JS如何根据另一个数组的排序对一个数组进行排序（）

Question

When A is.sorted(), it becomes 6, 7, 8, so those number get new index values, so do the same for B. get current index of the values inside B, then rearrange it based on A's new arrangement. So when A = 6,7,8, B would be u, z, h

var arrA = [8, 6, 7] // B follows this arr (A)
var arrB = ['h', 'u', 'z'] // B follows A

arrA.sort()
// output: 6, 7, 8

// arrB.followA’s arrangement somehow 
// output: u, z, h


arrA.reverse()
// output: 8, 7, 6

// arrB.follow A’s arrangement
// output: h, z, u


console.log(arrA);
console.log(arrB)

Answer 1

Use a brute force sort routine. Swap elements in "ary", and if those swap, then swap the same indexes in "bry". This way the index swapping will mirror each other.

var ary = [8, 7, 4, 3, 5, 1, 6];
var bry = ['u', 'z', 'y', 'a', 'b', 'f', 'r'];
var i = 0;
var j = 0;

for (i = 0; i < ary.length-1; i++){
    for (j = 0; j < ary.length-1; j++){
      if (ary[j] > ary[j+1]){
          let bigger = ary[j];
          ary[j] = ary[j+1];
          ary[j+1] = bigger;

     // make bry swap the same indexes as ary above
          bigger = bry[j];
          bry[j] = bry[j+1];
          bry[j+1] = bigger;
      }
    }
}

Answer 2

It's not possible if it is two different arrays. the simple approach to solve ur problem is to convert your second array into an object.

let b = {
    8: 'H',
    6: 'U',
    7: 'Z'
}

arrB = arrA.map((val) => return b[val]);

and use this object to get the value for array B based on the values of array A.

correct me if I'm wrong

Answer 3

Create a two-dimensional working array, whose elements are pairs of values from arrA and arrB :

var work = [];
arrA.forEach(function( v, i ) {
    work[ i ] = [ arrA[i], arrB[i] ];
});

Now you can arrange work in any order, and the values from arrA and arrB will stay in lockstep:

work.sort(function(x, y) {
    return Math.sign( x[0], y[0] );
});

(In the above example, we are ordering by the element in slot 0, which is from arrA . To order by the elements in arrB , change 0 to 1.)

Now you can alter work , eg:

work.reverse();

And extract the corresponding elements that were originally from arrA :

let newAarrA = work.map(function(x) {
    return x[0]; // from arrA
});
console.log(newArrA);

(Change 0 to 1 to get the corresponding elements from arrB instead.)

Answer 4

You can use a "Sorting with map" technique, you basically use a temporary array that maps arrB 's elements to values of arrA elements and then sort it.

 var arrA = [8, 6, 7] // B follows this arr (A) var arrB = ['h', 'u', 'z'] // B follows A var tmpMap = arrB.map((el, i) => ({ index: i, value: arrA[i] })); tmpMap.sort((a, b) => a.value - b.value); var arrB = tmpMap.map(el => arrB[el.index]); console.log(arrB);