Question

我有一个数组Arr1 = [1,1,2,2,3,8,4,6]。

如何根据元素位置的奇数/偶数将其拆分为两个数组？

subArr1 = [1,2,3,4]
subArr2 = [1,2,8,6]

Answer 1

odd  = arr.filter (v) -> v % 2
even = arr.filter (v) -> !(v % 2)

或者更加惯用的CoffeeScript：

odd  = (v for v in arr by 2)
even = (v for v in arr[1..] by 2)

Answer 2

你可以尝试：

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

使用嵌套数组会更容易：

result = [ [], [] ]

for (var i = 0; i < yourArray.length; i++)
    result[i & 1].push(yourArray[i])

如果您定位的是现代浏览器，则可以使用forEach替换循环：

yourArray.forEach(function(val, i) { 
    result[i & 1].push(val)
})

Answer 4

使用underscore的功能方法：

xs = [1, 1, 2, 2, 3, 8, 4, 6]
partition = _(xs).groupBy((x, idx) -> idx % 2 == 0)
[xs1, xs2] = [partition[true], partition[false]]

[edit]现在有_.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

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 7

没有模运算符的方法：

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 8

只是为了好玩，两行，因为它被标记为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 ]

Answer 9

作为使用下划线链接功能的tokland解决方案的单线改进：

xs = [1, 1, 2, 2, 3, 8, 4, 6]
_(xs).chain().groupBy((x, i) -> i % 2 == 0).values().value()

Answer 10

filters是非静态的＆amp;非内置Array方法，接受literal object过滤器功能＆amp;返回literal object个数组，其中输入和输入输出由对象键映射。

    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)
      })
    )

基于奇数/偶数位置将数组拆分为两个数组

10 个答案: