Question

提供了一个列表

[
{id:'i1',value:'v1'},
{id:'i3',value:'v3'},
{id:'i4',value:'v4'},
{id:'i2',value:'v2'},
{id:'i5',value:'v5'},
]

和另一个订购的ID列表

['i1','i2','i3','i4','i5']

获取

[
{id:'i1',value:'v1'},
{id:'i2',value:'v2'},
{id:'i3',value:'v3'},
{id:'i4',value:'v4'},
{id:'i5',value:'v5'},
]

如何在lodash中做到这一点？

Answer 1

使用sortBy方法并从iteratee函数返回ID的索引位置

let sortedArray = _.sortBy(input, (v) => order.indexOf(v.id))

其中

input = [
  {id:'i1',value:'v1'},
  {id:'i3',value:'v3'},
  {id:'i4',value:'v4'},
  {id:'i2',value:'v2'},
  {id:'i5',value:'v5'},
]
order = ['i1','i2','i3','i4','i5']

Answer 2

将数组转换为键为id的对象，以实现更好的查找。用实际值替换列表中的每个元素。

边缘情况：两个数组具有相同的元素

const foo = [
  {id:'i1',value:'v1'},
  {id:'i3',value:'v3'},
  {id:'i4',value:'v4'},
  {id:'i2',value:'v2'},
  {id:'i5',value:'v5'},
];

const ordered = ['i1','i2','i3','i4','i5'];

const fooMap = _.keyBy(f, 'id');

const orderedFoo = ordered.map(id => fooMap[id])

Answer 3

const arr = ['i1','i2','i3','i4','i5'];
const arrObj = [
{id:'i1',value:'v1'},
{id:'i3',value:'v3'},
{id:'i4',value:'v4'},
{id:'i2',value:'v2'},
{id:'i5',value:'v5'}
]
const sortedArr = [];

_.map(arr, a => sortedArr.push(_.find(arrObj, ['id', a])));
console.log(sortedArr)

Answer 4

let arr = ['i1','i2','i3','i4','i5'];
let arrObj = [
    {id:'i1',value:'v1'},   
    {id:'i3',value:'v3'},
    {id:'i4',value:'v4'},
    {id:'i2',value:'v2'},
    {id:'i5',value:'v5'}
]

console.table(arrObj);

arr.map(v => {
    arrObj.sort( v1 => { return v1.id == v ? true : false} )
})

console.table(arrObj);

如何在lodash中对另一个列表进行排序

4 个答案: