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

面向对象,可表达,可调试,无突变,已测试。

class Sorter {
    sortItem(array, fromIndex, toIndex) {
        const reduceItems = () => {
            const startingItems = array.slice(0, fromIndex);
            const endingItems = array.slice(fromIndex + 1);
            return startingItems.concat(endingItems);
        }
        const addMovingItem = (movingItem, reducedItems) => {
            const startingNewItems = reducedItems.slice(0, toIndex);
            const endingNewItems = reducedItems.slice(toIndex);
            const newItems = startingNewItems.concat([movingItem]).concat(endingNewItems);
            return newItems;
        }
        const movingItem = array[fromIndex];
        const reducedItems = reduceItems();
        const newItems = addMovingItem(movingItem, reducedItems);
        return newItems;
    }
}

const sorter = new Sorter();
export default sorter;
import sorter from 'src/common/Sorter';

test('sortItem first item forward', () => {
    const startingArray = ['a', 'b', 'c', 'd'];
    const expectedArray = ['b', 'a', 'c', 'd'];
    expect(sorter.sortItem(startingArray, 0, 1)).toStrictEqual(expectedArray);
});
test('sortItem middle item forward', () => {
    const startingArray = ['a', 'b', 'c', 'd'];
    const expectedArray = ['a', 'c', 'b', 'd'];
    expect(sorter.sortItem(startingArray, 1, 2)).toStrictEqual(expectedArray);
});
test('sortItem middle item backward', () => {
    const startingArray = ['a', 'b', 'c', 'd'];
    const expectedArray = ['a', 'c', 'b', 'd'];
    expect(sorter.sortItem(startingArray, 2, 1)).toStrictEqual(expectedArray);
});
test('sortItem last item backward', () => {
    const startingArray = ['a', 'b', 'c', 'd'];
    const expectedArray = ['a', 'b', 'd', 'c'];
    expect(sorter.sortItem(startingArray, 3, 2)).toStrictEqual(expectedArray);
});

其他回答

var ELEMS = ['a', 'b', 'c', 'd', 'e']; /* Source item will remove and it will be placed just after destination */ function moveItemTo(sourceItem, destItem, elements) { var sourceIndex = elements.indexOf(sourceItem); var destIndex = elements.indexOf(destItem); if (sourceIndex >= -1 && destIndex > -1) { elements.splice(destIndex, 0, elements.splice(sourceIndex, 1)[0]); } return elements; } console.log('Init: ', ELEMS); var result = moveItemTo('a', 'c', ELEMS); console.log('BeforeAfter: ', result);

我已经实现了一个不可变的ECMAScript 6解决方案,基于@Merc的答案在这里:

const moveItemInArrayFromIndexToIndex = (array, fromIndex, toIndex) => {
  if (fromIndex === toIndex) return array;

  const newArray = [...array];

  const target = newArray[fromIndex];
  const inc = toIndex < fromIndex ? -1 : 1;

  for (let i = fromIndex; i !== toIndex; i += inc) {
    newArray[i] = newArray[i + inc];
  }

  newArray[toIndex] = target;

  return newArray;
};

变量名可以缩短,只使用长变量名,这样代码就可以解释自己。

不复制数组的不可变版本:

const moveInArray = (arr, fromIndex, toIndex) => {
  if (toIndex === fromIndex || toIndex >= arr.length) return arr;

  const toMove = arr[fromIndex];
  const movedForward = fromIndex < toIndex;

  return arr.reduce((res, next, index) => {
    if (index === fromIndex) return res;
    if (index === toIndex) return res.concat(
      movedForward ? [next, toMove] : [toMove, next]
    );

    return res.concat(next);
  }, []);
};

一种方法是使用splice()从数组中删除项,然后再次使用splice()方法将删除的项插入到目标索引中。

Const array = ['a', 'b', 'c', 'd', 'e'] const newArray = moveItem(array, 3,1) //将元素从索引3移动到索引1 函数moveItem(arr, fromIndex, toIndex){ let itemRemoved = arr。splice(fromIndex, 1) //将删除的项赋值为数组 加勒比海盗。splice(toIndex, 0, itemRemoved[0]) //将itemRemoved插入目标索引 返回加勒比海盗 } console.log (newArray)

作为里德精彩回答的补充(因为我无法评论); 你可以使用模数来让负标和过大的下标“滚动”:

函数array_move(arr, old_index, new_index) { New_index =((New_index % arr.length) + arr.length) % arr.length; 加勒比海盗。拼接(new_index, 0, arr.)拼接(old_index, 1) [0]); 返回arr;//用于测试 } //返回[2,1,3] Console.log (array_move([1,2,3], 0,1));