我很难弄清楚如何移动数组中的一个元素。例如,给定以下条件:

var array = [ 'a', 'b', 'c', 'd', 'e'];

我怎么能写一个函数来移动元素'd'到'b'的左边?

还是c右边的a ?

移动元素之后,应该更新其余元素的索引。结果数组将是:

array = ['a', 'd', 'b', 'c', 'e']

这看起来应该很简单,但我无法理解它。


当前回答

我的2摄氏度。易读,有效,快速,不需要创建新数组。

function move(array, from, to) {
  if( to === from ) return array;

  var target = array[from];                         
  var increment = to < from ? -1 : 1;

  for(var k = from; k != to; k += increment){
    array[k] = array[k + increment];
  }
  array[to] = target;
  return array;
}

其他回答

在2022年,这个typescript实用程序将与单元测试一起工作。

export const arrayMove = <T>(arr: T[], fromIndex: number, toIndex: number) => {
  const newArr = [...arr];
  newArr.splice(toIndex, 0, newArr.splice(fromIndex, 1)[0]);
  return newArr;
};

const testArray = ['1', '2', '3', '4']; description ('arrayMove', () => { it('应该将数组项移动到toIndex', () => { expect(arrayMove(testArray, 2,0))。toEqual(['3', '1', '2', '4']); expect(arrayMove(testArray, 3,1))。toEqual(['1', '4', '2', '3']); expect(arrayMove(testArray, 1,2))。toEqual(['1', '3', '2', '4']); expect(arrayMove(testArray, 0,2))。toEqual(['2', '3', '1', '4']); }); });

我使用不可变性助手库解决了我的问题。

import update from 'immutability-helper';

const move = (arr: any[], from: number, to: number) => update(arr, {
  $splice: [
    [from, 1],
    [to, 0, arr[from] as string],
  ],
});

const testArray = ['a', 'b', 'c', 'd', 'e'];
console.log(move(testArray, 1, 3)); // [ 'c', 'b', 'c', 'd', 'e' ]
console.log(move(testArray, 4, 0)); // [ 'e', 'b', 'c', 'd', 'a' ]

另一个纯JS变体使用ES6数组展开运算符,没有突变

const reorder = (array, sourceIndex, destinationIndex) => { const smallerIndex = Math.min(sourceIndex, destinationIndex); const largerIndex = Math.max(sourceIndex, destinationIndex); return [ ...array.slice(0, smallerIndex), ...(sourceIndex < destinationIndex ? array.slice(smallerIndex + 1, largerIndex + 1) : []), array[sourceIndex], ...(sourceIndex > destinationIndex ? array.slice(smallerIndex, largerIndex) : []), ...array.slice(largerIndex + 1), ]; } // returns ['a', 'c', 'd', 'e', 'b', 'f'] console.log(reorder(['a', 'b', 'c', 'd', 'e', 'f'], 1, 4))

我最终将这两种方法结合起来,以便在移动小距离和大距离时更好地工作。我得到了相当一致的结果,但这可能会被比我更聪明的人稍微调整一下,以不同的大小工作,等等。

在小距离移动对象时,使用其他一些方法明显比使用拼接快(x10)。这可能会根据数组的长度而改变,但对于大型数组是正确的。

function ArrayMove(array, from, to) {
    if ( Math.abs(from - to) > 60) {
        array.splice(to, 0, array.splice(from, 1)[0]);
    } else {
        // works better when we are not moving things very far
        var target = array[from];
        var inc = (to - from) / Math.abs(to - from);
        var current = from;
        for (; current != to; current += inc) {
            array[current] = array[current + inc];
        }
        array[to] = target;    
    }
}

https://web.archive.org/web/20181026015711/https://jsperf.com/arraymove-many-sizes

    Array.prototype.moveUp = function (value, by) {
        var index = this.indexOf(value),
            newPos = index - (by || 1);

        if (index === -1)
            throw new Error("Element not found in array");

        if (newPos < 0)
            newPos = 0;

        this.splice(index, 1);
        this.splice(newPos, 0, value);
    };

    Array.prototype.moveDown = function (value, by) {
        var index = this.indexOf(value),
            newPos = index + (by || 1);

        if (index === -1)
            throw new Error("Element not found in array");

        if (newPos >= this.length)
            newPos = this.length;

        this.splice(index, 1);
        this.splice(newPos, 0, value);
    };



    var arr = ['banana', 'curyWurst', 'pc', 'remembaHaruMembaru'];

    alert('withiout changes= '+arr[0]+' ||| '+arr[1]+' ||| '+arr[2]+' ||| '+arr[3]);
    arr.moveDown(arr[2]);


    alert('third word moved down= '+arr[0] + ' ||| ' + arr[1] + ' ||| ' + arr[2] + ' ||| ' + arr[3]);
    arr.moveUp(arr[2]);
    alert('third word moved up= '+arr[0] + ' ||| ' + arr[1] + ' ||| ' + arr[2] + ' ||| ' + arr[3]);

http://plnkr.co/edit/JaiAaO7FQcdPGPY6G337?p=preview