How to remove all the odd indexes (eg: a[1],a[3]..) value from the array

Question

I have an array like var aa = ["a","b","c","d","e","f","g","h","i","j","k","l"]; I wanted to remove element which is place on even index. so ouput will be line aa = ["a","c","e","g","i","k"];

I tried in this way

for (var i = 0; aa.length; i = i++) {
if(i%2 == 0){
    aa.splice(i,0);
}
};

But it is not working.

Answer 1

Use Array#filter method

var aa = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l"];

var res = aa.filter(function(v, i) {
  // check the index is odd
  return i % 2 == 0;
});

console.log(res);

---

If you want to update existing array then do it like.

var aa = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l"],
    // variable for storing delete count
  dCount = 0,
    // store array length
  len = aa.length;

for (var i = 0; i < len; i++) {
  // check index is odd
  if (i % 2 == 1) {
    // remove element based on actual array position 
    // with use of delete count
    aa.splice(i - dCount, 1);
    // increment delete count
    // you combine the 2 lines as `aa.splice(i - dCount++, 1);`
    dCount++;
  }
}


console.log(aa);

---

Another way to iterate for loop in reverse order( from last element to first ).

var aa = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l"];

// iterate from last element to first
for (var i = aa.length - 1; i >= 0; i--) {
  // remove element if index is odd
  if (i % 2 == 1)
    aa.splice(i, 1);
}


console.log(aa);