Delete even/odd occurance numbers in an array with For Loop

Question

I have an array as shown below, containing 6 numbers. I'm trying to create a function that deletes a value if it's an even/odd occurance. For example, if I type the function deleteData(myArr,"even"), it will remove data number 0th, 2nd, and 4th. And the other way around if it's "odd". But when I tried running my code, it didn't work the way i wanted it to go. Can anybody tell me why?

const myArr=[90,34,28,19,26,22];


function deleteData (array,occuranceType){
    switch (occuranceType){
        case "odd":
            for (let i=0;i<array.length;i++){
                if (i%2!==0){
                    array.splice(i,1);
                };
            };
            break;
        case "even":
            for (let i=0;i<array.length;i++){
                if (i%2==0){
                    array.splice(i,1);
                };
            };
            break;
    };
};

deleteData(arr,"odd")
console.log(arr)

This is the result in the console : [ 90, 28, 19, 22 ]

you need subtract 1 from `i` variable every time you remove an item. — vanowm, Jul 03 '22 at 14:04

score 0 · Answer 1 · edited Jul 03 '22 at 14:18

0

You can use filter for this since it's easier and you don't mutate the original array.

let myArr = [90, 34, 28, 19, 26, 22];

function deleteData(array, removeEven) {
  return array.filter((v, i) => removeEven ? (i % 2 !== 0) : (i % 2 === 0));
};

myArr = deleteData(myArr, true)
console.log(myArr)

edited Jul 03 '22 at 14:18

DecPK

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answered Jul 03 '22 at 14:11

Cristian-Florin Calina

928
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1

You can also do as `array.filter((o, i) => (occuranceType === 'odd' ? !(i % 2) : i % 2))` – DecPK Jul 03 '22 at 14:15
1

that was my first solution that I wanted to post, but I figured since it's only 2 choices, a `boolean` fits the scenario. – Cristian-Florin Calina Jul 03 '22 at 14:16
Yup, you are right – DecPK Jul 03 '22 at 14:16
Why do you explicitly compare it with `0`? We already know it's an integer. It doesn't bring any visual or performance benefit...in fact it looks much worse than `removeEven ? i % 2 : !(i % 2)` – vanowm Jul 03 '22 at 14:58
It's not a performance benefit, it's just easier to read in my opinion. – Cristian-Florin Calina Jul 03 '22 at 15:08

SSM · Answer 2 · 2022-07-03T14:31:22.223

0

You can filter the array based on the pos argument passed to the function.

const deleteData = (array, pos) =>
  array.filter((_, i) => i % 2 === Number(pos !== "even"));

console.log(deleteData([90, 34, 28, 19, 26, 22], "odd"));
console.log(deleteData([90, 34, 28, 19, 26, 22], "even"));

Note: Number(false) returns 0 and Number(true) returns 1.

edited Jul 03 '22 at 14:31

answered Jul 03 '22 at 14:25

SSM

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Delete even/odd occurance numbers in an array with For Loop

2 Answers2