This JavaScript code tries to get the indices of a given value 5 in a new array. Any idea how to do it in an elegant way. I cold use for loop but I was hoping to use map or reduce. thx.
console.log( [1, 2, 3, 5, 6, 5].map((y, i) => {
if (y === 5) return i
}))
// gives --> [undefined, undefined, undefined, 3, undefined, 5]
// expected [3,5]Unfortunately map and reduce are not lazily-evaluated in JavaScript (i.e. they're not generators/iterators), but if you don't mind the double array allocation, you can do this:
var indices = [ 1, 2, 3, 5, 6, 5 ]
.map( ( e, i ) => e == 5 ? i : -1 )
.filter( e => e > -1 );
// indicies == [3,5]
Another approach, that's cheaper:
var indices = [];
[ 1, 2, 3, 5, 6, 5 ].forEach( (e, i) => e == 5 ? indices.push( i ) : null );
This takes advantage of the fact you can use a void expression inside the ?: ternary operator.
You can use reduce. The third argument of the callback is the index.
[1, 2, 3, 5, 6, 5].reduce((indexes, n, index) => {
if (n === 5) indexes.push(index)
return indexes
}, [])
You can either use map() or reduce().
reduce() will be able to do it in a single loop:
[1,2,3,4,5].reduce((result, num, index) => result.concat(num === 5 ? index : []), []);
Using concat() instead of push() is a tiny bit slower, but cleaner, and in reasonably small sets, the difference will be negligible.
map() will need filter() to remove extras:
[1,2,3,4,5].map((num, index) => num === 5 && index)
.filter(e => e !== false);
.filter(Boolean) is a clean, short-hand way of casting whatever value to a Boolean, which filter will then use to determine what it needs to do. num === 5 && index will be either false or the index. Another clean way to go through.
You can use reduce operator on your array as:
[1,2,3,5,6,5].reduce(function(a, e, i) {
if (e === 5)
a.push(i);
return a;
}, []);
You could run a map to test the value and write either a null (no match) or the position (match) then a filter to remove the nulls.
The map would look like this:
map( (value, index) => {
return value === 5 ? index : null;
})
Then you'd filter it like this:
filter( (value) => {
return value != null;
})
console.log( [1, 2, 3, 5, 6, 5].map((y, i) => {
if (y === 5) return i
}).filter((x) => {return /\d/.test(x);}));