如何将具有重复值的数组映射到 Javascript 中的唯一数组？

Question

我有以下数组：

var tst = 
[
 {"topicId":1,"subTopicId":1,"topicName":"a","subTopicName":"w"},
 {"topicId":1,"subTopicId":2,"topicName":"b","subTopicName":"x"},
 {"topicId":1,"subTopicId":3,"topicName":"c","subTopicName":"y"},
 {"topicId":2,"subTopicId":4,"topicName":"c","subTopicName":"z"}
]

有没有一种简单的方法可以将它映射到这种类型的数组，其中 topicId > id 和 topicName > name:

var t = 
[
  {"id":1,"name":"a"},
  {"id":2,"name":"c"}
]

我使用的是现代浏览器，如果有帮助，我也有 _lodash。请注意，tst 数组中将有大约 100 行，因此我不需要非常优化的解决方案。一个简单且易于维护的解决方案将更为重要。

Answer 1

最近的

_.uniqBy is now preferable

Full working example here

var tst = [
 {"topicId":1,"subTopicId":1,"topicName":"a","subTopicName1":"w"},
 {"topicId":2,"subTopicId":2,"topicName":"b","subTopicName2":"x"},
 {"topicId":3,"subTopicId":3,"topicName":"c","subTopicName3":"y"},
 {"topicId":1,"subTopicId":4,"topicName":"c","subTopicName4":"z"}
];

var result = _.map(_.uniqBy(tst, 'topicId'), function (item) {
    return {
        id: item.topicId,
        name: item.topicName
    };  
});

console.log(result);

---

遗留

http://lodash.com/docs#uniq 是一个好的开始

_.uniq([{ 'x': 1 }, { 'x': 2 }, { 'x': 1 }], 'x');

您的代码将如下所示以获取唯一的主题 id

var t = _.uniq(tst, 'topicId');

编辑

我做了一个 jsfiddle

http://jsfiddle.net/q5HNw/

更新

删除了不必要的名称唯一性

http://jsfiddle.net/q5HNw/1/

Answer 2

我是那些使用原生函数的人之一 :)

var results = tst.reduce(function(res,topic){
var exists = res.some(function(t){ return (t.id === topic.topicId && t.name === topic.topicName);});        
     if (!exists){
        res.push({"id": topic.topicId, "name": topic.topicName});
     }
return res; },[]);

Lodash 版本

我不是使用 lodash 的专家，可能我会尝试这样的事情：

var results = _.reduce(tst, function(res, topic){       
    var exists = _.findIndex(res, function(t){
        return (t.id === topic.topicId && t.name === topic.topicName);
    });
    if (exists === -1){
      res.push({"id": topic.topicId, "name": topic.topicName});
    }
    return res; 
},[]);

Answer 3

使用 ECMAScript 2015 Array.prototype.find()

find() 方法返回数组中满足提供的测试函数的第一个元素的值。否则返回 undefined。

let tst = [
     {"topicId":1,"subTopicId":1,"topicName":"a","subTopicName":"w"},
     {"topicId":1,"subTopicId":2,"topicName":"b","subTopicName":"x"},
     {"topicId":1,"subTopicId":3,"topicName":"c","subTopicName":"y"},
     {"topicId":2,"subTopicId":4,"topicName":"c","subTopicName":"z"},
];

let t = [];
tst.forEach(obj => {
  // Check if the id already exists in the array 't'
  if (!t.find((self) => self.id === obj.topicId)) {
    // If not, pushes obj to t
    t.push({
      id: obj.topicId,
      name: obj.topicName
    });
  }
});

console.log(t);

您还可以比较多个属性：

let tst = [
         {"topicId":1,"subTopicId":1,"topicName":"a","subTopicName":"w"},
         {"topicId":1,"subTopicId":2,"topicName":"b","subTopicName":"x"},
         {"topicId":1,"subTopicId":3,"topicName":"c","subTopicName":"y"},
         {"topicId":2,"subTopicId":4,"topicName":"c","subTopicName":"z"},
];


let t = [];
tst.forEach(obj => {
  // Check if the 'id' and 'subId' already exist in t 
  if (!t.find((self) => self.id === obj.topicId && self.subId === obj.subTopicId)) {
    // If not, pushes obj to t
    t.push({
      id: obj.topicId,
      subId: obj.subTopicId,
      name: obj.topicName
    });
  }
});

console.log(t);