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]
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)
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);
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)
})
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)
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)
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.
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];
}
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 ]
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()
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)
})
)
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