Question

在undercore.js的帮助下，有没有办法通过user_id和assess_id的比较获得最大时间值，所以user_id和assess_id是一种复合

列表将是

var list = [
      {user_id: 316,asess_id:1,time:1418109860},
      {user_id: 316,asess_id:1,time:1417708379},
      {user_id: 316,asess_id:1,time:1417708231},
      {user_id: 316,asess_id:2,time:1417708379},
      {user_id: 316,asess_id:2,time:1417692420},
      {user_id: 5092,asess_id:2,time:1417692141},
      {user_id: 316,asess_id:1,time:1417633200},
      {user_id: 5092,asess_id:1,time:1417633200},
      {user_id: 448,asess_id:1,time:1417633200}
    ];

必要的答案是

var answer = [
      {user_id: 316,asess_id:1,time:1418109860},
      {user_id: 316,asess_id:2,time:1417708379},
      {user_id: 5092,asess_id:2,time:1417692141},
      {user_id: 5092,asess_id:1,time:1417633200},
      {user_id: 448,asess_id:1,time:1417633200}
    ];

虽然我知道一个解决方案，我循环遍历整个列表并开始填写答案，并在回答中输入数据检查重复值我已经实现了此解决方案。想要在Underscore.js

的帮助下学习

Answer 1

您可以使用groupBy和map / max方法的组合：

var result = _.chain(list)
    .groupBy(function(el) {
        return el.user_id + ',' + el.asess_id;
    })
    .map(function(obj) {
        return _.max(obj, 'time');
    })
    .value();

检查下面的测试演示。

＆＃13;

var list = [
    {user_id: 316,asess_id:1,time:1418109860},
    {user_id: 316,asess_id:1,time:1417708379},
    {user_id: 316,asess_id:1,time:1417708231},
    {user_id: 316,asess_id:2,time:1417708379},
    {user_id: 316,asess_id:2,time:1417692420},
    {user_id: 5092,asess_id:2,time:1417692141},
    {user_id: 316,asess_id:1,time:1417633200},
    {user_id: 5092,asess_id:1,time:1417633200},
    {user_id: 448,asess_id:1,time:1417633200}
];

var result = _.chain(list)
    .groupBy(function(el) {
        return el.user_id + ',' + el.asess_id;
    })
    .map(function(obj) {
        return _.max(obj, 'time')
    })
    .value();

alert(JSON.stringify(result, null, 4));

＆＃13;

<script src="//cdnjs.cloudflare.com/ajax/libs/underscore.js/1.7.0/underscore-min.js"></script>

＆＃13;

Answer 2

首先尝试_.groupBy(list, 'user_id')。然后根据您在asess_id和time上定义的任何条件对子数组进行排序，并再次减少数组。

var grouped = _.groupBy(list, 'user_id');

var selected = _.reduce(grouped, function(result, value, key) {
  result[key] = _.max(value, 'time');
  return result;
}, {});

如果你打电话给_.values(selected)，你会得到：

[
  {
    "user_id": 316,
    "asess_id": 1,
    "time": 1418109860
  },
  {
    "user_id": 448,
    "asess_id": 1,
    "time": 1417633200
  },
  {
    "user_id": 5092,
    "asess_id": 2,
    "time": 1417692141
  }
]

尝试一下：http://jsfiddle.net/687f6suh/

通过比较获取最大的Underscore.js

2 个答案: