How to split an Array into Odd Array and Even Array

Question

I have a problem with splitting the array into to separate arrays Odd and Even. What I did was:

componentDidMount() {
 axios.get( `${this.props.cmsUrl}types/genres`)
 .then(response => { if ( response.data !== undefined ) {

  let even = [];
  let odd = [];
  let allData = ["a", "b", "c", "d", "e", "f", "g", "h", ];

  for (var i = 0; i < allData.length; ++i) {
    if ( ( allData[i] % 2 ) === 0) { 
      even.push(allData[i]);
    } 
    else { 
      odd.push(allData[i]); 
    }
  };
  console.log("allData : ",allData);
  console.log("even : ",even);
  console.log("even : ",odd);
 }}
)
}

What I really was expecting is

allData = [a, b, c, d]
odd = [a , c]
even = [b, d]

But what really happened is

allData = even 
odd = empty array

is this my problem => ( allData[i] % 2 ) === 1

The console.log is :

  allData :  (8) ["a", "b", "c", "d", "e", "f", "g", "h"]0: "a"1: "b"2: "c"3: 
  "d"4: "e"5: "f"6: "g"7: "h"length: 8__proto__: Array(0)
    details.js:56 even :  []length: 0__proto__: Array(0)
    details.js:57 odd :  (8) ["a", "b", "c", "d", "e", "f", "g", "h"]0: 
   "a"1: "b"2: "c"3: "d"4: "e"5: "f"6: "g"7: "h"length: 8__proto__: Array(0)

Thanks

Mr Soroush_Neshat : This is the console log allData : (8) ["a", "b", "c", "d", "e", "f", "g", "h"]0: "a"1: "b"2: "c"3: "d"4: "e"5: "f"6: "g"7: "h"length: 8__proto__: Array(0) details.js:56 even : []length: 0__proto__: Array(0) details.js:57 odd : (8) ["a", "b", "c", "d", "e", "f", "g", "h"]0: "a"1: "b"2: "c"3: "d"4: "e"5: "f"6: "g"7: "h"length: 8__proto__: Array(0) — Ranny
– Ranny, Commented Sep 4, 2018 at 7:38
@Ranny tnx for infos . i posted the answer to solve your problem — SoroushNeshat
– SoroushNeshat, Commented Sep 4, 2018 at 7:49

SoroushNeshat · Accepted Answer · 2018-09-04 08:46:29Z

2

You've to use charCodeAt method because allData is an array of string not array of numbers.

It's the solution of your problem:

if ( ( allData[i].charCodeAt(0) % 2 ) === 0) { your code }

=================

as Thomas Scheffer mentioned it seems u want to sort based on character index not its value .

so if u want to sort based on index u have to write :

if ( ( i % 2 ) === 0) { your code }

it converts like this :

allData = ["b","f","z","w"] => { odd=["f","w"], even=["b","z"]  }

but if u want to sort based on character value u have to write :

if ( ( allData[i].charCodeAt(0) % 2 ) === 0) { your code }

it converts like this :

allData = ["b","f","z","w"] => { odd=["w"] , even=["b","f","z"] }

edited Sep 4, 2018 at 8:46

answered Sep 4, 2018 at 7:48

SoroushNeshat

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9 Comments

Thomas Scheffer Over a year ago

Or just use if( ( i % 2 ) === 0 ){ } as its about the index number not really the data in allData.

Ranny Over a year ago

@Soroush_Neshat Thaaaaaaaaaank you very much it is working now

SoroushNeshat Over a year ago

@ThomasScheffer if characters are shuffled u cant do that anymore

Thomas Scheffer Over a year ago

@Soroush_Neshat charCodeAt() returns the ASCII code for the first character in that value. Not sure if thats what @Ranny was asking for.

Thomas Scheffer Over a year ago

@Ranny Just remember that Arrays start at 0, so this will skew your expected results. As "a" would be in even instead of odd.

|

subramanian · Accepted Answer · 2018-09-04 08:03:18Z

1

instead of for loops you can use filter method

/*ES6*/

let evenArray = result.filter((a,i) => i%2);
let oddArray = result.filter((a,i) => !(i%2));

/*ES5*/

let evenArray = result.filter(function(a,i){return i%2});
let oddArray = result.filter(function(a,i){return !(i%2)});

edited Sep 4, 2018 at 8:03

answered Sep 4, 2018 at 7:38

subramanian

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Comments

Artem Mirchenko · Accepted Answer · 2018-09-04 07:31:43Z

0

Right condition is if ( ( allData[i] % 2 ) === 0)

Try in console:

4 % 2
4 % 3

answered Sep 4, 2018 at 7:31

Artem Mirchenko

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3 Comments

Ranny Over a year ago

That flip my empty array to be same allData and second one is empty

Artem Mirchenko Over a year ago

Can you print your allData array?

Ranny Over a year ago

if i made : if ( ( allData[i] % 2 ) === 0) and else if ( ( allData[i] % 2 ) === 1) both arrays are empty now

Teneff · Accepted Answer · 2018-09-04 07:52:30Z

0

You can use Array.prototype.reduce

let allData = [ "a", "b", "c", "d", "e", "f", "g", "h" ];

const { even, odd } = allData.reduce((acc, val, index) => {
    const key = index % 2 ? 'odd' : 'even';
    return {
       ...acc,
       [key]: acc[key].concat(val),
    };
}, { odd: [], even: [] });

console.log( 'even', even );
console.log( 'odd', odd );

edited Sep 4, 2018 at 7:52

answered Sep 4, 2018 at 7:42

Teneff

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Comments

Mustafa Poya · Accepted Answer · 2022-11-13 08:28:17Z

0

let allData = ["a", "b", "c", "d", "e", "f", "g", "h", ];

let even = [];
let odd = [];

for (let i = 0; i < allData.length; i++) {
if (i % 2 == 0) {
    even.push(allData[i]);
} else {
    odd.push(allData[i]);
}
}

console.log("Even array is : " + even);
console.log("Even array is : " + odd);

edited Nov 13, 2022 at 8:28

Mustafa Poya

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answered Nov 8, 2022 at 9:29

Sunny More

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