按分数排序数组，可能使用lodash。 （的NodeJS）

Question

我有点困在这一个：我有一个根据层次结构（ parent_id ）排序的对象数组，如下所示：

let myArray = [
 { id: 1, parent_id: null, score: 20, type: 1 },
 { id: 12, parent_id: 1, score: 25, type: 2 },
 { id: 23, parent_id: 12, score: 55, type: 3 },
 { id: 35, parent_id: 12, score: 25, type: 3 },
 { id: 10, parent_id: null, score: 75, type: 1 },
 { id: 25, parent_id: 10, score: 15, type: 2 },
 { id: 100, parent_id: 25, score: 88, type: 3 }
]

现在我想维护层次结构顺序，还要按分数对元素进行排序以获得如下内容：

let expected = [
 { id: 10, parent_id: null, score: 75, type: 1 },
 { id: 25, parent_id: 10, score: 15, type: 2 },
 { id: 100, parent_id: 25, score: 88, type: 3 },
 { id: 1, parent_id: null, score: 20, type: 1 },
 { id: 12, parent_id: 1, score: 25, type: 2 },
 { id: 23, parent_id: 12, score: 55, type: 3 },
 { id: 35, parent_id: 12, score: 25, type: 3 },
]

我正在编写非常低效的代码，嵌套的foreach几乎可以工作，但还没有完成。 我想知道是否有更整洁的解决方案。 （很确定有，但对我来说太聪明了）。 同样在我的代码中我转发了type属性，但理想情况下我不会用它来进行排序。

注意：这个数据只是一个例子，真实数组更大，每个父项的子数变化。

由于我的解释不是很好，我们可以用这种方式思考层次结构type:1 - >国家type:2 - >州type:3 - >城市

所以我需要通过这样的得分desc来订购

- Country
  - State
    - City 
    - City
  - State
    - City
 - Country and so on...

感谢愿意帮助我的人，

Answer 1

单个排序不起作用，因为在排序数据时不遵守父子关系。

这种方法分为三个部分：

1. 按score对数据排序，因为以插入顺序构建了以下树。
2. 用给定的关系构建一棵树。
3. 遍历树并返回已排序的平面数据。

 var data = [{ id: 1, parent_id: null, score: 20, type: 1 }, { id: 12, parent_id: 1, score: 25, type: 2 }, { id: 23, parent_id: 12, score: 55, type: 3 }, { id: 35, parent_id: 12, score: 25, type: 3 }, { id: 10, parent_id: null, score: 75, type: 1 }, { id: 25, parent_id: 10, score: 15, type: 2 }, { id: 100, parent_id: 25, score: 88, type: 3 }] .sort(function (a, b) { return b.score - a.score; }), tree = function (data, root) { var r = [], o = {}; data.forEach(function (a) { o[a.id] = { data: a, children: o[a.id] && o[a.id].children }; if (a.parent_id === root) { r.push(o[a.id]); } else { o[a.parent_id] = o[a.parent_id] || {}; o[a.parent_id].children = o[a.parent_id].children || []; o[a.parent_id].children.push(o[a.id]); } }); return r; }(data, null), // null is the root value of parent_id sorted = tree.reduce(function traverse(r, a) { return r.concat(a.data, (a.children || []).reduce(traverse, [])); }, []) console.log(sorted); console.log(tree); 

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Answer 2

我更改了您的数据集以使示例更易于理解。 我使用了几个函数来对数据进行排序：

 let arr = [ {country: 'USA', state:'Washington', city:'Washington DC'}, {country: 'USA', state:'Washington', city:'Boston'}, {country: 'USA', state:'California', city:'San Diego'}, {country: 'Brazil', state:'North', city:'C'}, {country: 'Brazil', state:'North', city:'B'}, {country: 'Brazil', state:'South', city:'A'}, {country: 'Turkey', state:'East', city:'f'}, {country: 'Turkey', state:'East', city:'e'}, {country: 'Turkey', state:'West', city:'d'}, ]; let expected = [ {country: 'Brazil', state:'North', city:'B'}, {country: 'Brazil', state:'North', city:'C'}, {country: 'Brazil', state:'South', city:'A'}, {country: 'Turkey', state:'East', city:'e'}, {country: 'Turkey', state:'East', city:'f'}, {country: 'Turkey', state:'West', city:'d'}, {country: 'USA', state:'California', city:'San Diego'}, {country: 'USA', state:'Washington', city:'Boston'}, {country: 'USA', state:'Washington', city:'Washington DC'}, ]; const sortByCountry = arr => { return arr.sort((city1,city2) => city1.country > city2.country); }; const sortByState = arr => arr.sort( (city1,city2) => { // Don't compare when countries differ. if (city1.country !== city2.country) return 0; return city1.state > city2.state; } ); const sortByCity = arr => arr.sort( (city1,city2) => { // Don't compare when countries or states differ. if (city1.country !== city2.country) return 0; if (city1.state !== city2.state) return 0; return city1.city > city2.city; } ); let result = sortByCity(sortByState(sortByCountry(arr))); console.log('------expected-----'); console.log(expected); console.log('------result-----'); console.log(result); 

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