divide an array into two separate arrays depending on whether the index position is odd or even

Question

divide an array into two separate arrays depending on whether the index position is odd or even

Here is an example array: Arr1 = [1,1,2,2,3,8,4,6].

I'm trying to divide this array into two new arrays based on the odd or even position of elements. Can anyone provide a solution?

New Array 1 (odd positions): [1,2,3,4]
New Array 2 (even positions): [1,2,8,6]

javascript coffeescript

Answer 1

Answer №1

nums = list.select (n) -> n % 7 == 0
otherNums = list.filter (n) -> n % 7 != 0

In a more stylistic CoffeeScript approach:

nums = (n for n in list if n % 7 == 0)
otherNums = (n for n in list when n % 7 != 0)

Answer 2

nums = list.select (n) -> n % 7 == 0
otherNums = list.filter (n) -> n % 7 != 0

In a more stylistic CoffeeScript approach:

nums = (n for n in list if n % 7 == 0)
otherNums = (n for n in list when n % 7 != 0)

Answer 3

Answer №2

If you're looking for a solution, you might want to consider the following approach:

var Array1 = [1,1,2,2,3,8,4,6],
    Array2 = [],
    Array3 = [];

for (var index=0;index<Array1.length;index++){
    if ((index+2)%2==0) {
        Array3.push(Array1[index]);
    }
    else {
        Array2.push(Array1[index]);
    }
}

console.log(Array2);

Check out this JS Fiddle demo.

Answer 4

If you're looking for a solution, you might want to consider the following approach:

var Array1 = [1,1,2,2,3,8,4,6],
    Array2 = [],
    Array3 = [];

for (var index=0;index<Array1.length;index++){
    if ((index+2)%2==0) {
        Array3.push(Array1[index]);
    }
    else {
        Array2.push(Array1[index]);
    }
}

console.log(Array2);

Check out this JS Fiddle demo.

Answer 5

Answer №3

To make things simpler, consider using nested arrays:

finalResult = [ [], [] ]

for (let j = 0; j < inputArray.length; j++)
    finalResult[j & 1].push(inputArray[j])

If your audience mainly consists of users with modern browsers, you could swap the loop for forEach:

inputArray.forEach(function(value, index) { 
    finalResult[index & 1].push(value)
})

Answer 6

To make things simpler, consider using nested arrays:

finalResult = [ [], [] ]

for (let j = 0; j < inputArray.length; j++)
    finalResult[j & 1].push(inputArray[j])

If your audience mainly consists of users with modern browsers, you could swap the loop for forEach:

inputArray.forEach(function(value, index) { 
    finalResult[index & 1].push(value)
})

Answer 7

Answer №4

Implementing a practical method utilizing underscore:

numbers = [1, 1, 2, 2, 3, 8, 4, 6]
split = _(numbers).groupBy((num, index) -> index % 2 == 0)
[numbers1, numbers2] = [split[true], split[false]]

[edit] An updated solution is available using _.partition:

[numbers1, numbers2] = _(numbers).partition((num, index) -> index % 2 == 0)

Answer 8

Implementing a practical method utilizing underscore:

numbers = [1, 1, 2, 2, 3, 8, 4, 6]
split = _(numbers).groupBy((num, index) -> index % 2 == 0)
[numbers1, numbers2] = [split[true], split[false]]

[edit] An updated solution is available using _.partition:

[numbers1, numbers2] = _(numbers).partition((num, index) -> index % 2 == 0)

Answer 9

Answer №5

let numbers = [1, 1, 2, 2, 3, 8, 4, 6]
let evenNumbers = []; 
let oddNumbers = []

let index;
for (index = 0; index <= numbers.length; index = index + 2) {
    if (numbers[index] !== undefined) {
        evenNumbers.push(numbers[index]);
        oddNumbers.push(numbers[index + 1]);
    }
}
console.log(evenNumbers, oddNumbers)

Answer 10

let numbers = [1, 1, 2, 2, 3, 8, 4, 6]
let evenNumbers = []; 
let oddNumbers = []

let index;
for (index = 0; index <= numbers.length; index = index + 2) {
    if (numbers[index] !== undefined) {
        evenNumbers.push(numbers[index]);
        oddNumbers.push(numbers[index + 1]);
    }
}
console.log(evenNumbers, oddNumbers)

Answer 11

Answer №6

An idea would be to create two separate for loops that increase by 2 each time. In the initial loop, you could begin at 0, while in the subsequent loop, you could start at 1.

Answer 12

An idea would be to create two separate for loops that increase by 2 each time. In the initial loop, you could begin at 0, while in the subsequent loop, you could start at 1.

Answer 13

Answer №7

An alternative approach eliminating the need for modulo operator:

let evenArray = [];
let oddArray = [];
let index = 0;
let j;

for (j = 1; j < mainArray.length; j = j + 2){
    // Assign values to even and odd arrays based on index
    evenArray[index] = mainArray[j];
    oddArray[index] = mainArray[j - 1];
    index++;
}

// Handle last remaining number if array length is odd
if((j - 1) < mainArray.length){
    oddArray[index] = mainArray[j - 1];
}

Answer 14

An alternative approach eliminating the need for modulo operator:

let evenArray = [];
let oddArray = [];
let index = 0;
let j;

for (j = 1; j < mainArray.length; j = j + 2){
    // Assign values to even and odd arrays based on index
    evenArray[index] = mainArray[j];
    oddArray[index] = mainArray[j - 1];
    index++;
}

// Handle last remaining number if array length is odd
if((j - 1) < mainArray.length){
    oddArray[index] = mainArray[j - 1];
}

Answer 15

Answer №8

Just for kicks, here is a little coffeescript snippet in two concise lines:

Numbers = [3,5,7,9,10,13]
[odd, even] = [a, b] = [[], []]
([b,a]=[a,b])[0].push num for num in Numbers

console.log odd, even
# [ 3, 5, 7, 9, 13 ] [ 10 ]

Answer 16

Just for kicks, here is a little coffeescript snippet in two concise lines:

Numbers = [3,5,7,9,10,13]
[odd, even] = [a, b] = [[], []]
([b,a]=[a,b])[0].push num for num in Numbers

console.log odd, even
# [ 3, 5, 7, 9, 13 ] [ 10 ]

Answer 17

Answer №9

Here's a more concise version of tokland's proposed solution, utilizing underscore chaining function:

numbers = [8, 4, 21, 15, 10, 3, 6, 9]
_(numbers).chain().groupBy((num, index) -> index % 2 == 0).values().value()

Answer 18

Here's a more concise version of tokland's proposed solution, utilizing underscore chaining function:

numbers = [8, 4, 21, 15, 10, 3, 6, 9]
_(numbers).chain().groupBy((num, index) -> index % 2 == 0).values().value()

Answer 19

Answer №10

CustomFilter is a unique Array method that allows users to input a collection of filter functions and receive an output containing arrays where the input and output are associated by keys within an object.


    Array.prototype.CustomFilter = function (filters) {
      let results = {};
      Object.keys(filters).forEach((key)=>{
         results[key] = this.filter(filters[key])   
      }); 
      return results;
    }
    //---- example : 
    
    console.log(
      [12,2,11,7,92,14,5,5,3,0].CustomFilter({
        odd:  (e) => (e%2),
        even: (e) => !(e%2)
      })
    )

Answer 20

CustomFilter is a unique Array method that allows users to input a collection of filter functions and receive an output containing arrays where the input and output are associated by keys within an object.


    Array.prototype.CustomFilter = function (filters) {
      let results = {};
      Object.keys(filters).forEach((key)=>{
         results[key] = this.filter(filters[key])   
      }); 
      return results;
    }
    //---- example : 
    
    console.log(
      [12,2,11,7,92,14,5,5,3,0].CustomFilter({
        odd:  (e) => (e%2),
        even: (e) => !(e%2)
      })
    )

divide an array into two separate arrays depending on whether the index position is odd or even

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

Answer №8

Answer №9

Answer №10

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