基于奇数/偶数位置将数组拆分为两个数组

Question

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]

Answer 1

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)

Answer 2

You 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);

JS Fiddle demo .

Answer 3

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)
})

Answer 4

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)

Answer 5

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)

Answer 6

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()

Answer 7

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) }) ) 

Answer 8

我猜你可以为2增加2的循环，在第一个循环中以0开始，在第二个循环中以1开始

Answer 9

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];
}

Answer 10

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 ]