基于奇/偶位置将一个数组拆分为两个数组

split an array into two arrays based on odd/even position

本文关键字：数组 一个 拆分 两个 于奇 位置      更新时间：2023-09-26

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/偶数将其分成两个数组?

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

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

或者在更习惯的CoffeeScript中:

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

你可以试试:

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提琴演示

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])

如果您针对的是现代浏览器，您可以将循环替换为forEach:

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

使用下划线:

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

[edit]现在有一个__partition:

[xs1, xs2] = _(xs).partition((x, idx) -> idx % 2 == 0)

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)

我猜你可以创建两个以2为增量的for循环第一个循环从0开始第二个循环从1开始

不带模算子的方法:

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];
}

只是为了好玩，在两行中，假设它被标记为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 ]

使用下划线链函数对tokland的解决方案进行一行改进:

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

filters为非静态&非内置数组方法，接受滤波器函数&amp的literal object;返回一个literal object数组，其中输入&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)
      })
    )

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