javascript 根据奇数/偶数位置将数组拆分为两个数组
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split an array into two arrays based on odd/even position
提问by Running Turtle
I have an array Arr1 = [1,1,2,2,3,8,4,6].
我有一个数组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?
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]
采纳答案by David says reinstate Monica
You could try:
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);
回答by Ricardo Tomasi
odd = arr.filter (v) -> v % 2
even = arr.filter (v) -> !(v % 2)
Or in more idiomatic CoffeeScript:
或者在更惯用的 CoffeeScript 中：
odd = (v for v in arr by 2)
even = (v for v in arr[1..] by 2)
回答by georg
It would be easier using nested arrays:
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])
if you're targeting modern browsers, you can replace the loop with forEach:
如果您的目标是现代浏览器，则可以将循环替换为forEach：
yourArray.forEach(function(val, i) {
result[i & 1].push(val)
})
回答by tokland
A functional approach using underscore:
使用下划线的功能方法：
xs = [1, 1, 2, 2, 3, 8, 4, 6]
partition = _(xs).groupBy((x, idx) -> idx % 2 == 0)
[xs1, xs2] = [partition[true], partition[false]]
[edit] Now there is _.partition:
[编辑] 现在有_.partition：
[xs1, xs2] = _(xs).partition((x, idx) -> idx % 2 == 0)
回答by Ashish Yadav
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)
回答by Abdennour TOUMI
filtersis a non-static & non-built-in Array method , which accepts literal objectof filters functions & returns a literal objectof arrays where the input & the output are mapped by object keys.
filters是一个非静态和非内置数组方法，它接受literal object过滤器函数并返回一个literal object数组，其中输入和输出由对象键映射。
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)
})
)回答by Adrien
Just for fun, in two lines, given that it's been tagged coffeescript :
只是为了好玩，在两行中，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 ]
回答by kmARC
As a one-liner improvement to tokland's solution using underscore chaining function:
作为使用下划线链接函数对 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()
回答by Samir Adel
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
我猜你可以做 2 个 for 循环，以 2 为增量，在第一个循环中从 0 开始，在第二个循环中从 1 开始
回答by lauhub
A method without modulo operator:
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];
}
