Question

假设我有一个数组：[0,3,4,2,5,1]。

我想要做的是对数组进行排序，例如：

["one", "two", "three", "four", "five", "six"]

这样顺序对应第一个数组。

这将是输出：

["one", "four", "five", "three", "six", "two"]

有没有一种简单的方法可以实现这一目标？

Answer 1

您可以这样做：

function getSorted(arr, sortArr) {
  var result = [];
  for (var i = 0; i < arr.length; i++) {
    console.log(sortArr[i], arr[i]);
    result[i] = arr[sortArr[i]];
  }
  return result;
}
var arr = ["one", "two", "three", "four", "five", "six"];
var sortArr = [0, 3, 4, 2, 5, 1];
alert(getSorted(arr, sortArr));

注意：这假设您传入的数组大小相同，如果情况不是这样，您需要添加一些额外的检查。

Answer 2

orderedArray= function(arr,order){
    return  order.map(function(itm){return arr[itm]});
}

var sequence= [0, 3, 4, 2, 5, 1],arr=["one","two","three","four","five","six"]

arr=new orderedArray(arr,sequence);

/*  returned value: (Array)
one,four,five,three,six,two
*/

//您可以使订单成为数组的无索引属性， //并调用array.ordered（）

Array.prototype.ordered= function(order){
    var arr= this;
    order=order || this.order;
    return order.map(function(itm){
        return arr[itm];
    });
}


var arr= ["one","two","three","four","five","six"],
sequence= [0, 3, 4, 2, 5, 1];

arr.order=sequence;

arr.ordered()

/*  returned value: (Array)
one,four,five,three,six,two
*/

Answer 3

我在接受电话采访时被问到这个问题。然后在不创建另一个数组的情况下进行，假设数组非常大。我不知道这是否是答案，因为我无法接听电话（该死的！），但这就是我想出来的。

var my_obj_array = ['a', 'b', 'c', 'd'];
var my_indicies = [3, 1, 0, 2];
// desired result ['d', 'b', 'a', 'c']

var temp = {};
for (var i = 0; i < my_indicies.length; i++) {
    temp[i] = my_obj_array[i];  // preserve
    var j = my_indicies[i];
    if (j in temp) {
        my_obj_array[i] = temp[j];
        delete temp[j];
    } else {
        my_obj_array[i] = my_obj_array[j];
    }
}

http://jsfiddle.net/innerb/RENjW/

Answer 4

不是你如何得到你的第一个数组，但你可以使用一组对象而不是[0,3,4,2,5,1]：

var arr = [
  {n:0, s:'one'},
  {n:3, s:'four'},
  {n:4, s:'five'},
  {n:2, s:'three'},
  {n:5, s:'six'},
  {n:1, s:'two'}
]

避免处理它。

Answer 5

You can use first array as directory for sorting second one using map method:

const firstArray = [0, 3, 4, 2, 5, 1];
const secondArray = ['one', 'two', 'three', 'four', 'five', 'six'];

const result = firstArray.map((item) => {
 return secondArray[item];
});
// result = ["one", "four", "five", "three", "six", "two"]

Answer 6

let inds = [0,3,4,2,5,1];
let x = ["one", "two", "three", "four", "five", "six"];
let x2 = [];  //Output
inds.forEach(ind => x2.push(x[ind]));
x2
//(6) ["one", "four", "five", "three", "six", "two"]

Answer 7

class test1
{
  public static String[] sort(int[] array,String[] str)
  {
    String[] out=new String[str.length];
    for(int i=0;i<str.length;i++)
    {
      out[i]=str[array[i]];
    }
    return out;
  }
}

根据另一个整数数组对数组进行排序

7 个答案: