Get the indexes of javascript array elements that satisfy condition

Question

I have an array like this:

[{prop1:"abc",prop2:"qwe"},{prop1:"abc",prop2:"yutu"},{prop1:"xyz",prop2:"qwrq"}]

I'd like to get the index of all elements that satisfy a condition; in this case, when prop1 == "abc". So the desired output is something like [0,1].

Struggling to find a clean way of doing this? indexes = a.findIndex(x => x.prop1==="abc") will return 0 for the above array, because it stops at the first successful find.

I feel like I want something like this: indexes = a.filtered(x.index() => x.prop1==="abc")

Answer 1

You can use Array#reduce method.

var data = [{prop1:"abc",prop2:"qwe"},{prop1:"abc",prop2:"yutu"},{prop1:"xyz",prop2:"qwrq"}];

console.log(
  data.reduce(function(arr, e, i) {
    if (e.prop1 == 'abc') arr.push(i);
    return arr;
  }, [])
)

// with ES6 arrow syntax
console.log(
  data.reduce((arr, e, i) => ((e.prop1 == 'abc') && arr.push(i), arr), [])
)

---

With simple for loop

var data = [{prop1:"abc",prop2:"qwe"},{prop1:"abc",prop2:"yutu"},{prop1:"xyz",prop2:"qwrq"}];

var res = [];

for (var i = 0; i < data.length; i++) {
  if (data[i].prop1 == 'abc') res.push(i);
}

console.log(res)

---

Or use Array#filter with Array#map method(Not an efficient way since it needs to iterate twice).

var data = [{prop1:"abc",prop2:"qwe"},{prop1:"abc",prop2:"yutu"},{prop1:"xyz",prop2:"qwrq"}];


console.log(
  data.map(function(e, i) {
    return i;
  }).filter(function(e) {
    return data[e].prop1 == 'abc';
  })
)

// With ES6 arrow syntax
console.log(
  data.map((_, i) => i).filter(e => data[e].prop1 == 'abc')
)

Answer 2

You can use array.prototype.reduce with shorthand if to make it with a single line of code:

var arr = [ {prop1:"abc",prop2:"qwe"}, {prop1:"abc",prop2:"yutu"},{prop1:"xyz",prop2:"qwrq"} ];

var indexes = arr.reduce((m, e, i) => (e.prop1 === 'abc' && m.push(i), m), []);
console.log(indexes);