entityMap|blocks|key|fvn96|text|根据您的性能/可读性需求，可能的解决方案包括：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|b8rur|arrayOfObj.filter(a+=>+arrayOfObj.filter(obj+=>+obj.style+===+a.style).length+===+1)|code-block|syntax|javascript|77s1b^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

<p>On of the possible solutions depending on your performance / readability needs can be:</p>
<pre><code>arrayOfObj.filter(a =&gt; arrayOfObj.filter(obj =&gt; obj.style === a.style).length === 1)
</code></pre>


entityMap|blocks|key|bv6pl|text|找到现有项并将其移除时，请使用拼接|type|unstyled|depth|inlineStyleRanges|entityRanges|data|fv42v|​|cj9am|const+arrayOfObj+=+[{
++name:+'a',
++style:+'p'
},+{
++name:+'b',
++style:+'q'
},+{
++name:+'c',
++style:+'q'
}]

const+result+=+arrayOfObj.reduce((acc,+x)+=>+{
++const+index+=+acc.findIndex(y+=>+y.style+===+x.style);
++if+(index+>=+0)+{
++++acc.splice(index,+1);
++}+else+{
++++acc.push(x);
++}
++return+acc;

},+[])

console.log(result)|code-block|syntax|javascript|6u3aj|71hiq^0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|M|9|@]|A|@]|B|$]]|$2|E|4|F|6|G|8|N|9|@]|A|@]|B|$H|I]]|$2|J|4|D|6|7|8|O|9|@]|A|@]|B|$]]|$2|K|4|-4|6|7|8|P|9|@]|A|@]|B|$]]]]

<p>Use splice when you find the existing item and remove it</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 arrayOfObj = [{
  name: 'a',
  style: 'p'
}, {
  name: 'b',
  style: 'q'
}, {
  name: 'c',
  style: 'q'
}]

const result = arrayOfObj.reduce((acc, x) =&gt; {
  const index = acc.findIndex(y =&gt; y.style === x.style);
  if (index &gt;= 0) {
    acc.splice(index, 1);
  } else {
    acc.push(x);
  }
  return acc;

}, [])

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


entityMap|blocks|key|9l5ru|text|你减少了它。检查数组中是否已经存在具有相同样式的元素，并将其从累加器中移除，否则将其推送到累加器|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7knqm|​|6s225|const+arr+=+[
++{+name:+"a",+style:+"p"+},
++{+name:+"b",+style:+"q"+},
++{+name:+"c",+style:+"q"+}
];

++let+result+=+arr.reduce((a,v)+=>+{
+++++let+i+=+a.findIndex(el+=>+el.style+===+v.style);
+++++if(i+!==+-1)+{
++++++++a.splice(i,1);
++++++++return+a;
+++++}
+++++a.push(v)
+++++return+a;
++},[])

console.log(result);|code-block|syntax|javascript|21jom|35h1t^0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|M|9|@]|A|@]|B|$]]|$2|E|4|F|6|G|8|N|9|@]|A|@]|B|$H|I]]|$2|J|4|D|6|7|8|O|9|@]|A|@]|B|$]]|$2|K|4|-4|6|7|8|P|9|@]|A|@]|B|$]]]]

<p>You reduce it. Check if in the array already an element with the same style exists and remove it from the accumulator otherwise push it to the accumulator</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 arr = [
  { name: "a", style: "p" },
  { name: "b", style: "q" },
  { name: "c", style: "q" }
];

  let result = arr.reduce((a,v) =&gt; {
     let i = a.findIndex(el =&gt; el.style === v.style);
     if(i !== -1) {
        a.splice(i,1);
        return a;
     }
     a.push(v)
     return a;
  },[])

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


entityMap|blocks|key|6urg2|text|这里有一个O(n)时间复杂度的解决方案。您可以迭代所有条目以跟踪条目出现的频率。然后使用filter()函数来过滤那些只出现一次的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|ck496|​|3vdnm|const+arrayOfObj+=+[
++{+name:+"a",+style:+"p"+},
++{+name:+"b",+style:+"q"+},
++{+name:+"c",+style:+"q"+},
]

const+styleCount+=+{}

arrayOfObj.forEach((obj)+=>+{
++styleCount[obj.style]+=+(styleCount[obj.style]+%7C%7C+0)+%2B+1
})

const+res+=+arrayOfObj.filter((obj)+=>+styleCount[obj.style]+===+1)

console.log(res)|code-block|syntax|javascript|4vetb|7pp74^0|18|8|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|P|9|@$A|Q|B|R|C|D]]|E|@]|F|$]]|$2|G|4|H|6|7|8|S|9|@]|E|@]|F|$]]|$2|I|4|J|6|K|8|T|9|@]|E|@]|F|$L|M]]|$2|N|4|H|6|7|8|U|9|@]|E|@]|F|$]]|$2|O|4|-4|6|7|8|V|9|@]|E|@]|F|$]]]]

<p>Here is a solution in O(n) time complexity. You can iterate all entries to track how often an entry occurs. And then use the <code>filter()</code> function to filter the ones that occur only once.</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 arrayOfObj = [
  { name: "a", style: "p" },
  { name: "b", style: "q" },
  { name: "c", style: "q" },
]

const styleCount = {}

arrayOfObj.forEach((obj) =&gt; {
  styleCount[obj.style] = (styleCount[obj.style] || 0) + 1
})

const res = arrayOfObj.filter((obj) =&gt; styleCount[obj.style] === 1)

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


entityMap|blocks|key|73rav|text|这里有一个O(n)时间复杂度的解决方案。您可以迭代所有条目以跟踪条目出现的频率。然后使用filter()函数来过滤那些只出现一次的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|dff59|​|9gmpi|const+arrayOfObj+=+[+{name:'a'+,+style:'p'}+,+{name:'b'+,+style:'q'}+,+{name:'c'+,+style:'q'}];

let+count+=+{};

arrayOfObj.forEach(({style})+=>+{
++++count[style]+=+(count[style]+%7C%7C+0)+%2B+1;
});

let+result+=+arrayOfObj.filter(({style})+=>+count[style]+===+1);
console.log(result);|code-block|syntax|javascript|2a2k1|dht2k^0|18|8|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|P|9|@$A|Q|B|R|C|D]]|E|@]|F|$]]|$2|G|4|H|6|7|8|S|9|@]|E|@]|F|$]]|$2|I|4|J|6|K|8|T|9|@]|E|@]|F|$L|M]]|$2|N|4|H|6|7|8|U|9|@]|E|@]|F|$]]|$2|O|4|-4|6|7|8|V|9|@]|E|@]|F|$]]]]

<p>Here is a solution in O(n) time complexity. You can iterate all entries to track how often an entry occurs. And then use the <code>filter()</code> function to filter the ones that occur only once.</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 arrayOfObj = [ {name:'a' , style:'p'} , {name:'b' , style:'q'} , {name:'c' , style:'q'}];

let count = {};

arrayOfObj.forEach(({style}) =&gt; {
    count[style] = (count[style] || 0) + 1;
});

let result = arrayOfObj.filter(({style}) =&gt; count[style] === 1);
console.log(result);</code></pre>
</div>
</div>
</p>


entityMap|blocks|key|af103|text|如果你正在使用lodash库(uniqBy(数组，迭代器))，也有一个线性的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|6mmp1|const+arr+=+[
++{+name:+"a",+style:+"p"+},
++{+name:+"b",+style:+"q"+},
++{+name:+"c",+style:+"q"+}
];
let+result+=+_.uniqBy(arrayOfObj,'style')
console.log(result)|code-block|syntax|javascript|ehhuh^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

<p>There is a one liner answer too if you are using lodash library
(uniqBy(array, iteratee))</p>
<pre><code>const arr = [
  { name: &quot;a&quot;, style: &quot;p&quot; },
  { name: &quot;b&quot;, style: &quot;q&quot; },
  { name: &quot;c&quot;, style: &quot;q&quot; }
];
let result = _.uniqBy(arrayOfObj,'style')
console.log(result)
</code></pre>


<p>I have an array of objects i want to filter only the unique style and is not repeated .</p>
<pre><code>const arrayOfObj = [ {name:'a' , style:'p'} , {name:'b' , style:'q'} , {name:'c' , style:'q'}]

result expected : [ {name:'a' , style:'p'}]

</code></pre>


Filter only unique values from an array of object 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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 我有一个对象数组，我只想过滤唯一的样式，并且不会重复。 const arrayOfObj = [ {name:'a' , style:'p'} , {name:'b' , style:'q'} , {name:'c' , style:'q'}]result expected : [ {name:'a' , style...

问只过滤对象javascript数组中的唯一值
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