blocks|key|56587|text|如果你想节省一些周期，我会使用一个对象和一个数组：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|56588|var+lookup+=+{};
var+items+=+json.DATA;
var+result+=+[];

for+(var+item,+i+=+0;+item+=+items[i%2B%2B];)+{
++var+name+=+item.name;

++if+(!(name+in+lookup))+{
++++lookup[name]+=+1;
++++result.push(name);
++}
}|code-block|syntax|javascript|56589|这样基本上就避免了indexOf+/+inArray调用，并且您将得到一个迭代速度比迭代对象属性更快的数组-这也是因为在第二种情况下，您需要检查hasOwnProperty。|offset|length|style|CODE|56590|当然，如果你只需要一个对象，你就完全可以避免检查和result.push，如果使用Object.keys(lookup)来获取数组，但它不会比这更快。|56591|entityMap^0|0|0|9|7|J|7|20|E|0|P|B|15|J|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@$I|10|J|11|K|L]|$I|12|J|13|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$]]

I would use one Object and one Array, if you want to save some cycle:

<pre><code>var lookup = {};
var items = json.DATA;
var result = [];

for (var item, i = 0; item = items[i++];) {
 var name = item.name;

 if (!(name in lookup)) {
 lookup[name] = 1;
 result.push(name);
 }
}
</code></pre>

In this way you're basically avoiding the <code>indexOf</code> / <code>inArray</code> call, and you will get an Array that can be iterate quicker than iterating object's properties – also because in the second case you need to check <code>hasOwnProperty</code>.

Of course if you're fine with just an Object you can avoid the check and the <code>result.push</code> at all, and in case get the array using <code>Object.keys(lookup)</code>, but it won't be faster than that.

blocks|key|273157|text|这是一个减少的好地方。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|273158|var+uniqueArray+=+o.DATA.reduce(function+(a,+d)+{
+++++++if+(a.indexOf(d.name)+===+-1)+{
+++++++++a.push(d.name);
+++++++}
+++++++return+a;
++++},+[]);|code-block|syntax|javascript|273159|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|$]]

This is a great spot for a reduce

<pre><code>var uniqueArray = o.DATA.reduce(function (a, d) {
 if (a.indexOf(d.name) === -1) {
 a.push(d.name);
 }
 return a;
 }, []);
</code></pre>

blocks|key|56644|text|使用唯一的Jquery方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|56645|var+UniqueNames=+$.unique(data.DATA.map(function+(d)+{return+d.name;}));

alert($.unique(names));|code-block|syntax|javascript|56646|JSFiddle|offset|length|56647|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/hardiksondagar/QCx4a/^0|0|0|0|8|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]]]]

Use Jquery Method unique.

<pre><code>var UniqueNames= $.unique(data.DATA.map(function (d) {return d.name;}));

alert($.unique(names));
</code></pre>

<a href="http://jsfiddle.net/hardiksondagar/QCx4a/" rel="noreferrer">JSFiddle</a>

blocks|key|56676|text|对于这类事情，Underscore.js非常好用。您可以使用_.countBy()来获取每个name的计数|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|56677|data+=+[{"id":11,"name":"ajax","subject":"OR","mark":63},
++++++++{"id":12,"name":"javascript","subject":"OR","mark":63},
++++++++{"id":13,"name":"jquery","subject":"OR","mark":63},
++++++++{"id":14,"name":"ajax","subject":"OR","mark":63},
++++++++{"id":15,"name":"jquery","subject":"OR","mark":63},
++++++++{"id":16,"name":"ajax","subject":"OR","mark":63},
++++++++{"id":20,"name":"ajax","subject":"OR","mark":63}]

_.countBy(data,+function(data)+{+return+data.name;+});|code-block|syntax|javascript|56678|提供：|56679|{ajax:+4,+javascript:+1,+jquery:+2}+|56680|对于键的数组，只需使用_.keys()|56681|_.keys(_.countBy(data,+function(data)+{+return+data.name;+}));|56682|56683|["ajax",+"javascript",+"jquery"]|56684|entityMap|0|LINK|mutability|MUTABLE|url|http://underscorejs.org/|1|http://underscorejs.org/#countBy|2|http://underscorejs.org/#keys^0|U|B|1A|4|7|D|0|U|B|1|0|0|0|0|B|8|B|8|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@$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]]|$1|K|3|L|5|6|7|1J|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1K|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1L|8|@$9|1M|A|1N|B|C]]|D|@$9|1O|A|1P|1|1Q]]|E|$]]|$1|Q|3|R|5|H|7|1R|8|@]|D|@]|E|$I|J]]|$1|S|3|L|5|6|7|1S|8|@]|D|@]|E|$]]|$1|T|3|U|5|H|7|1T|8|@]|D|@]|E|$I|J]]|$1|V|3|-4|5|6|7|1U|8|@]|D|@]|E|$]]]|W|$X|$5|Y|Z|10|E|$11|12]]|13|$5|Y|Z|10|E|$11|14]]|15|$5|Y|Z|10|E|$11|16]]]]

<a href="http://underscorejs.org/" rel="noreferrer">Underscore.js</a> is great for this kind of thing. You can use <a href="http://underscorejs.org/#countBy" rel="noreferrer"><code>_.countBy()</code></a> to get the counts per <code>name</code>:

<pre><code>data = [{"id":11,"name":"ajax","subject":"OR","mark":63},
 {"id":12,"name":"javascript","subject":"OR","mark":63},
 {"id":13,"name":"jquery","subject":"OR","mark":63},
 {"id":14,"name":"ajax","subject":"OR","mark":63},
 {"id":15,"name":"jquery","subject":"OR","mark":63},
 {"id":16,"name":"ajax","subject":"OR","mark":63},
 {"id":20,"name":"ajax","subject":"OR","mark":63}]

_.countBy(data, function(data) { return data.name; });
</code></pre>

Gives:

<pre><code>{ajax: 4, javascript: 1, jquery: 2} 
</code></pre>

For an array of the keys just use <a href="http://underscorejs.org/#keys" rel="noreferrer"><code>_.keys()</code></a>

<pre><code>_.keys(_.countBy(data, function(data) { return data.name; }));
</code></pre>

Gives:

<pre><code>["ajax", "javascript", "jquery"]
</code></pre>

blocks|key|56725|text|首先，我们可以只运行map()函数来获得新的数组，该数组的结果是在varjson.DATA中的每个元素上调用所提供的函数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|56726|varjson.DATA.map(({name})=>name))|code-block|syntax|javascript|56727|在从varjson.DATA获得name数组之后。我们可以将其转换为一个集合，该集合将丢弃数组的所有重复条目，并应用spread+operator来获得唯一名称的数组：|56728|[...new+Set(varjson.DATA.map(({name})=>name))]|56729|​|56730|const+varjson+=+{
++"DATA":+[{
++++++"id":+11,
++++++"name":+"ajax",
++++++"subject":+"OR",
++++++"mark":+63
++++},
++++{
++++++"id":+12,
++++++"name":+"javascript",
++++++"subject":+"OR",
++++++"mark":+63
++++},
++++{
++++++"id":+13,
++++++"name":+"jquery",
++++++"subject":+"OR",
++++++"mark":+63
++++},
++++{
++++++"id":+14,
++++++"name":+"ajax",
++++++"subject":+"OR",
++++++"mark":+63
++++},
++++{
++++++"id":+15,
++++++"name":+"jquery",
++++++"subject":+"OR",
++++++"mark":+63
++++},
++++{
++++++"id":+16,
++++++"name":+"ajax",
++++++"subject":+"OR",
++++++"mark":+63
++++},
++++{
++++++"id":+20,
++++++"name":+"ajax",
++++++"subject":+"OR",
++++++"mark":+63
++++}
++],
++"COUNT":+"120"
}

console.log(+[...new+Set(varjson.DATA.map(({name})=>name))]);|56731|56732|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax^0|A|5|X|C|0|0|2|C|G|4|1M|F|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@$9|1C|A|1D|1|1E]]|E|$]]|$1|M|3|N|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1G|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1H|8|@]|D|@]|E|$I|J]]|$1|S|3|P|5|6|7|1I|8|@]|D|@]|E|$]]|$1|T|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|U|$V|$5|W|X|Y|E|$Z|10]]]]

First we can just run <code>map()</code> function to get the new array with the results of calling a provided function on every element in the <code>varjson.DATA</code>.

<pre><code>varjson.DATA.map(({name})=&gt;name))
</code></pre>

After getting the array of <code>name</code> from the <code>varjson.DATA</code>. We can convert it into a set that will discard all duplicate entries of array and apply <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax" rel="nofollow noreferrer">spread operator</a> to get a array of unique names:

<pre><code>[...new Set(varjson.DATA.map(({name})=&gt;name))]
</code></pre>

<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>const varjson = {
 "DATA": [{
 "id": 11,
 "name": "ajax",
 "subject": "OR",
 "mark": 63
 },
 {
 "id": 12,
 "name": "javascript",
 "subject": "OR",
 "mark": 63
 },
 {
 "id": 13,
 "name": "jquery",
 "subject": "OR",
 "mark": 63
 },
 {
 "id": 14,
 "name": "ajax",
 "subject": "OR",
 "mark": 63
 },
 {
 "id": 15,
 "name": "jquery",
 "subject": "OR",
 "mark": 63
 },
 {
 "id": 16,
 "name": "ajax",
 "subject": "OR",
 "mark": 63
 },
 {
 "id": 20,
 "name": "ajax",
 "subject": "OR",
 "mark": 63
 }
 ],
 "COUNT": "120"
}

console.log( [...new Set(varjson.DATA.map(({name})=&gt;name))]);</code></pre>
</div>
</div>

blocks|key|273307|text|试试这个，MYJSON将成为你的json数据。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|273308|var+mytky=[];
mytky=DistinctRecords(MYJSON,"mykeyname");

function+DistinctRecords(MYJSON,prop)+{
++return+MYJSON.filter((obj,+pos,+arr)+=>+{
++++return+arr.map(mapObj+=>+mapObj[prop]).indexOf(obj[prop])+===+pos;
+})
}|code-block|syntax|javascript|273309|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|$]]

try this, MYJSON will be your json data.

<pre><code>var mytky=[];
mytky=DistinctRecords(MYJSON,"mykeyname");

function DistinctRecords(MYJSON,prop) {
 return MYJSON.filter((obj, pos, arr) =&gt; {
 return arr.map(mapObj =&gt; mapObj[prop]).indexOf(obj[prop]) === pos;
 })
}
</code></pre>

blocks|key|56795|text|试一试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|56796|var+distinct_list+

++=+data.DATA.map(function+(d)+{return+d[x];}).filter((v,+i,+a)+=>+a.indexOf(v)+===+i)|code-block|syntax|javascript|56797|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|$]]

Give this a go:
<pre><code>var distinct_list 

 = data.DATA.map(function (d) {return d[x];}).filter((v, i, a) =&gt; a.indexOf(v) === i)
</code></pre>

I have my JSON as follows

<pre><code>{"DATA": [{"id":11,"name":"ajax","subject":"OR","mark":63},
{"id":12,"name":"javascript","subject":"OR","mark":63},
{"id":13,"name":"jquery","subject":"OR","mark":63},
{"id":14,"name":"ajax","subject":"OR","mark":63},
{"id":15,"name":"jquery","subject":"OR","mark":63},
{"id":16,"name":"ajax","subject":"OR","mark":63},
{"id":20,"name":"ajax","subject":"OR","mark":63}],"COUNT":"120"}
</code></pre>

Is there any good method to find out the <code>distinct name</code> from this JSON

Result <code>javascript,jquery,ajax</code>

I can do this using following methode 

<pre><code>var arr=[''];
var j=0;
for (var i = 0; i &lt; varjson.DATA.length; i++) {
 if($.inArray(varjson.DATA[i]["name"],arr)&lt;0){
 arr[j]=varjson.DATA[i]["name"];
 j++;
 }
}
</code></pre>

Is there any <code>better method</code> which gave me better performance?

Selecting distinct values from a JSON

JSON

我的JSON如下所示{"DATA": [{"id":11,"name":"ajax","subject":"OR","mark":63},{"id":12,"name":"javascript","subject":"OR","mark":63},{"id":13,"name":"jquery","subject":"...

问从JSON中选择不同的值
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