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
该方法将保留原始数组，并检查边界错误。
const move = (from, to, arr) => {
to = Math.max(to,0)
from > to
? [].concat(
arr.slice(0,to),
arr[from],
arr.filter((x,i) => i != from).slice(to))
: to > from
? [].concat(
arr.slice(0, from),
arr.slice(from + 1, to + 1),
arr[from],
arr.slice(to + 1))
: arr}
其他回答
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))
打印稿版本
摘自@Merc的回答。我最喜欢这个，因为它没有创建新的数组，而是在适当的地方修改数组。我所做的只是更新到ES6并添加类型。
export function moveItemInArray<T>(workArray: T[], fromIndex: number, toIndex: number): T[] {
if (toIndex === fromIndex) {
return workArray;
}
const target = workArray[fromIndex];
const increment = toIndex < fromIndex ? -1 : 1;
for (let k = fromIndex; k !== toIndex; k += increment) {
workArray[k] = workArray[k + increment];
}
workArray[toIndex] = target;
return workArray;
}
该方法将保留原始数组，并检查边界错误。
const move = (from, to, arr) => {
to = Math.max(to,0)
from > to
? [].concat(
arr.slice(0,to),
arr[from],
arr.filter((x,i) => i != from).slice(to))
: to > from
? [].concat(
arr.slice(0, from),
arr.slice(from + 1, to + 1),
arr[from],
arr.slice(to + 1))
: arr}
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
下面是我在JSPerf....上找到的一行代码
Array.prototype.move = function(from, to) {
this.splice(to, 0, this.splice(from, 1)[0]);
};
这是很棒的阅读，但如果你想要性能(在小数据集)尝试……
Array.prototype.move2 = function(pos1, pos2) {
// local variables
var i, tmp;
// cast input parameters to integers
pos1 = parseInt(pos1, 10);
pos2 = parseInt(pos2, 10);
// if positions are different and inside array
if (pos1 !== pos2 && 0 <= pos1 && pos1 <= this.length && 0 <= pos2 && pos2 <= this.length) {
// save element from position 1
tmp = this[pos1];
// move element down and shift other elements up
if (pos1 < pos2) {
for (i = pos1; i < pos2; i++) {
this[i] = this[i + 1];
}
}
// move element up and shift other elements down
else {
for (i = pos1; i > pos2; i--) {
this[i] = this[i - 1];
}
}
// put element from position 1 to destination
this[pos2] = tmp;
}
}
这不是我的功劳，这都应该归功于理查德·斯卡洛特。在这个性能测试中，它在较小的数据集上击败了基于拼接的方法。然而，正如Darwayne指出的那样，在较大的数据集上，它要慢得多。
