查找具有多次匹配属性的 JSON 对象答案

【问题标题】：Find a JSON object with property matched more than once查找具有多次匹配属性的 JSON 对象
【发布时间】：2016-07-04 18:09:24
【问题描述】：

我需要在 json 数组中找到具有相同名称属性的每个元素，例如这里 Alaska 是两次，然后我需要比较两个对象的 lastupdate 并选择具有最新更新时间的那个。采用 stackoverflow 中的答案（对不起，我丢失了链接）我可以删除具有相同名称属性的对象，但我怎样才能保留最新更新时间的对象？

[{
    "name": "Alaska",
    "Republican_fre": 3,
    "Democrats_fre": 0,
    "winner": "R",
    "iso_2": "AK",
    "electoral_vote": 3,
    "totalComponents": 3,
    "date": "29.06.2016",
    "lastupdate": "1467233426"
}, {
    "name": "Alabama",
    "Republican_fre": 3,
    "Democrats_fre": 0,
    "winner": "R",
    "iso_2": "AL",
    "electoral_vote": 9,
    "totalComponents": 3,
    "date": "29.06.2016",
    "lastupdate": "1467233426"
}, {
    "name": "Arkansas",
    "Republican_fre": 2,
    "Democrats_fre": 0,
    "winner": "R",
    "iso_2": "AR",
    "electoral_vote": 6,
    "totalComponents": 2,
    "date": "29.06.2016",
    "lastupdate": "1467233426"
},
{
    "name": "Alaska",
    "Republican_fre": 5,
    "Democrats_fre": 0,
    "winner": "R",
    "iso_2": "AK",
    "electoral_vote": 3,
    "totalComponents": 5,
    "date": "29.06.2016",
    "lastupdate": "1467282133"                 
}]

代码：

function arrUnique(arr) {
    var cleaned = [];
    data.forEach(function(itm) {
        var unique = true;
        cleaned.forEach(function(itm2) {
         var minValue = Math.min(itm.lastupdate, itm2.lastupdate)
            if (_.isEqual(itm.name, itm2.name)){
            unique = false;
            } 
        });
        if (unique)  cleaned.push(itm);
    });
    return cleaned;
}

var uniqueStandards = arrUnique(data);

jsfiddle:

预期输出 预期的输出是它保留了具有最新“lastupdate”值的 Alsaka 对象之一。因此，它首先检查具有相同名称属性的对象，然后比较 lastupdate 值并保留具有最新值的对象

【问题讨论】：

你应该为此使用 lodash 库。
“选择更新时间最近的”是什么意思？你想从 JSON 中删除其他人还是想要别的东西？
预期结果是什么？
@developer033 预期的输出是它保留了具有最新“lastupdate”值的 Alsaka 对象之一。因此，它首先检查具有相同名称属性的对象，然后比较 lastupdate 值并保留具有最新值的对象

标签： javascript angularjs json underscore.js

【解决方案1】：

您可以使用下划线的sortBy() 对集合中的项目按其lastupdate 键排序，reverse() 使所有项目按lastupdate 降序排列，然后使用uniq() 仅保留唯一的name 项目。

var uniqueStandards = _.uniq(_.sortBy(data, 'lastupdate').reverse(), 'name');

var data = [{
  "name": "Alaska",
  "Republican_fre": 3,
  "Democrats_fre": 0,
  "winner": "R",
  "iso_2": "AK",
  "electoral_vote": 3,
  "totalComponents": 3,
  "date": "29.06.2016",
  "lastupdate": "1467233426"
}, {
  "name": "Alabama",
  "Republican_fre": 3,
  "Democrats_fre": 0,
  "winner": "R",
  "iso_2": "AL",
  "electoral_vote": 9,
  "totalComponents": 3,
  "date": "29.06.2016",
  "lastupdate": "1467233426"
}, {
  "name": "Arkansas",
  "Republican_fre": 2,
  "Democrats_fre": 0,
  "winner": "R",
  "iso_2": "AR",
  "electoral_vote": 6,
  "totalComponents": 2,
  "date": "29.06.2016",
  "lastupdate": "1467233426"
}, {
  "name": "Alaska",
  "Republican_fre": 5,
  "Democrats_fre": 0,
  "winner": "R",
  "iso_2": "AK",
  "electoral_vote": 3,
  "totalComponents": 5,
  "date": "29.06.2016",
  "lastupdate": "1467282133"
}];

var uniqueStandards = _.uniq(_.sortBy(data, 'lastupdate').reverse(), 'name');

document.body.innerHTML = '<pre>' + JSON.stringify(uniqueStandards, 0, 4) + '</pre>';

&lt;script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"&gt;&lt;/script&gt;

一个普通的 JS 解决方案是：

var uniqueStandards = data
.slice() // this makes sure that we're not mutating the original array
.sort(function(x, y) { return y.lastupdate - x.lastupdate; }) // sort in descending order
.filter(function(x) {  // this ensure items with unique names
  return (this[x.name]? false: (this[x.name] = true));
}, {});

var data = [{
  "name": "Alaska",
  "Republican_fre": 3,
  "Democrats_fre": 0,
  "winner": "R",
  "iso_2": "AK",
  "electoral_vote": 3,
  "totalComponents": 3,
  "date": "29.06.2016",
  "lastupdate": "1467233426"
}, {
  "name": "Alabama",
  "Republican_fre": 3,
  "Democrats_fre": 0,
  "winner": "R",
  "iso_2": "AL",
  "electoral_vote": 9,
  "totalComponents": 3,
  "date": "29.06.2016",
  "lastupdate": "1467233426"
}, {
  "name": "Arkansas",
  "Republican_fre": 2,
  "Democrats_fre": 0,
  "winner": "R",
  "iso_2": "AR",
  "electoral_vote": 6,
  "totalComponents": 2,
  "date": "29.06.2016",
  "lastupdate": "1467233426"
}, {
  "name": "Alaska",
  "Republican_fre": 5,
  "Democrats_fre": 0,
  "winner": "R",
  "iso_2": "AK",
  "electoral_vote": 3,
  "totalComponents": 5,
  "date": "29.06.2016",
  "lastupdate": "1467282133"
}];

var uniqueStandards = data
.slice() // this makes sure that we're not mutating the original array
.sort(function(x, y) { return y.lastupdate - x.lastupdate; }) // sort in descending order
.filter(function(x) {  // this ensure items with unique names
  return (this[x.name]? false: (this[x.name] = true));
}, {});

document.body.innerHTML = '<pre>' + JSON.stringify(uniqueStandards, 0, 4) + '</pre>';

或者，您可以尝试lodash：

var uniqueStandards = _(data).orderBy('lastupdate', 'desc').uniqBy('name').value();

上面的代码片段使用orderBy() 按lastupdate 降序排列集合，并使用uniqBy() 确保集合只有唯一名称。

var data = [{
  "name": "Alaska",
  "Republican_fre": 3,
  "Democrats_fre": 0,
  "winner": "R",
  "iso_2": "AK",
  "electoral_vote": 3,
  "totalComponents": 3,
  "date": "29.06.2016",
  "lastupdate": "1467233426"
}, {
  "name": "Alabama",
  "Republican_fre": 3,
  "Democrats_fre": 0,
  "winner": "R",
  "iso_2": "AL",
  "electoral_vote": 9,
  "totalComponents": 3,
  "date": "29.06.2016",
  "lastupdate": "1467233426"
}, {
  "name": "Arkansas",
  "Republican_fre": 2,
  "Democrats_fre": 0,
  "winner": "R",
  "iso_2": "AR",
  "electoral_vote": 6,
  "totalComponents": 2,
  "date": "29.06.2016",
  "lastupdate": "1467233426"
}, {
  "name": "Alaska",
  "Republican_fre": 5,
  "Democrats_fre": 0,
  "winner": "R",
  "iso_2": "AK",
  "electoral_vote": 3,
  "totalComponents": 5,
  "date": "29.06.2016",
  "lastupdate": "1467282133"
}];

var uniqueStandards = _(data).orderBy('lastupdate', 'desc').uniqBy('name').value();

document.body.innerHTML = '<pre>' + JSON.stringify(uniqueStandards, 0, 4) + '</pre>';

&lt;script src="https://cdn.jsdelivr.net/lodash/4.13.1/lodash.min.js"&gt;&lt;/script&gt;

【讨论】：

【解决方案2】：

这是一种方法。以名称为键创建对象并根据lastUpdate更新对象，然后将对象映射到数组

function arrUnique(arr){
   var tmp={};
   arr.forEach(function(item) {
      if(!tmp[item.name] || +item.lastupdate > +tmp[item.name].lastupdate){         
           tmp[item.name] = item ;        
       }
   });
   return Object.keys(tmp).map(function(key){
      return tmp[key]
   });
}

请注意，您的 lastUpdate 的字符串比较可能不会返回正确的结果，这就是我转换为数字的原因

DEMO

【讨论】：

Object.keys 可能会更改顺序。您可以通过简单地使用过滤器return arr.filter(item => tmp[item.name] === item) 使其健壮
@yury 或在这种情况下 Object.keys(tmp).sort().map( 将按州名的顺序返回
为什么要将O(n log n)操作添加到O(n)算法中？
好点...不知道 OP 数组是否以开始排序
是的，使用filter 将保持相同的相对顺序。不管arr 是否以某种方式排序。

【解决方案3】：

希望我理解正确。

您想获得按 max(lastupdate) 排序的不同值的输出数组。

这段代码和我描述的一样工作。它称为数组分组

var group = [];
arr.forEach(function(val, key)
    {
        if(!group[val.name])
            group[val.name] = val;
        else{
            if(group[val.name].lastupdate < val.lastupdate)
                group[val.name] = val;
        }
    }
);
console.log(group);

【讨论】：

【解决方案4】：

我严重怀疑 JSON 是应用该操作的最佳格式。

如果你真的需要使用 JSON，最好在输入时检查它，并覆盖属性（或只是日期）。在这种情况下，您可以确保不存在重复项。

如果您在字符串中有任意值并且您搜索重复项，这将是一项非常棘手的任务。显而易见的解决方案是订购它，然后在 O(nlogn) 时间内搜索骗子。如果我们使用散列，这个问题可以在 O(n) 复杂度内解决。

但是知道您有已知数量的状态，您应该为每个状态迭代槽数组。

foreach state in states
    var choosenOne = {}
    foreach item in array
        if(choosenOne == {}) {
            choosenOne = item;
        } else {
            if(item.name == state) {
                if(choosenOne.lastupdate > item.lastupdate)
                    delete item;
            } else {
                delete choosenOne
                choosenOne = item;
            }
        }

这只是应该为您提供 O(50*n) ~ O(n) 解决方案的算法

【讨论】：

OP 只是将 javascript 对象数组称为 JSON。检查@charlietfl 解决方案。它的操作量比您的伪代码少 25 倍。
是的，地图解决方案在操作方面当然是最便宜的解决方案，但它不会在原地工作，它会使用n倍的内存。我同意他的解决方案很好。

【解决方案5】：

这是带有forEach 和map 的普通javascript 解决方案，用于检查索引并通过lastupdate 更新新对象。

var data = [{"name":"Alaska","Republican_fre":3,"Democrats_fre":0,"winner":"R","iso_2":"AK","electoral_vote":3,"totalComponents":3,"date":"29.06.2016","lastupdate":"1467233426"},{"name":"Alabama","Republican_fre":3,"Democrats_fre":0,"winner":"R","iso_2":"AL","electoral_vote":9,"totalComponents":3,"date":"29.06.2016","lastupdate":"1467233426"},{"name":"Arkansas","Republican_fre":2,"Democrats_fre":0,"winner":"R","iso_2":"AR","electoral_vote":6,"totalComponents":2,"date":"29.06.2016","lastupdate":"1467233426"},{"name":"Alaska","Republican_fre":5,"Democrats_fre":0,"winner":"R","iso_2":"AK","electoral_vote":3,"totalComponents":5,"date":"29.06.2016","lastupdate":"1467282133"}]
var result = [];

data.forEach(function(e) {
  if(!this[e.name]) {
    this[e.name] = e;
     result.push(e);
  } else {
    var index = result.map(function(a) { return a.name}).indexOf(e.name);
    if(e.lastupdate > result[index].lastupdate) result[index] = e;
  }
}, {});

console.log(result)

【讨论】：