blocks|key|4384626|text|在普通的Javascript中，您可以对对象使用Array#reduce|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4384627|​|4384628|var+cars+=+[{+make:+'audi',+model:+'r8',+year:+'2012'+},+{+make:+'audi',+model:+'rs5',+year:+'2013'+},+{+make:+'ford',+model:+'mustang',+year:+'2012'+},+{+make:+'ford',+model:+'fusion',+year:+'2015'+},+{+make:+'kia',+model:+'optima',+year:+'2012'+}],
++++result+=+cars.reduce(function+(r,+a)+{
++++++++r[a.make]+=+r[a.make]+%7C%7C+[];
++++++++r[a.make].push(a);
++++++++return+r;
++++},+Object.create(null));

console.log(result);|code-block|syntax|javascript|4384629|.as-console-wrapper+{+max-height:+100%25+!important;+top:+0;+}|4384630|4384631|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce^0|O|C|O|C|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@$9|10|A|11|1|12]]|E|$]]|$1|F|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|O|3|G|5|6|7|16|8|@]|D|@]|E|$]]|$1|P|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|Q|$R|$5|S|T|U|E|$V|W]]]]

<p>In plain Javascript, you could use <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce" rel="noreferrer"><code>Array#reduce</code></a> with an object</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 cars = [{ make: 'audi', model: 'r8', year: '2012' }, { make: 'audi', model: 'rs5', year: '2013' }, { make: 'ford', model: 'mustang', year: '2012' }, { make: 'ford', model: 'fusion', year: '2015' }, { make: 'kia', model: 'optima', year: '2012' }],
    result = cars.reduce(function (r, a) {
        r[a.make] = r[a.make] || [];
        r[a.make].push(a);
        return r;
    }, Object.create(null));

console.log(result);</code></pre>
<pre class="snippet-code-css lang-css prettyprint-override"><code>.as-console-wrapper { max-height: 100% !important; top: 0; }</code></pre>
</div>
</div>
</p>


blocks|key|4384115|text|我会用Timo's+answer做这件事。简单的_.groupBy，并允许在分组结构中的对象中进行一些复制。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4384116|但是，OP还要求删除重复的make密钥。如果你想走完全程：|4384117|var+grouped+=+_.mapValues(_.groupBy(cars,+'make'),
++++++++++++++++++++++++++clist+=>+clist.map(car+=>+_.omit(car,+'make')));

console.log(grouped);|code-block|syntax|javascript|4384118|收益率：|4384119|{+audi:
+++[+{+model:+'r8',+year:+'2012'+},
+++++{+model:+'rs5',+year:+'2013'+}+],
++ford:
+++[+{+model:+'mustang',+year:+'2012'+},
+++++{+model:+'fusion',+year:+'2015'+}+],
++kia:+
+++[+{+model:+'optima',+year:+'2012'+}+]+
}|4384120|如果您想使用Underscore.js来实现这一点，请注意它的_.mapValues版本称为_.mapObject。|4384121|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/40774759/240490^0|O|9|3|D|0|0|D|4|0|0|0|0|V|B|1A|B|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|6|7|16|8|@$9|17|A|18|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|19|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1A|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1B|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1C|8|@$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

<p><a href="https://stackoverflow.com/a/40774759/240490">Timo's answer</a> is how I would do it. Simple <code>_.groupBy</code>, and allow some duplications in the objects in the grouped structure.</p>
<p>However the OP also asked for the duplicate <code>make</code> keys to be removed. If you wanted to go all the way:</p>

<pre class="lang-js prettyprint-override"><code>var grouped = _.mapValues(_.groupBy(cars, 'make'),
                          clist =&gt; clist.map(car =&gt; _.omit(car, 'make')));

console.log(grouped);
</code></pre>
<p>Yields:</p>
<pre class="lang-js prettyprint-override"><code>{ audi:
   [ { model: 'r8', year: '2012' },
     { model: 'rs5', year: '2013' } ],
  ford:
   [ { model: 'mustang', year: '2012' },
     { model: 'fusion', year: '2015' } ],
  kia: 
   [ { model: 'optima', year: '2012' } ] 
}
</code></pre>
<p>If you wanted to do this using Underscore.js, note that its version of <code>_.mapValues</code> is called <code>_.mapObject</code>.</p>


blocks|key|4384710|text|您可以尝试修改_.groupBy函数每次迭代调用的函数中的对象。注意源数组改变了他的元素！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4384711|var+res+=+_.groupBy(cars,(car)=>{
++++const+makeValue=car.make;
++++delete+car.make;
++++return+makeValue;
})
console.log(res);
console.log(cars);|code-block|syntax|javascript|4384712|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 try to modify the object inside the function called per iteration by _.groupBy func. 
Notice that the source array change his elements!</p>

<pre><code>var res = _.groupBy(cars,(car)=&gt;{
    const makeValue=car.make;
    delete car.make;
    return makeValue;
})
console.log(res);
console.log(cars);
</code></pre>


blocks|key|4384770|text|完全没有理由下载第三方库来实现这个简单的问题，就像上面的解决方案所建议的那样。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4384771|根据es6中的某个list对对象的key进行分组的单行版本：|offset|length|style|CODE|4384772|const+groupByKey+=+(list,+key)+=>+list.reduce((hash,+obj)+=>+({...hash,+[obj[key]]:(+hash[obj[key]]+%7C%7C+[]+).concat(obj)}),+{})|code-block|syntax|javascript|4384773|过滤掉没有key的对象的较长版本|4384774|​|4384775|function+groupByKey(array,+key)+{
+++return+array
+++++.reduce((hash,+obj)+=>+{
+++++++if(obj[key]+===+undefined)+return+hash;+
+++++++return+Object.assign(hash,+{+[obj[key]]:(+hash[obj[key]]+%7C%7C+[]+).concat(obj)})
+++++},+{})
}


var+cars+=+[{'make':'audi','model':'r8','year':'2012'},{'make':'audi','model':'rs5','year':'2013'},{'make':'ford','model':'mustang','year':'2012'},{'make':'ford','model':'fusion','year':'2015'},{'make':'kia','model':'optima','year':'2012'}];

console.log(groupByKey(cars,+'make'))|4384776|4384777|注意:原来的问题似乎是问如何按制造商对汽车进行分组，但省略了每组中的制造商。因此，如果没有第三方库，简短的答案将如下所示：|4384778|4384779|const+groupByKey+=+(list,+key,+{omitKey=false})+=>+list.reduce((hash,+{[key]:value,+...rest})+=>+({...hash,+[value]:(+hash[value]+%7C%7C+[]+).concat(omitKey+?+{...rest}+:+{[key]:value,+...rest})}+),+{})

var+cars+=+[{'make':'audi','model':'r8','year':'2012'},{'make':'audi','model':'rs5','year':'2013'},{'make':'ford','model':'mustang','year':'2012'},{'make':'ford','model':'fusion','year':'2015'},{'make':'kia','model':'optima','year':'2012'}];

console.log(groupByKey(cars,+'make',+{omitKey:true}))|4384780|4384781|entityMap^0|0|9|4|H|3|0|0|5|3|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|12|8|@$D|13|E|14|F|G]|$D|15|E|16|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|17|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|18|8|@$D|19|E|1A|F|G]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1B|8|@]|9|@]|A|$]]|$1|Q|3|R|5|J|7|1C|8|@]|9|@]|A|$K|L]]|$1|S|3|P|5|6|7|1D|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1E|8|@]|9|@]|A|$]]|$1|V|3|P|5|6|7|1F|8|@]|9|@]|A|$]]|$1|W|3|X|5|J|7|1G|8|@]|9|@]|A|$K|L]]|$1|Y|3|P|5|6|7|1H|8|@]|9|@]|A|$]]|$1|Z|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|10|$]]

<p>There is absolutely no reason to download a 3rd party library to achieve this simple problem, like the above solutions suggest.</p>
<p>The one line version to group a <code>list</code> of objects by a certain <code>key</code> in es6:</p>
<pre><code>const groupByKey = (list, key) =&gt; list.reduce((hash, obj) =&gt; ({...hash, [obj[key]]:( hash[obj[key]] || [] ).concat(obj)}), {})
</code></pre>
<p>The longer version that filters out the objects without the <code>key</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>function groupByKey(array, key) {
   return array
     .reduce((hash, obj) =&gt; {
       if(obj[key] === undefined) return hash; 
       return Object.assign(hash, { [obj[key]]:( hash[obj[key]] || [] ).concat(obj)})
     }, {})
}


var cars = [{'make':'audi','model':'r8','year':'2012'},{'make':'audi','model':'rs5','year':'2013'},{'make':'ford','model':'mustang','year':'2012'},{'make':'ford','model':'fusion','year':'2015'},{'make':'kia','model':'optima','year':'2012'}];

console.log(groupByKey(cars, 'make'))</code></pre>
</div>
</div>
</p>
<p>NOTE: It appear the original question asks how to group cars by make, but omit the make in each group. So the short answer, without 3rd party libraries, 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>const groupByKey = (list, key, {omitKey=false}) =&gt; list.reduce((hash, {[key]:value, ...rest}) =&gt; ({...hash, [value]:( hash[value] || [] ).concat(omitKey ? {...rest} : {[key]:value, ...rest})} ), {})

var cars = [{'make':'audi','model':'r8','year':'2012'},{'make':'audi','model':'rs5','year':'2013'},{'make':'ford','model':'mustang','year':'2012'},{'make':'ford','model':'fusion','year':'2015'},{'make':'kia','model':'optima','year':'2012'}];

console.log(groupByKey(cars, 'make', {omitKey:true}))</code></pre>
</div>
</div>
</p>


blocks|key|4384246|text|下面是您自己的groupBy函数，它是以下代码的泛化：https://github.com/you-dont-need/You-Dont-Need-Lodash-Underscore|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4384247|​|4384248|function+groupBy(xs,+f)+{
++return+xs.reduce((r,+v,+i,+a,+k+=+f(v))+=>+((r[k]+%7C%7C+(r[k]+=+[])).push(v),+r),+{});
}

const+cars+=+[{+make:+'audi',+model:+'r8',+year:+'2012'+},+{+make:+'audi',+model:+'rs5',+year:+'2013'+},+{+make:+'ford',+model:+'mustang',+year:+'2012'+},+{+make:+'ford',+model:+'fusion',+year:+'2015'+},+{+make:+'kia',+model:+'optima',+year:+'2012'+}];

const+result+=+groupBy(cars,+(c)+=>+c.make);
console.log(result);|code-block|syntax|javascript|4384249|4384250|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/you-dont-need/You-Dont-Need-Lodash-Underscore^0|7|7|R|1S|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@$9|Y|A|Z|1|10]]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]]]

<p>Here is your very own <code>groupBy</code> function which is a generalization of the code from: <a href="https://github.com/you-dont-need/You-Dont-Need-Lodash-Underscore" rel="noreferrer">https://github.com/you-dont-need/You-Dont-Need-Lodash-Underscore</a></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 groupBy(xs, f) {
  return xs.reduce((r, v, i, a, k = f(v)) =&gt; ((r[k] || (r[k] = [])).push(v), r), {});
}

const cars = [{ make: 'audi', model: 'r8', year: '2012' }, { make: 'audi', model: 'rs5', year: '2013' }, { make: 'ford', model: 'mustang', year: '2012' }, { make: 'ford', model: 'fusion', year: '2015' }, { make: 'kia', model: 'optima', year: '2012' }];

const result = groupBy(cars, (c) =&gt; c.make);
console.log(result);</code></pre>
</div>
</div>
</p>


blocks|key|4384271|text|使用一个简单的for循环也是可能的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4384272|+const+result+=+{};

+for(const+{make,+model,+year}+of+cars)+{
+++if(!result[make])+result[make]+=+[];
+++result[make].push({+model,+year+});
+}|code-block|syntax|javascript|4384273|entityMap^0|7|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

<p>Its also possible with a simple <code>for</code> loop:</p>

<pre><code> const result = {};

 for(const {make, model, year} of cars) {
   if(!result[make]) result[make] = [];
   result[make].push({ model, year });
 }
</code></pre>


blocks|key|4384866|text|我会让JS数组的REAL+GROUP+BY示例与这个任务here完全相同|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4384867|​|4384868|const+inputArray+=+[+
++++{+Phase:+"Phase+1",+Step:+"Step+1",+Task:+"Task+1",+Value:+"5"+},
++++{+Phase:+"Phase+1",+Step:+"Step+1",+Task:+"Task+2",+Value:+"10"+},
++++{+Phase:+"Phase+1",+Step:+"Step+2",+Task:+"Task+1",+Value:+"15"+},
++++{+Phase:+"Phase+1",+Step:+"Step+2",+Task:+"Task+2",+Value:+"20"+},
++++{+Phase:+"Phase+2",+Step:+"Step+1",+Task:+"Task+1",+Value:+"25"+},
++++{+Phase:+"Phase+2",+Step:+"Step+1",+Task:+"Task+2",+Value:+"30"+},
++++{+Phase:+"Phase+2",+Step:+"Step+2",+Task:+"Task+1",+Value:+"35"+},
++++{+Phase:+"Phase+2",+Step:+"Step+2",+Task:+"Task+2",+Value:+"40"+}
];

var+outObject+=+inputArray.reduce(function(a,+e)+{
++//+GROUP+BY+estimated+key+(estKey),+well,+may+be+a+just+plain+key
++//+a+--+Accumulator+result+object
++//+e+--+sequentally+checked+Element,+the+Element+that+is+tested+just+at+this+itaration

++//+new+grouping+name+may+be+calculated,+but+must+be+based+on+real+value+of+real+field
++let+estKey+=+(e['Phase']);+

++(a[estKey]+?+a[estKey]+:+(a[estKey]+=+null+%7C%7C+[])).push(e);
++return+a;
},+{});

console.log(outObject);|code-block|syntax|javascript|4384869|4384870|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/52797477/4882514^0|8|D|S|4|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@$9|Y|A|Z|1|10]]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]]]

<p>I'd leave <code>REAL GROUP BY</code> for JS Arrays example exactly the same this task <a href="https://stackoverflow.com/a/52797477/4882514">here</a></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>const inputArray = [ 
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" },
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" }
];

var outObject = inputArray.reduce(function(a, e) {
  // GROUP BY estimated key (estKey), well, may be a just plain key
  // a -- Accumulator result object
  // e -- sequentally checked Element, the Element that is tested just at this itaration

  // new grouping name may be calculated, but must be based on real value of real field
  let estKey = (e['Phase']); 

  (a[estKey] ? a[estKey] : (a[estKey] = null || [])).push(e);
  return a;
}, {});

console.log(outObject);</code></pre>
</div>
</div>
</p>


blocks|key|4384325|text|创建一个可重用的方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4384326|Array.prototype.groupBy+=+function(prop)+{
++++++return+this.reduce(function(groups,+item)+{
++++++++const+val+=+item[prop]
++++++++groups[val]+=+groups[val]+%7C%7C+[]
++++++++groups[val].push(item)
++++++++return+groups
++++++},+{})
++++};|code-block|syntax|javascript|4384327|然后在下面你可以按任何标准进行分组|4384328|const+groupByMake+=+cars.groupBy('make');
++++++++console.log(groupByMake);|4384329|​|4384330|var+cars+=+[
++++{
++++++++'make':+'audi',
++++++++'model':+'r8',
++++++++'year':+'2012'
++++},+{
++++++++'make':+'audi',
++++++++'model':+'rs5',
++++++++'year':+'2013'
++++},+{
++++++++'make':+'ford',
++++++++'model':+'mustang',
++++++++'year':+'2012'
++++},+{
++++++++'make':+'ford',
++++++++'model':+'fusion',
++++++++'year':+'2015'
++++},+{
++++++++'make':+'kia',
++++++++'model':+'optima',
++++++++'year':+'2012'
++++},
];
++//re-usable+method
Array.prototype.groupBy+=+function(prop)+{
	++return+this.reduce(function(groups,+item)+{
		const+val+=+item[prop]
		groups[val]+=+groups[val]+%7C%7C+[]
		groups[val].push(item)
		return+groups
	++},+{})
	};
++
+//+initiate+your+groupBy.+Notice+the+recordset+Cars+and+the+field+Make....
++const+groupByMake+=+cars.groupBy('make');
		console.log(groupByMake);
++++
++++//At+this+point+we+have+objects.+You+can+use+Object.keys+to+return+an+array|4384331|4384332|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|T|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|V|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|O|3|L|5|6|7|X|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|Q|$]]

<p>Create a method which can be re-used</p>

<pre><code>Array.prototype.groupBy = function(prop) {
      return this.reduce(function(groups, item) {
        const val = item[prop]
        groups[val] = groups[val] || []
        groups[val].push(item)
        return groups
      }, {})
    };
</code></pre>

<p>Then below you can group by any criteria</p>

<pre><code>const groupByMake = cars.groupBy('make');
        console.log(groupByMake);
</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>var cars = [
    {
        'make': 'audi',
        'model': 'r8',
        'year': '2012'
    }, {
        'make': 'audi',
        'model': 'rs5',
        'year': '2013'
    }, {
        'make': 'ford',
        'model': 'mustang',
        'year': '2012'
    }, {
        'make': 'ford',
        'model': 'fusion',
        'year': '2015'
    }, {
        'make': 'kia',
        'model': 'optima',
        'year': '2012'
    },
];
  //re-usable method
Array.prototype.groupBy = function(prop) {
	  return this.reduce(function(groups, item) {
		const val = item[prop]
		groups[val] = groups[val] || []
		groups[val].push(item)
		return groups
	  }, {})
	};
  
 // initiate your groupBy. Notice the recordset Cars and the field Make....
  const groupByMake = cars.groupBy('make');
		console.log(groupByMake);
    
    //At this point we have objects. You can use Object.keys to return an array</code></pre>
</div>
</div>
</p>


blocks|key|4384334|text|function+groupBy(data,+property)+{
++return+data.reduce((acc,+obj)+=>+{
++++const+key+=+obj[property];
++++if+(!acc[key])+{
++++++acc[key]+=+[];
++++}
++++acc[key].push(obj);
++++return+acc;
++},+{});
}
groupBy(people,+'age');|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4384335|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 groupBy(data, property) {
  return data.reduce((acc, obj) =&gt; {
    const key = obj[property];
    if (!acc[key]) {
      acc[key] = [];
    }
    acc[key].push(obj);
    return acc;
  }, {});
}
groupBy(people, 'age');
</code></pre>


blocks|key|4384349|text|试试这件吧，它对我来说很好。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4384350|let+grouped+=+_.groupBy(cars,+'make');|offset|length|style|CODE|4384351|注意:使用lodash+lib，所以要包含它。|4384352|entityMap^0|0|0|12|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

<p>Just try this one it works fine for me.</p>
<p><code>let grouped = _.groupBy(cars, 'make');</code></p>
<p>Note: Using lodash lib, so include it.</p>


blocks|key|4384358|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4384359|var+cars+=+[{
++make:+'audi',
++model:+'r8',
++year:+'2012'
},+{
++make:+'audi',
++model:+'rs5',
++year:+'2013'
},+{
++make:+'ford',
++model:+'mustang',
++year:+'2012'
},+{
++make:+'ford',
++model:+'fusion',
++year:+'2015'
},+{
++make:+'kia',
++model:+'optima',
++year:+'2012'
}].reduce((r,+car)+=>+{

++const+{
++++model,
++++year,
++++make
++}+=+car;

++r[make]+=+[...r[make]+%7C%7C+[],+{
++++model,
++++year
++}];

++return+r;
},+{});

console.log(cars);|code-block|syntax|javascript|4384360|4384361|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>var cars = [{
  make: 'audi',
  model: 'r8',
  year: '2012'
}, {
  make: 'audi',
  model: 'rs5',
  year: '2013'
}, {
  make: 'ford',
  model: 'mustang',
  year: '2012'
}, {
  make: 'ford',
  model: 'fusion',
  year: '2015'
}, {
  make: 'kia',
  model: 'optima',
  year: '2012'
}].reduce((r, car) =&gt; {

  const {
    model,
    year,
    make
  } = car;

  r[make] = [...r[make] || [], {
    model,
    year
  }];

  return r;
}, {});

console.log(cars);</code></pre>
</div>
</div>
</p>


blocks|key|4384940|text|使用lodash/fp，您可以使用_.flow()创建一个函数，第一个按键分组，然后映射每个组，并从每个项目中省略一个键：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4384941|​|4384942|const+{+flow,+groupBy,+mapValues,+map,+omit+}+=+_;

const+groupAndOmitBy+=+key+=>+flow(
++groupBy(key),
++mapValues(map(omit(key)))
);

const+cars+=+[{+make:+'audi',+model:+'r8',+year:+'2012'+},+{+make:+'audi',+model:+'rs5',+year:+'2013'+},+{+make:+'ford',+model:+'mustang',+year:+'2012'+},+{+make:+'ford',+model:+'fusion',+year:+'2015'+},+{+make:+'kia',+model:+'optima',+year:+'2012'+}];

const+groupAndOmitMake+=+groupAndOmitBy('make');

const+result+=+groupAndOmitMake(cars);

console.log(result);|code-block|syntax|javascript|4384943|.as-console-wrapper+{+max-height:+100%25+!important;+top:+0;+}|4384944|<script+src='https://cdn.jsdelivr.net/g/lodash@4(lodash.min.js%2Blodash.fp.min.js)'></script>|4384945|4384946|entityMap^0|H|8|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|X|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|O|3|P|5|J|7|Z|8|@]|D|@]|E|$K|L]]|$1|Q|3|G|5|6|7|10|8|@]|D|@]|E|$]]|$1|R|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|S|$]]

<p>With lodash/fp you can create a function with <code>_.flow()</code> that 1st groups by a key, and then map each group, and omits a key from each item:</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>const { flow, groupBy, mapValues, map, omit } = _;

const groupAndOmitBy = key =&gt; flow(
  groupBy(key),
  mapValues(map(omit(key)))
);

const cars = [{ make: 'audi', model: 'r8', year: '2012' }, { make: 'audi', model: 'rs5', year: '2013' }, { make: 'ford', model: 'mustang', year: '2012' }, { make: 'ford', model: 'fusion', year: '2015' }, { make: 'kia', model: 'optima', year: '2012' }];

const groupAndOmitMake = groupAndOmitBy('make');

const result = groupAndOmitMake(cars);

console.log(result);</code></pre>
<pre class="snippet-code-css lang-css prettyprint-override"><code>.as-console-wrapper { max-height: 100% !important; top: 0; }</code></pre>
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;script src='https://cdn.jsdelivr.net/g/lodash@4(lodash.min.js+lodash.fp.min.js)'&gt;&lt;/script&gt;</code></pre>
</div>
</div>
</p>


blocks|key|4384426|text|如果您不想键入所有字段，请在@Jonas_Wilms的answer上构建：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4384427|++++var+result+=+{};

++++for+(+let+{+first_field,+...fields+}+of+your_data+)+
++++{+
+++++++result[first_field]+=+result[first_field]+%7C%7C+[];
+++++++result[first_field].push({+...fields+});+
++++}|code-block|syntax|javascript|4384428|我没有做任何基准测试，但我相信使用for循环也会比this+answer中建议的任何东西更有效。|4384429|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/51563349/2842348|1|https://stackoverflow.com/questions/14446511/most-efficient-method-to-groupby-on-an-array-of-objects?page=2&tab=votes#tab-top^0|R|6|0|0|0|P|B|1|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]]|$1|I|3|J|5|6|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

<p>Building on the <a href="https://stackoverflow.com/a/51563349/2842348">answer</a> by @Jonas_Wilms if you do not want to type in all your fields:</p>

<pre><code>    var result = {};

    for ( let { first_field, ...fields } of your_data ) 
    { 
       result[first_field] = result[first_field] || [];
       result[first_field].push({ ...fields }); 
    }

</code></pre>

<p>I didn't make any benchmark but I believe using a for loop would be more efficient than anything suggested in <a href="https://stackoverflow.com/questions/14446511/most-efficient-method-to-groupby-on-an-array-of-objects?page=2&amp;tab=votes#tab-top">this answer</a> as well.</p>


blocks|key|4384431|text|const+reGroup+=+(list,+key)+=>+{
++++const+newGroup+=+{};
++++list.forEach(item+=>+{
++++++++const+newItem+=+Object.assign({},+item);
++++++++delete+newItem[key];
++++++++newGroup[item[key]]+=+newGroup[item[key]]+%7C%7C+[];
++++++++newGroup[item[key]].push(newItem);
++++});
++++return+newGroup;
};|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4384432|const+animals+=+[
++{
++++type:+'dog',
++++breed:+'puddle'
++},
++{
++++type:+'dog',
++++breed:+'labradoodle'
++},
++{
++++type:+'cat',
++++breed:+'siamese'
++},
++{
++++type:+'dog',
++++breed:+'french+bulldog'
++},
++{
++++type:+'cat',
++++breed:+'mud'
++}
];
console.log(reGroup(animals,+'type'));|4384433|const+cars+=+[
++{
++++++'make':+'audi',
++++++'model':+'r8',
++++++'year':+'2012'
++},+{
++++++'make':+'audi',
++++++'model':+'rs5',
++++++'year':+'2013'
++},+{
++++++'make':+'ford',
++++++'model':+'mustang',
++++++'year':+'2012'
++},+{
++++++'make':+'ford',
++++++'model':+'fusion',
++++++'year':+'2015'
++},+{
++++++'make':+'kia',
++++++'model':+'optima',
++++++'year':+'2012'
++},
];

console.log(reGroup(cars,+'make'));|4384434|unstyled|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$B|C]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|H|3|-4|5|I|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>const reGroup = (list, key) =&gt; {
    const newGroup = {};
    list.forEach(item =&gt; {
        const newItem = Object.assign({}, item);
        delete newItem[key];
        newGroup[item[key]] = newGroup[item[key]] || [];
        newGroup[item[key]].push(newItem);
    });
    return newGroup;
};
</code></pre>

<pre><code>const animals = [
  {
    type: 'dog',
    breed: 'puddle'
  },
  {
    type: 'dog',
    breed: 'labradoodle'
  },
  {
    type: 'cat',
    breed: 'siamese'
  },
  {
    type: 'dog',
    breed: 'french bulldog'
  },
  {
    type: 'cat',
    breed: 'mud'
  }
];
console.log(reGroup(animals, 'type'));
</code></pre>

<pre><code>const cars = [
  {
      'make': 'audi',
      'model': 'r8',
      'year': '2012'
  }, {
      'make': 'audi',
      'model': 'rs5',
      'year': '2013'
  }, {
      'make': 'ford',
      'model': 'mustang',
      'year': '2012'
  }, {
      'make': 'ford',
      'model': 'fusion',
      'year': '2015'
  }, {
      'make': 'kia',
      'model': 'optima',
      'year': '2012'
  },
];

console.log(reGroup(cars, 'make'));
</code></pre>


blocks|key|4384485|text|原型版本也使用了ES6。基本上，这使用reduce函数传入一个累加器和当前项，然后使用它根据传入的键构建“分组”数组。reduce的内部可能看起来很复杂，但本质上它是在测试传入对象的键是否存在，如果不存在，则创建一个空数组并将当前项附加到新创建的数组中，否则使用扩展操作符传递当前键数组的所有对象并附加当前项。希望这对某些人有帮助！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4384486|Array.prototype.groupBy+=+function(k)+{
++return+this.reduce((acc,+item)+=>+((acc[item[k]]+=+[...(acc[item[k]]+%7C%7C+[]),+item]),+acc),{});
};

const+projs+=+[
++{
++++project:+"A",
++++timeTake:+2,
++++desc:+"this+is+a+description"
++},
++{
++++project:+"B",
++++timeTake:+4,
++++desc:+"this+is+a+description"
++},
++{
++++project:+"A",
++++timeTake:+12,
++++desc:+"this+is+a+description"
++},
++{
++++project:+"B",
++++timeTake:+45,
++++desc:+"this+is+a+description"
++}
];

console.log(projs.groupBy("project"));|code-block|syntax|javascript|4384487|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>Prototype version using ES6 as well.
Basically this uses the reduce function to pass in an accumulator and current item, which then uses this to build your "grouped" arrays based on the passed in key. the inner part of the reduce may look complicated but essentially it is testing to see if the key of the passed in object exists and if it doesn't then create an empty array and append the current item to that newly created array otherwise using the spread operator pass in all the objects of the current key array and append current item. Hope this helps someone!.</p>

<pre><code>Array.prototype.groupBy = function(k) {
  return this.reduce((acc, item) =&gt; ((acc[item[k]] = [...(acc[item[k]] || []), item]), acc),{});
};

const projs = [
  {
    project: "A",
    timeTake: 2,
    desc: "this is a description"
  },
  {
    project: "B",
    timeTake: 4,
    desc: "this is a description"
  },
  {
    project: "A",
    timeTake: 12,
    desc: "this is a description"
  },
  {
    project: "B",
    timeTake: 45,
    desc: "this is a description"
  }
];

console.log(projs.groupBy("project"));
</code></pre>


blocks|key|4385136|text|以下是一个受Java中Collectors.groupingBy()启发的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4385137|​|4385138|function+groupingBy(list,+keyMapper)+{
++return+list.reduce((accummalatorMap,+currentValue)+=>+{
++++const+key+=+keyMapper(currentValue);
++++if(!accummalatorMap.has(key))+{
++++++accummalatorMap.set(key,+[currentValue]);
++++}+else+{
++++++accummalatorMap.set(key,+accummalatorMap.get(key).push(currentValue));
++++}
++++return+accummalatorMap;
++},+new+Map());
}|code-block|syntax|javascript|4385139|4385140|这将给出一个Map对象。|4385141|4385142|//+Usage

const+carMakers+=+groupingBy(cars,+car+=>+car.make);|4385143|4385144|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>Here is a solution inspired from Collectors.groupingBy() in Java:</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 groupingBy(list, keyMapper) {
  return list.reduce((accummalatorMap, currentValue) =&gt; {
    const key = keyMapper(currentValue);
    if(!accummalatorMap.has(key)) {
      accummalatorMap.set(key, [currentValue]);
    } else {
      accummalatorMap.set(key, accummalatorMap.get(key).push(currentValue));
    }
    return accummalatorMap;
  }, new Map());
}</code></pre>
</div>
</div>
</p>

<p>This will give a Map object.</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>// Usage

const carMakers = groupingBy(cars, car =&gt; car.make);</code></pre>
</div>
</div>
</p>


blocks|key|4384583|text|typescript中对象的分组数组，其中包含以下内容：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4384584|groupBy+(list:+any[],+key:+string):+Map<string,+Array<any>>+{
++++let+map+=+new+Map();
++++list.map(val=>+{
++++++++if(!map.has(val[key])){
++++++++++++map.set(val[key],list.filter(data+=>+data[key]+==+val[key]));
++++++++}
++++});
++++return+map;
});|code-block|syntax|javascript|4384585|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>Grouped Array of Object in typescript with this:</p>

<pre><code>groupBy (list: any[], key: string): Map&lt;string, Array&lt;any&gt;&gt; {
    let map = new Map();
    list.map(val=&gt; {
        if(!map.has(val[key])){
            map.set(val[key],list.filter(data =&gt; data[key] == val[key]));
        }
    });
    return map;
});
</code></pre>


blocks|key|4385272|text|我喜欢用简单的js编写它，没有依赖性/复杂性。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4385273|​|4385274|const+mp+=+{}
const+cars+=+[
++{
++++model:+'Imaginary+space+craft+SpaceX+model',
++++year:+'2025'
++},
++{
++++make:+'audi',
++++model:+'r8',
++++year:+'2012'
++},
++{
++++make:+'audi',
++++model:+'rs5',
++++year:+'2013'
++},
++{
++++make:+'ford',
++++model:+'mustang',
++++year:+'2012'
++},
++{
++++make:+'ford',
++++model:+'fusion',
++++year:+'2015'
++},
++{
++++make:+'kia',
++++model:+'optima',
++++year:+'2012'
++}
]

cars.forEach(c+=>+{
++if+(!c.make)+return+//+exit+(maybe+add+them+to+a+"no_make"+category)

++if+(!mp[c.make])+mp[c.make]+=+[{+model:+c.model,+year:+c.year+}]
++else+mp[c.make].push({+model:+c.model,+year:+c.year+})
})

console.log(mp)|code-block|syntax|javascript|4385275|4385276|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

<p>I love to write it with no dependency/complexity just pure simple js.</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>const mp = {}
const cars = [
  {
    model: 'Imaginary space craft SpaceX model',
    year: '2025'
  },
  {
    make: 'audi',
    model: 'r8',
    year: '2012'
  },
  {
    make: 'audi',
    model: 'rs5',
    year: '2013'
  },
  {
    make: 'ford',
    model: 'mustang',
    year: '2012'
  },
  {
    make: 'ford',
    model: 'fusion',
    year: '2015'
  },
  {
    make: 'kia',
    model: 'optima',
    year: '2012'
  }
]

cars.forEach(c =&gt; {
  if (!c.make) return // exit (maybe add them to a "no_make" category)

  if (!mp[c.make]) mp[c.make] = [{ model: c.model, year: c.year }]
  else mp[c.make].push({ model: c.model, year: c.year })
})

console.log(mp)</code></pre>
</div>
</div>
</p>


blocks|key|4384675|text|我做了一个基准测试来测试不使用外部库的每个解决方案的性能。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4384676|JSBen.ch|offset|length|4384677|由@Nina+Scholz发布的reduce()选项似乎是最好的。|style|CODE|4384678|entityMap|0|LINK|mutability|MUTABLE|url|https://jsben.ch/ZQyZo^0|0|0|8|0|0|G|8|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@$D|T|E|U|1|V]]|A|$]]|$1|F|3|G|5|6|7|W|8|@$D|X|E|Y|H|I]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]]]

<p>I made a benchmark to test the performance of each solution that don't use external libraries.</p>
<p><a href="https://jsben.ch/ZQyZo" rel="nofollow noreferrer">JSBen.ch</a></p>
<p>The <code>reduce()</code> option, posted by @Nina Scholz seems to be the optimal one.</p>


blocks|key|4384764|text|同意除非您经常使用这些库，否则不需要外部库。虽然有类似的解决方案可用，但我看到其中一些很难遵循here+is+a+gist，如果你想了解发生了什么，它有一个带有注释的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4384765|​|4384766|const+cars+=+[{
++'make':+'audi',
++'model':+'r8',
++'year':+'2012'
},+{
++'make':+'audi',
++'model':+'rs5',
++'year':+'2013'
},+{
++'make':+'ford',
++'model':+'mustang',
++'year':+'2012'
},+{
++'make':+'ford',
++'model':+'fusion',
++'year':+'2015'
},+{
++'make':+'kia',
++'model':+'optima',
++'year':+'2012'
},+];

/**
+*+Groups+an+array+of+objects+by+a+key+an+returns+an+object+or+array+grouped+by+provided+key.
+*+@param+array+-+array+to+group+objects+by+key.
+*+@param+key+-+key+to+group+array+objects+by.
+*+@param+removeKey++-+remove+the+key+and+it's+value+from+the+resulting+object.
+*+@param+outputType+-+type+of+structure+the+output+should+be+contained+in.
+*/
const+groupBy+=+(
++inputArray,
++key,
++removeKey+=+false,
++outputType+=+{},
)+=>+{
++return+inputArray.reduce(
++++(previous,+current)+=>+{
++++++//+Get+the+current+value+that+matches+the+input+key+and+remove+the+key+value+for+it.
++++++const+{
++++++++[key]:+keyValue
++++++}+=+current;
++++++//+remove+the+key+if+option+is+set
++++++removeKey+&&+keyValue+&&+delete+current[key];
++++++//+If+there+is+already+an+array+for+the+user+provided+key+use+it+else+default+to+an+empty+array.
++++++const+{
++++++++[keyValue]:+reducedValue+=+[]
++++++}+=+previous;

++++++//+Create+a+new+object+and+return+that+merges+the+previous+with+the+current+object
++++++return+Object.assign(previous,+{
++++++++[keyValue]:+reducedValue.concat(current)
++++++});
++++},
++++//+Replace+the+object+here+to+an+array+to+change+output+object+to+an+array
++++outputType,
++);
};

console.log(groupBy(cars,+'make',+true))|code-block|syntax|javascript|4384767|4384768|entityMap|0|LINK|mutability|MUTABLE|url|https://gist.github.com/wattry/de36aa0fdecbab93adbaf4ff7818ef94^0|1B|E|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|D|3|E|5|6|7|X|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|Y|8|@]|9|@]|C|$I|J]]|$1|K|3|E|5|6|7|Z|8|@]|9|@]|C|$]]|$1|L|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|M|$N|$5|O|P|Q|C|$R|S]]]]

<p>Agree that unless you use these  often there is no need for an external library. Although similar solutions are available, I see that some of them are tricky to follow <a href="https://gist.github.com/wattry/de36aa0fdecbab93adbaf4ff7818ef94" rel="nofollow noreferrer">here is a gist</a> that has a solution with comments if you're trying to understand what is happening.</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>const cars = [{
  'make': 'audi',
  'model': 'r8',
  'year': '2012'
}, {
  'make': 'audi',
  'model': 'rs5',
  'year': '2013'
}, {
  'make': 'ford',
  'model': 'mustang',
  'year': '2012'
}, {
  'make': 'ford',
  'model': 'fusion',
  'year': '2015'
}, {
  'make': 'kia',
  'model': 'optima',
  'year': '2012'
}, ];

/**
 * Groups an array of objects by a key an returns an object or array grouped by provided key.
 * @param array - array to group objects by key.
 * @param key - key to group array objects by.
 * @param removeKey  - remove the key and it's value from the resulting object.
 * @param outputType - type of structure the output should be contained in.
 */
const groupBy = (
  inputArray,
  key,
  removeKey = false,
  outputType = {},
) =&gt; {
  return inputArray.reduce(
    (previous, current) =&gt; {
      // Get the current value that matches the input key and remove the key value for it.
      const {
        [key]: keyValue
      } = current;
      // remove the key if option is set
      removeKey &amp;&amp; keyValue &amp;&amp; delete current[key];
      // If there is already an array for the user provided key use it else default to an empty array.
      const {
        [keyValue]: reducedValue = []
      } = previous;

      // Create a new object and return that merges the previous with the current object
      return Object.assign(previous, {
        [keyValue]: reducedValue.concat(current)
      });
    },
    // Replace the object here to an array to change output object to an array
    outputType,
  );
};

console.log(groupBy(cars, 'make', true))</code></pre>
</div>
</div>
</p>


blocks|key|4385360|text|另一种解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4385361|​|4385362|var+cars+=+[
++++{'make':+'audi','model':+'r8','year':+'2012'},+{'make':+'audi','model':+'rs5','year':+'2013'},+
++++{'make':+'ford','model':+'mustang','year':+'2012'},+{'make':+'ford','model':+'fusion','year':+'2015'},+
++++{'make':+'kia','model':+'optima','year':+'2012'},
];


const+reducedCars+=+cars.reduce((acc,+{+make,+model,+year+})+=>+(
++++{+
++++++...acc,+
++++++[make]:+acc[make]+?+[+...acc[make],+{+model,+year+}]+:+[+{+model,+year+}+],
++++}
+),+{});

console.log(reducedCars);|code-block|syntax|javascript|4385363|4385364|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

<p>Another one solution:</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 cars = [
    {'make': 'audi','model': 'r8','year': '2012'}, {'make': 'audi','model': 'rs5','year': '2013'}, 
    {'make': 'ford','model': 'mustang','year': '2012'}, {'make': 'ford','model': 'fusion','year': '2015'}, 
    {'make': 'kia','model': 'optima','year': '2012'},
];


const reducedCars = cars.reduce((acc, { make, model, year }) =&gt; (
    { 
      ...acc, 
      [make]: acc[make] ? [ ...acc[make], { model, year }] : [ { model, year } ],
    }
 ), {});

console.log(reducedCars);</code></pre>
</div>
</div>
</p>


<p>Does anyone know of a (lodash if possible too) way to group an array of objects by an object key then create a new array of objects based on the grouping? For example, I have an array of car objects:</p>

<pre><code>var cars = [
    {
        'make': 'audi',
        'model': 'r8',
        'year': '2012'
    }, {
        'make': 'audi',
        'model': 'rs5',
        'year': '2013'
    }, {
        'make': 'ford',
        'model': 'mustang',
        'year': '2012'
    }, {
        'make': 'ford',
        'model': 'fusion',
        'year': '2015'
    }, {
        'make': 'kia',
        'model': 'optima',
        'year': '2012'
    },
];
</code></pre>

<p>I want to make a new array of car objects that's grouped by <code>make</code>:</p>

<pre><code>var cars = {
    'audi': [
        {
            'model': 'r8',
            'year': '2012'
        }, {
            'model': 'rs5',
            'year': '2013'
        },
    ],

    'ford': [
        {
            'model': 'mustang',
            'year': '2012'
        }, {
            'model': 'fusion',
            'year': '2015'
        }
    ],

    'kia': [
        {
            'model': 'optima',
            'year': '2012'
        }
    ]
}
</code></pre>


How to group an array of objects by key

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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有谁知道(如果可能的话)有一种方法可以通过对象键对对象数组进行分组，然后根据分组创建一个新的对象数组？例如，我有一个汽车对象数组：var cars = [    {        'make': 'audi',        'model': 'r8',        'year': '2012'    }, {   ...

问如何按键对对象数组进行分组
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