Let's say I have an array with the following values -
var arr = [true, true, false, false, false, true, false];
I'm looking for logic which will give me the following output -
[0,1,5]
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Possible duplicate of How to find index of all occurrences of element in array?Sudipta Mondal– Sudipta Mondal2018-06-22 06:23:02 +00:00Commented Jun 22, 2018 at 6:23
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You should also post your own effort with the post.vibhor1997a– vibhor1997a2018-06-22 06:35:21 +00:00Commented Jun 22, 2018 at 6:35
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6 Answers
You can use .reduce() to do this in one pass:
const arr = [true, true, false, false, false, true, false]
const indices = arr.reduce(
(out, bool, index) => bool ? out.concat(index) : out,
[]
)
console.log(indices)You start by passing an empty array [] as the initialValue to .reduce() and use a ternary ? operator to determine whether to .concat() the index or not.
Alternatively, you can use the more recent .flatMap() method:
const arr = [true, true, false, false, false, true, false]
const indices = arr.flatMap((bool, index) => bool ? index : [])
console.log(indices)If your browser does not yet support it, you'll get an Uncaught TypeError: arr.flatMap is not a function. In that case, you can use my polyfill definition from here.
Comments
A potentially more elegant solution is:
[...arr.keys()].filter(i => arr[i])
or alternatively
[...arr.entries()].filter(([, v]) => v).map(([i]) => i)
Comments
You could map the indices or for falsy values -1 and then filter all indices which are not -1.
var array = [true, true, false, false, false, true, false],
result = array
.map((b, i) => b ? i : -1)
.filter(i => i !== -1);
console.log(result);
Comments
You can simply use a loop and check the boolean value true in your if condition:
var arr = [true, true, false, false, false, true, false];
var indexArray = [];
arr.forEach((val, index) => {
if(val){
indexArray.push(index);
}
});
console.log(indexArray);
Comments
try this
var arr = [true, true, false, false, false, true, false];
var indexArray = [];
arr.forEach(function(value, index) {
if (value) {
indexArray.push(index);
}
})
console.log(indexArray);
Comments
Different solution based on logic itself and not JS paradigms
let newArr = [];
for (let i = 0 ; i<indexArray.length;i++){
if (indexArray[i]) newArr.push(i)
}