Question

我的数组看起来像这样：

var arr = [
{user: 50, id: 70, time: '14:30'}, // ignore this user 50
{user: 70 id: 50, time: '14:50'}, // output this time is higher id 50
{user: 83, id: 50, time: '18:30'}
];

我想要的结果是按时间从阵列中获取唯一的用户或id对象：

var result = [
{user: 83, id: 50, time: 18:30},
{user: 70, id: 50, time: 14:50}
];

或

var arr = [
    {user: 70, id: 50, time: '14:30'}, // ignores this // id 50
    {user: 50 id: 70, time: '14:50'}, // output this // user 50
    {user: 83, id: 50, time: '18:30'}
    ];

然后结果应该降低

var result = [
    {user: 83, id: 50, time: 18:30},
    {user: 50, id: 70, time: 14:50}
    ];

进一步解释我想要的东西。考虑到obj1用户是50并且id是70并且我们还具有obj2用户70和id 50，只要在阵列中存在共享用户＆lt;＆gt;的另一个对象。 id - id＆lt;＆gt;用户将最近的时间价值降低。

Answer 1

您可以使用lodash#orderBy按降序排列时间，以确保我们按照最高时间顺序唯一删除每个项目的相同user和id。最后，我们使用lodash#uniqWith来执行比较。

var result = _(arr)
  .orderBy('time', 'desc')
  .uniqWith((v1, v2) => v1.id === v2.user && v1.user === v2.id)
  .value();

// First data set
var arr = [
  {user: 50, id: 70, time: '14:30'}, // ignore this user 50
  {user: 70, id: 50, time: '14:50'}, // output this time is higher id 50
  {user: 83, id: 50, time: '18:30'}
];

var result = _(arr)
  .orderBy('time', 'desc')
  .uniqWith((v1, v2) => v1.id === v2.user && v1.user === v2.id)
  .value();
  
console.log('First data set');
console.log(result);

// Second data set
arr = [
  {user: 70, id: 50, time: '14:30'}, // ignores this // id 50
  {user: 50, id: 70, time: '14:50'}, // output this // user 50
  {user: 83, id: 50, time: '18:30'}
];

result = _(arr)
  .orderBy('time', 'desc')
  .uniqWith((v1, v2) => v1.id === v2.user && v1.user === v2.id)
  .value();
  
console.log('Second data set');
console.log(result);

.as-console-wrapper{min-height: 100%;top: 0;}

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>

Answer 2

您需要按id

进行过滤

_.filter(arr, obj => obj.id === 50);

您可以使用核心js

arr.filter(item => item.id === 50)

如何在lodash中实现这一目标

2 个答案: