,
++++obj+=+{},
++++res+=+[],
++++i,+j,+k;

++++while+(l--)+{
++++++++k+=+args[l];
++++++++i+=+k.length;

++++++++while+(i--)+{
++++++++++++j+=+k[i];
++++++++++++if+(!obj[j])+{
++++++++++++++++obj[j]+=+1;
++++++++++++++++res.push(j);
++++++++++++}
++++++++}+++
++++}

++++return+res;
}
var+unionArr+=+unionArrays([34,+35,+45,+48,+49],+[44,+55]);
console.log(unionArr);|code-block|syntax|javascript|4334291|4334292|在方法上与alejandro的方法有点类似，但略短一些，应该适用于任何数量的数组。|4334293|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|M|8|@]|9|@]|A|$E|F]]|$1|G|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

<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 unionArrays() {
    var args = arguments,
    l = args.length,
    obj = {},
    res = [],
    i, j, k;

    while (l--) {
        k = args[l];
        i = k.length;

        while (i--) {
            j = k[i];
            if (!obj[j]) {
                obj[j] = 1;
                res.push(j);
            }
        }   
    }

    return res;
}
var unionArr = unionArrays([34, 35, 45, 48, 49], [44, 55]);
console.log(unionArr);</code></pre>
</div>
</div>
</p>

<p>Somewhat similar in approach to alejandro's method, but a little shorter and should work with any number of arrays.</p>


blocks|key|1724732|text|function+unite(arr1,+arr2,+arr3)+{
+newArr=arr1.concat(arr2).concat(arr3);

+a=newArr.filter(function(value){
+++return+!arr1.some(function(value2){
++++++return+value+==+value2;
+++});
+});

console.log(arr1.concat(a));

}//This+is+for+Sorted+union+following+the+order+:)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1724733|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>function unite(arr1, arr2, arr3) {
 newArr=arr1.concat(arr2).concat(arr3);

 a=newArr.filter(function(value){
   return !arr1.some(function(value2){
      return value == value2;
   });
 });

console.log(arr1.concat(a));

}//This is for Sorted union following the order :)
</code></pre>


blocks|key|1728315|text|出于同样的原因，我之前写过(适用于任何数量的数组)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1728316|/**
+*+Returns+with+the+union+of+the+given+arrays.
+*
+*+@param+Any+amount+of+arrays+to+be+united.
+*+@returns+{array}+The+union+array.
+*/
function+uniteArrays()
{
++++var+union+=+[];
++++for+(var+argumentIndex+=+0;+argumentIndex+<+arguments.length;+argumentIndex%2B%2B)
++++{
++++++++eachArgument+=+arguments[argumentIndex];
++++++++if+(typeof+eachArgument+!==+'array')
++++++++{
++++++++++++eachArray+=+eachArgument;
++++++++++++for+(var+index+=+0;+index+<+eachArray.length;+index%2B%2B)
++++++++++++{
++++++++++++++++eachValue+=+eachArray[index];
++++++++++++++++if+(arrayHasValue(union,+eachValue)+==+false)
++++++++++++++++union.push(eachValue);
++++++++++++}
++++++++}
++++}

++++return+union;
}++++

function+arrayHasValue(array,+value)
{+return+array.indexOf(value)+!=+-1;+}|code-block|syntax|javascript|1728317|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>Just wrote before for the same reason (works with any amount of arrays):</p>

<pre><code>/**
 * Returns with the union of the given arrays.
 *
 * @param Any amount of arrays to be united.
 * @returns {array} The union array.
 */
function uniteArrays()
{
    var union = [];
    for (var argumentIndex = 0; argumentIndex &lt; arguments.length; argumentIndex++)
    {
        eachArgument = arguments[argumentIndex];
        if (typeof eachArgument !== 'array')
        {
            eachArray = eachArgument;
            for (var index = 0; index &lt; eachArray.length; index++)
            {
                eachValue = eachArray[index];
                if (arrayHasValue(union, eachValue) == false)
                union.push(eachValue);
            }
        }
    }

    return union;
}    

function arrayHasValue(array, value)
{ return array.indexOf(value) != -1; }
</code></pre>


blocks|key|979617|text|处理合并单个数组值的简单方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|979618|var+values[0]+=+{"id":1235,"name":"value+1"}
values[1]+=+{"id":4323,"name":"value+2"}

var+object=null;
var+first=values[0];
for+(var+i+in+values)+
+if(i>0)++++
+object=+$.merge(values[i],first)|code-block|syntax|javascript|979619|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>Simple way to deal with merging single array values.</p>

<pre><code>var values[0] = {"id":1235,"name":"value 1"}
values[1] = {"id":4323,"name":"value 2"}

var object=null;
var first=values[0];
for (var i in values) 
 if(i&gt;0)    
 object= $.merge(values[i],first)
</code></pre>


blocks|key|4334584|text|你可以尝试这些：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4334585|function+union(a,+b)+{
++++return+a.concat(b).reduce(function(prev,+cur)+{
++++++++if+(prev.indexOf(cur)+===+-1)+prev.push(cur);
++++++++return+prev;
++++},+[]);++++
}|code-block|syntax|javascript|4334586|或|4334587|function+union(a,+b)+{
++++return+a.concat(b.filter(function(el)+{
++++++++return+a.indexOf(el)+===+-1;
++++}));
}|4334588|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|$]]

<p>You can try these:</p>

<pre><code>function union(a, b) {
    return a.concat(b).reduce(function(prev, cur) {
        if (prev.indexOf(cur) === -1) prev.push(cur);
        return prev;
    }, []);    
}
</code></pre>

<p>or</p>

<pre><code>function union(a, b) {
    return a.concat(b.filter(function(el) {
        return a.indexOf(el) === -1;
    }));
}
</code></pre>


blocks|key|982056|text|ES2015版本|type|unstyled|depth|inlineStyleRanges|entityRanges|data|982057|Array.prototype.diff+=+function(a)+{return+this.filter(i+=>+a.indexOf(i)+<+0)};

Array.prototype.union+=+function(a)+{return+[...this.diff(a),+...a]}|code-block|syntax|javascript|982058|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>ES2015 version</p>

<pre><code>Array.prototype.diff = function(a) {return this.filter(i =&gt; a.indexOf(i) &lt; 0)};

Array.prototype.union = function(a) {return [...this.diff(a), ...a]}
</code></pre>


blocks|key|1724805|text|如果您希望自定义的equals函数与您的元素相匹配，您可以在ES2015中使用此函数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1724806|function+unionEquals(left,+right,+equals){
++++return+left.concat(right).reduce(+(acc,element)+=>+{
++++++++return+acc.some(elt+=>+equals(elt,+element))?+acc+:+acc.concat(element)
++++},+[]);
}|code-block|syntax|javascript|1724807|它遍历left%2Bright数组。然后，对于每个元素，如果在累加器中找不到该元素，将填充累加器。最后，没有equals函数指定的重复项。|CODE|1724808|很漂亮，但在处理数千个对象时效率可能不是很高。|1724809|entityMap^0|5|A|0|0|1G|6|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|U|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|V|8|@$9|W|A|X|B|M]]|D|@]|E|$]]|$1|N|3|O|5|6|7|Y|8|@]|D|@]|E|$]]|$1|P|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|Q|$]]

<p>If you want a <strong>custom equals</strong> function to match your elements, you can use this function in ES2015:</p>

<pre><code>function unionEquals(left, right, equals){
    return left.concat(right).reduce( (acc,element) =&gt; {
        return acc.some(elt =&gt; equals(elt, element))? acc : acc.concat(element)
    }, []);
}
</code></pre>

<p>It traverses the left+right array. Then for each element, will fill the accumulator if it does not find that element in the accumulator. At the end, there are no duplicate as specified by the <code>equals</code> function.</p>

<p>Pretty, but probably not very efficient with thousands of objects.</p>


blocks|key|2924818|text|我认为最简单的方法是创建一个新数组，只添加由indexOf确定的唯一值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2924819|在我看来，这似乎是最直接的解决方案，尽管我不知道它是否是最有效的。不保留排序规则。|2924820|var+a+=+[34,+35,+45,+48,+49],
++++b+=+[48,+55];

var+c+=+union(a,+b);

function+union(a,+b)+{+//+will+work+for+n+>=+2+inputs
++++var+newArray+=+[];

++++//cycle+through+input+arrays
++++for+(var+i+=+0,+l+=+arguments.length;+i+<+l;+i%2B%2B)+{

++++++++//cycle+through+each+input+arrays+elements
++++++++var+array+=+arguments[i];
++++++++for+(var+ii+=+0,+ll+=+array.length;+ii+<+ll;+ii%2B%2B)+{
++++++++++++var+val+=+array[ii];

++++++++++++//only+add+elements+to+the+new+array+if+they+are+unique
++++++++++++if+(newArray.indexOf(val)+<+0)+newArray.push(val);
++++++++}
++++}
++++return+newArray;
}|code-block|syntax|javascript|2924821|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>I think it would be simplest to create a new array, adding the unique values only as determined by indexOf.</p>

<p>This seems to me to be the most straightforward solution, though I don't know if it is the most efficient. Collation is not preserved.</p>

<pre><code>var a = [34, 35, 45, 48, 49],
    b = [48, 55];

var c = union(a, b);

function union(a, b) { // will work for n &gt;= 2 inputs
    var newArray = [];

    //cycle through input arrays
    for (var i = 0, l = arguments.length; i &lt; l; i++) {

        //cycle through each input arrays elements
        var array = arguments[i];
        for (var ii = 0, ll = array.length; ii &lt; ll; ii++) {
            var val = array[ii];

            //only add elements to the new array if they are unique
            if (newArray.indexOf(val) &lt; 0) newArray.push(val);
        }
    }
    return newArray;
}
</code></pre>


blocks|key|2924849|text|[i+for(+i+of+new+Set(array1.concat(array2)))]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2924850|让我为你把它分成几部分|unstyled|2924851|//+This+is+a+list+by+comprehension
//+Store+each+result+in+an+element+of+the+array
[i
//+will+be+placed+in+the+variable+"i",+for+each+element+of...
++++for(+i+of
++++//+...+the+Set+which+is+made+of...
++++++++++++new+Set(
++++++++++++++++//+...the+concatenation+of+both+arrays
++++++++++++++++array1.concat(array2)
++++++++++++)
++++)
]|2924852|换句话说，它首先连接两者，然后删除重复项(根据定义，一个集合不能有重复项)|2924853|但请注意，在这种情况下，不能保证元素的顺序。|2924854|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>[i for( i of new Set(array1.concat(array2)))]
</code></pre>

<p>Let me break this into parts for you</p>

<pre><code>// This is a list by comprehension
// Store each result in an element of the array
[i
// will be placed in the variable "i", for each element of...
    for( i of
    // ... the Set which is made of...
            new Set(
                // ...the concatenation of both arrays
                array1.concat(array2)
            )
    )
]
</code></pre>

<p>In other words, it first concatenates both and then it removes the duplicates (a Set, by definition cannot have duplicates)</p>

<p>Do note, though, that the order of the elements is not guaranteed, in this case.</p>


<p>Say I have an array of <code>[34, 35, 45, 48, 49]</code> and another array of <code>[48, 55]</code>. How can I get a resulting array of <code>[34, 35, 45, 48, 49, 55]</code>?</p>


Getting a union of two arrays 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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假设我有一个[34, 35, 45, 48, 49]数组和另一个[48, 55]数组。如何获得[34, 35, 45, 48, 49, 55]的结果数组

问在JavaScript中获取两个数组的并集
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回答 21

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回答 21

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问在JavaScript中获取两个数组的并集
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