blocks|key|44540|text|假设你有一个这样的Object：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|44541|var+obj+=+{a:+1,+b:+2,+undefined:+1}|code-block|syntax|javascript|44542|你可以做到的|44543|var+max+=+Math.max.apply(null,Object.keys(obj).map(function(x){+return+obj[x]+}));
console.log(Object.keys(obj).filter(function(x){+return+obj[x]+==+max;+})[0]);|44544|entityMap^0|9|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

Supposing you've an <code>Object</code> like this:

<pre><code>var obj = {a: 1, b: 2, undefined: 1}
</code></pre>

You can do this

<pre><code>var max = Math.max.apply(null,Object.keys(obj).map(function(x){ return obj[x] }));
console.log(Object.keys(obj).filter(function(x){ return obj[x] == max; })[0]);
</code></pre>

blocks|key|4551114|text|非常基本的方法。处理起来可能会很慢|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4551115|var+v+=+{a:+1,+b:+2,+undefined:+1};

function+geth(o){
++++var+vals+=+[];++++
++++for(var+i+in+o){
+++++++vals.push(o[i]);
++++}

++++var+max+=+Math.max.apply(null,+vals);

+++++for(var+i+in+o){
++++++++if(o[i]+==+max){
++++++++++++return+i;
++++++++}
++++}
}

console.log(geth(v));|code-block|syntax|javascript|4551116|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|$]]

Very basic method. might be slow to process

<pre><code>var v = {a: 1, b: 2, undefined: 1};

function geth(o){
 var vals = []; 
 for(var i in o){
 vals.push(o[i]);
 }

 var max = Math.max.apply(null, vals);

 for(var i in o){
 if(o[i] == max){
 return i;
 }
 }
}

console.log(geth(v));
</code></pre>

blocks|key|4551129|text|使用下划线或短划线：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4551130|var+maxKey+=+_.max(Object.keys(obj),+function+(o)+{+return+obj[o];+});|code-block|syntax|javascript|4551131|使用ES6箭头函数：|4551132|var+maxKey+=+_.max(Object.keys(obj),+o+=>+obj[o]);|4551133|jsFiddle+demo|offset|length|4551134|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/jt631Lcc/^0|0|0|0|0|0|D|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]]]]

Using Underscore or Lo-Dash:

<pre><code>var maxKey = _.max(Object.keys(obj), function (o) { return obj[o]; });
</code></pre>

With ES6 Arrow Functions:

<pre><code>var maxKey = _.max(Object.keys(obj), o =&gt; obj[o]);
</code></pre>

<a href="http://jsfiddle.net/jt631Lcc/" rel="noreferrer">jsFiddle demo</a>

blocks|key|4551144|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4551145|const+data+=+{
++abc:+'',
++myShortKey:+'',
++myVeryLongKey:+''
}


const+longestKeyLength+=+Math.max(...Object.keys(data).map((d)+=>+d.length));

console.log(longestKeyLength);

console.log(Object.entries(data).map(([key,+val])+=>+`${key}:+${key.length}`));|code-block|syntax|javascript|4551146|4551147|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|$]]

<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="true">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const data = {
 abc: '',
 myShortKey: '',
 myVeryLongKey: ''
}


const longestKeyLength = Math.max(...Object.keys(data).map((d) =&gt; d.length));

console.log(longestKeyLength);

console.log(Object.entries(data).map(([key, val]) =&gt; `${key}: ${key.length}`));</code></pre>
</div>
</div>

blocks|key|44604|text|let+data+=+{a:1,b:2,undefined:3}
let+maxValue+=+Object.entries(data).sort((x,y)=>y[1]-x[1])[0]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|44605|注意:这是一个非常昂贵的过程，如果与大尺寸对象(>=1000000)一起使用，会阻塞事件循环。对于大型数组，对条目进行切片，并使用setTimeout递归地调用上述方法。|unstyled|44606|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>let data = {a:1,b:2,undefined:3}
let maxValue = Object.entries(data).sort((x,y)=&gt;y[1]-x[1])[0]
</code></pre>
note: this is a very expensive process and would block the event loop if used with objects of large sizes(&gt;=1000000). with large array slice the entries and call the above method recurrently using setTimeout.

blocks|key|4551190|text|结合其他答案的一些想法。这将获得具有最大值的所有键，但使用using运算符来获得最大值，然后使用filter数组方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4551191|const+getMax+=+object+=>+{
++let+max+=+Math.max(...Object.values(object))
++return+Object.keys(object).filter(key+=>+object[key]==max)
}

let+obj+=+{a:+12,+b:+11,+c:+12};
getMax(obj)|code-block|syntax|javascript|4551192|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|$]]

Combination of some ideas from other answers. This will get all the keys with the highest value, but using the spread operator to get the maximum value and then filter array method:
<pre><code>const getMax = object =&gt; {
 let max = Math.max(...Object.values(object))
 return Object.keys(object).filter(key =&gt; object[key]==max)
}

let obj = {a: 12, b: 11, c: 12};
getMax(obj)
</code></pre>

I have a object like that one:

<pre><code>Object {a: 1, b: 2, undefined: 1} 
</code></pre>

How can I quickly pull the largest value identifier (here: <code>b</code>) from it? I tried converting it to array and then sorting, but it didn't work out, since it got sorted alphabetically (and it seems like a overkill to juggle data back and forth just for getting one value out of three).

Getting key with the highest value from object

我有一个类似的对象：Object {a: 1, b: 2, undefined: 1} 如何快速地从中提取最大的值标识符(这里：b)？我尝试将其转换为数组，然后进行排序，但没有成功，因为它是按字母顺序排序的(而且仅仅为了获得三个值中的一个值而来回处理数据似乎有点过分)。

问从对象中获取具有最高值的键
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