I have an array Arr1 = [1,1,2,2,3,8,4,6].
How can I split it into two arrays based on the odd/even-ness of element positions?
subArr1 = [1,2,3,4]
subArr2 = [1,2,8,6]
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And please read the formatting FAQ; you've had almost a year to do so.Lightness Races in Orbit– Lightness Races in Orbit2011-11-14 10:08:47 +00:00Commented Nov 14, 2011 at 10:08
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What's the point in this? I don't really get what your problem is or according to which criteria you want to split them. A few more infos would be great.Termi– Termi2011-11-14 10:09:16 +00:00Commented Nov 14, 2011 at 10:09
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@Termi: It's pretty clear (albeit mostly from the title)Lightness Races in Orbit– Lightness Races in Orbit2011-11-14 10:10:10 +00:00Commented Nov 14, 2011 at 10:10
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@TomalakGeret'kal: You're right, my bad. I just overflew it.Termi– Termi2011-11-14 10:13:31 +00:00Commented Nov 14, 2011 at 10:13
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That said... do you want the answer in JavaScript, or CoffeeScript? Learn to be precise!Lightness Races in Orbit– Lightness Races in Orbit2011-11-14 10:13:38 +00:00Commented Nov 14, 2011 at 10:13
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11 Answers
odd = arr.filter (v) -> v % 2
even = arr.filter (v) -> !(v % 2)
Or in more idiomatic CoffeeScript:
odd = (v for v in arr by 2)
even = (v for v in arr[1..] by 2)
1 Comment
Exis Zhang
should be
arr.filter((v, i) => i % 2) since OP's question is based on position not valueYou could try:
var Arr1 = [1,1,2,2,3,8,4,6],
Arr2 = [],
Arr3 = [];
for (var i=0;i<Arr1.length;i++){
if ((i+2)%2==0) {
Arr3.push(Arr1[i]);
}
else {
Arr2.push(Arr1[i]);
}
}
console.log(Arr2);
1 Comment
David Thomas
Man, I could be good at something, if I'd just avoid being distracted...sigh. Thanks! (And edited the link in..!) XD
It would be easier using nested arrays:
result = [ [], [] ]
for (var i = 0; i < yourArray.length; i++)
result[i & 1].push(yourArray[i])
if you're targeting modern browsers, you can replace the loop with forEach:
yourArray.forEach(function(val, i) {
result[i & 1].push(val)
})
Comments
A functional approach using underscore:
xs = [1, 1, 2, 2, 3, 8, 4, 6]
partition = _(xs).groupBy((x, idx) -> idx % 2 == 0)
[xs1, xs2] = [partition[true], partition[false]]
[edit] Now there is _.partition:
[xs1, xs2] = _(xs).partition((x, idx) -> idx % 2 == 0)
Comments
var Arr1 = [1, 1, 2, 2, 3, 8, 4, 6]
var evenArr=[];
var oddArr = []
var i;
for (i = 0; i <= Arr1.length; i = i + 2) {
if (Arr1[i] !== undefined) {
evenArr.push(Arr1[i]);
oddArr.push(Arr1[i + 1]);
}
}
console.log(evenArr, oddArr)
Comments
I guess you can make 2 for loops that increment by 2 and in the first loop start with 0 and in the second loop start with 1
Comments
A method without modulo operator:
var subArr1 = [];
var subArr2 = [];
var subArrayIndex = 0;
var i;
for (i = 1; i < Arr1.length; i = i+2){
//for even index
subArr1[subArrayIndex] = Arr1[i];
//for odd index
subArr2[subArrayIndex] = Arr1[i-1];
subArrayIndex++;
}
//For the last remaining number if there was an odd length:
if((i-1) < Arr1.length){
subArr2[subArrayIndex] = Arr1[i-1];
}
2 Comments
Lightness Races in Orbit
This is not the simplest way.
lauhub
well, I agree I should a change the first sentence and write: "a method without modulo calculation"
Just for fun, in two lines, given that it's been tagged coffeescript :
Arr1 = [1,1,2,2,3,8,4,6]
[even, odd] = [a, b] = [[], []]
([b,a]=[a,b])[0].push v for v in Arr1
console.log even, odd
# [ 1, 2, 3, 4 ] [ 1, 2, 8, 6 ]
Comments
As a one-liner improvement to tokland's solution using underscore chaining function:
xs = [1, 1, 2, 2, 3, 8, 4, 6]
_(xs).chain().groupBy((x, i) -> i % 2 == 0).values().value()
Comments
filters is a non-static & non-built-in Array method , which accepts literal object of filters functions & returns a literal object of arrays where the input & the output are mapped by object keys.
Array.prototype.filters = function (filters) {
let results = {};
Object.keys(filters).forEach((key)=>{
results[key] = this.filter(filters[key])
});
return results;
}
//---- then :
console.log(
[12,2,11,7,92,14,5,5,3,0].filters({
odd: (e) => (e%2),
even: (e) => !(e%2)
})
)
Comments
You can use Object.groupBy (since ECMAScript 2024) for this. With destructuring your example can be processed like this:
const arr = [1,1,2,2,3,8,4,6];
const { even, odd } = Object.groupBy(arr, (_, i) => i % 2 ? "odd" : "even");
console.log(even);
console.log(odd);