21

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其映射到 topicId > id 和 topicName > name 的这种数组：

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

我正在使用现代浏览器，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。

4

3 回答 3

33

最近

_.uniqBy 现在更可取

完整的工作示例在这里 

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);

遗产

http://lodash.com/docs#uniq是一个好的开始

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

您的代码看起来像这样，以通过 id 获取唯一的主题

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/

于 2014-01-28T19:04:59.927 回答
2

我是那些使用本机功能的人之一:)

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; 
},[]);
于 2014-01-28T19:11:28.280 回答
-1

使用 ECMAScript 2015Array.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。否则返回未定义。

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);

于 2017-05-11T05:33:08.150 回答