blocks|key|4041274|text|map()并不是很正确，因为您不会对原始数组中的每个元素都有相应的输出。在本例中，reduce()可能就是您想要的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4041275|​|4041276|function+pos(arr){
++++return+arr.reduce((ret_arr,+number,+index)+=>+{
++++++++if+(number+>=+0)+ret_arr.push(index)
++++++++return+ret_arr
++++},+[])

}

console.log(pos([1,-4,0,5,2]))|code-block|syntax|javascript|4041277|4041278|entityMap^0|0|5|15|8|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><code>map()</code> isn't quite right because you won't have corresponding output for every element in the original array. <code>reduce()</code> is probably what you want in this case:</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 pos(arr){
    return arr.reduce((ret_arr, number, index) =&gt; {
        if (number &gt;= 0) ret_arr.push(index)
        return ret_arr
    }, [])

}

console.log(pos([1,-4,0,5,2]))</code></pre>
</div>
</div>
</p>


blocks|key|4041311|text|如果值为正或-1，您可以首先映射索引，然后过滤数组中现有的索引。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4041312|Array#map返回一个带有元素索引的新数组，如果元素小于零，则返回-1。|4041313|Array#filter返回所有非-1的值。简短的加法只返回一个truthy值，这是一个可能的索引。|4041314|​|4041315|function+pos(array)+{
++++return+array+++++++++++++++++++++++//+[1,+-4,+0,+5,+2]
++++++++.map((v,+i)+=>+v+<+0+?+-1+:+i)+//+[0,+-1,+2,+3,+4]
++++++++.filter(i+=>+i+%2B+1);+++++++++++//+[0,+++++2,+3,+4]
}

console.log(pos([1,+-4,+0,+5,+2]));|code-block|syntax|javascript|4041316|4041317|您的映射回调|4041318|4041319|function+pos(arr)+{
++++return+arr.map(function+(ele)+{
++++++++return+arr.findIndex(ele+=>+ele+>+0);
++++});
}

console.log(pos([1,+-4,+0,+5,+2]));|4041320|4041321|不接受任何元素，因为以后永远不会使用第一个ele变量。然后总是返回arr的第一个元素的索引，即1。这个值大于零，并且find的回调返回true和find移交给实际的索引。|4041322|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map|1|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter|2|https://developer.mozilla.org/en-US/docs/Glossary/Truthy^0|6|2|0|0|9|Z|2|0|9|0|0|0|C|H|2|0|C|1|W|6|2|0|0|0|0|0|0|0|0|L|3|X|3|1B|1|1M|4|1V|4|0^^$0|@$1|2|3|4|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1E|8|@$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]]|D|@$9|1J|A|1K|1|1L]]|E|$]]|$1|H|3|I|5|6|7|1M|8|@$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]]|D|@$9|1R|A|1S|1|1T]|$9|1U|A|1V|1|1W]]|E|$]]|$1|J|3|K|5|6|7|1X|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|1Y|8|@]|D|@]|E|$O|P]]|$1|Q|3|K|5|6|7|1Z|8|@]|D|@]|E|$]]|$1|R|3|S|5|6|7|20|8|@]|D|@]|E|$]]|$1|T|3|K|5|6|7|21|8|@]|D|@]|E|$]]|$1|U|3|V|5|N|7|22|8|@]|D|@]|E|$O|P]]|$1|W|3|K|5|6|7|23|8|@]|D|@]|E|$]]|$1|X|3|Y|5|6|7|24|8|@$9|25|A|26|B|C]|$9|27|A|28|B|C]|$9|29|A|2A|B|C]|$9|2B|A|2C|B|C]|$9|2D|A|2E|B|C]]|D|@]|E|$]]|$1|Z|3|-4|5|6|7|2F|8|@]|D|@]|E|$]]]|10|$11|$5|12|13|14|E|$15|16]]|17|$5|12|13|14|E|$15|18]]|19|$5|12|13|14|E|$15|1A]]]]

<p>You could first map the index, if the value is positive or <code>-1</code> and then filter the array for existing indices.</p>

<p><a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map" rel="nofollow noreferrer"><code>Array#map</code></a> returns a new array with the index of the element or <code>-1</code> if the element is smaller than zero.</p>

<p><a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter" rel="nofollow noreferrer"><code>Array#filter</code></a> returns all values which are not <code>-1</code>. The short adding just one returns a <a href="https://developer.mozilla.org/en-US/docs/Glossary/Truthy" rel="nofollow noreferrer">truthy</a> value which is a possible index.</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 pos(array) {
    return array                       // [1, -4, 0, 5, 2]
        .map((v, i) =&gt; v &lt; 0 ? -1 : i) // [0, -1, 2, 3, 4]
        .filter(i =&gt; i + 1);           // [0,     2, 3, 4]
}

console.log(pos([1, -4, 0, 5, 2]));</code></pre>
</div>
</div>
</p>

<p>Your callback for mapping</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 pos(arr) {
    return arr.map(function (ele) {
        return arr.findIndex(ele =&gt; ele &gt; 0);
    });
}

console.log(pos([1, -4, 0, 5, 2]));</code></pre>
</div>
</div>
</p>

<p>takes no element, because the first <code>ele</code> variable is never used later. Then you return always the index of the first element of <code>arr</code>, which is <code>1</code>. This value is greater as zero and the callback for <code>find</code> returns <code>true</code> abd find hands over the actual index.</p>


blocks|key|4041759|text|不要使用findIndex。基本上你要找的是|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4041760|function+pos(arr)+{
++++var+res+=+[];
++++for+(const+[index,+el]+of+arr.entries())
++++++++if+(el+>+0)
++++++++++++res.push(index)
++++return+res;
}|code-block|syntax|javascript|4041761|如果您想使用高阶数组方法来完成此操作，您可以这样做|4041762|function+pos(arr)+{
++++return+arr.map((ele,+index)+=>+[ele,+index])
++++++++++++++.filter(([ele])+=>+ele+>+0)
++++++++++++++.map(([_,+origIndex])+=>+origIndex);
}|4041763|@尼纳斯霍尔茨也有一个很好的方法，尽管我把它写成|4041764|function+pos(arr)+{
++++return+arr.map((ele,+index)+=>+ele+>+0+?+index+:+null)
++++++++++++++.filter(index+=>+index+!==+null);
}|4041765|entityMap^0|4|9|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|T|$]]

<p>Don't use <code>findIndex</code>. Basically you are looking for</p>

<pre><code>function pos(arr) {
    var res = [];
    for (const [index, el] of arr.entries())
        if (el &gt; 0)
            res.push(index)
    return res;
}
</code></pre>

<p>If you want to do this using the higher-order array methods, you could do</p>

<pre><code>function pos(arr) {
    return arr.map((ele, index) =&gt; [ele, index])
              .filter(([ele]) =&gt; ele &gt; 0)
              .map(([_, origIndex]) =&gt; origIndex);
}
</code></pre>

<p>@NinaScholz has a good approach as well, though I'd write it as</p>

<pre><code>function pos(arr) {
    return arr.map((ele, index) =&gt; ele &gt; 0 ? index : null)
              .filter(index =&gt; index !== null);
}
</code></pre>


blocks|key|4041874|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4041875|[1,-4,0,5,2].reduce((acc,value,i)+=>+value+>=+0+?+[...acc,i]+:+acc,[]);|code-block|syntax|javascript|4041876|在任何情况下，当你同时使用过滤和其他东西的时候？reduce始终是最好的选择。对于像这样的基本算法，请尝试通过扩展返回一个包含先前值的新数组。|4041877|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>Try this:</p>

<pre><code>[1,-4,0,5,2].reduce((acc,value,i) =&gt; value &gt;= 0 ? [...acc,i] : acc,[]);
</code></pre>

<p>In any case where you're ding both filtering and ?something-else? reduce is always the best option. 
For basic algorithms like this try returning a new array containing the previous values via spread.</p>


blocks|key|4041879|text|const+possitive+=+[];
const+negative+=+[];
const+array+=+[-1,+-2,+0,+4,+1,+3,+2];

//+Separate+negative+from+possitive+values
array.map(number+=>
++number+>=+0+?+possitive.push(number)+:+negative.push(number)
);

//+Sort+accessing+order
possitive.sort((a,+b)+=>+a+-+b);
negative.sort((a,+b)+=>+a+-+b);

//+Join+both+arrays
const+completed+=+possitive.concat(negative);

//+Console+log+completed+array
console.log(completed);

//Result
[0,+1,+2,+3,+4,+-2,+-1];|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4041880|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>const possitive = [];
const negative = [];
const array = [-1, -2, 0, 4, 1, 3, 2];

// Separate negative from possitive values
array.map(number =&gt;
  number &gt;= 0 ? possitive.push(number) : negative.push(number)
);

// Sort accessing order
possitive.sort((a, b) =&gt; a - b);
negative.sort((a, b) =&gt; a - b);

// Join both arrays
const completed = possitive.concat(negative);

// Console log completed array
console.log(completed);

//Result
[0, 1, 2, 3, 4, -2, -1];
</code></pre>


<p>I want to return an array of indices of an array which has positive elements using pure functions in JS. EX:</p>

<p><code>pos([1,-4,0,5,2])</code></p>

<pre><code>&gt;&gt;[0,2,3,4]
</code></pre>

<p>So, I tried: </p>

<pre><code>  arr.map(function (ele){
   return arr.findIndex(ele =&gt; ele &gt; 0);
  })

But my output is :[1,1,1,1]
</code></pre>

<p>Can someone tell me what is wrong? I believe that my map function is correct but something is wrong with findIndex(). (Please don't tell me the code as it is a question of my assignment. Only need hints.)</p>


Find the indices of positive elements in an array in JavaScript

我希望返回一个数组的索引数组，该数组包含使用JS中的纯函数的正元素。例如：pos([1,-4,0,5,2])>>[0,2,3,4]所以，我试着：  arr.map(function (ele){   return arr.findIndex(ele => ele > 0);  })But my output is :[...

问在JavaScript中查找数组中正元素的索引
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