根据其他数组中的索引对对象数组进行排序

Sorting Array of Objects based on Index in other array

我有以下一组唯一 ID:

idArray = ["56f4cf96dd2ca7275feaf802",
"56f4cf96dd2ca7275feaf7b7",
"56f4cf96dd2ca7275feaf805",
"56f4cf96dd2ca7275feaf7ac"]

我还有另一个对象数组:

stories = [{"title": Story2, id = "56f4cf96dd2ca7275feaf7b7"},
{"title": Story4, id = "56f4cf96dd2ca7275feaf7ac"},
{"title": Story1, id = "56f4cf96dd2ca7275feaf802"},
{"title": Story3, id = "56f4cf96dd2ca7275feaf805"}]

如何根据第一个数组的索引对第二个数组进行排序?最好使用 lodash,因为数组可以变大一点。

到目前为止,我有以下方法从第一个数组中获取索引:

var sortArray = _.toPairs(idArray)

[ [ '0', 56f4cf96dd2ca7275feaf802 ],
[ '1', 56f4cf96dd2ca7275feaf7b7 ],
[ '2', 56f4cf96dd2ca7275feaf805 ],
[ '3', 56f4cf96dd2ca7275feaf7ac ] ]

尝试 _.map() 和 _.sortBy() 的不同组合,我似乎无法正确获得我想要的结果:

desiredResult = [{"title": Story1, id = "56f4cf96dd2ca7275feaf802"},
          {"title": Story2, id = "56f4cf96dd2ca7275feaf7b7"},
          {"title": Story3, id = "56f4cf96dd2ca7275feaf805"},
          {"title": Story4, id = "56f4cf96dd2ca7275feaf7ac"}]

试试这个

var idsToIndexes = {};

for (var i = 0; i < idArray.length; i++)
    idsToIndexes[idArray[i]] = i;

stories.sort(function(a, b){return idsToIndexes[a.id] - idsToIndexes[b.id];});

这可以在没有任何库的情况下使用 Array.sort()

var stories = [{"title": 'Story2', id : "56f4cf96dd2ca7275feaf7b7"},
{"title": 'Story4', id : "56f4cf96dd2ca7275feaf7ac"},
{"title": 'Story1', id : "56f4cf96dd2ca7275feaf802"},
{"title": 'Story3', id : "56f4cf96dd2ca7275feaf805"}];

var idArray = ["56f4cf96dd2ca7275feaf802",
"56f4cf96dd2ca7275feaf7b7",
"56f4cf96dd2ca7275feaf805",
"56f4cf96dd2ca7275feaf7ac"];

var ordered = stories.sort(function(a, b){
 return idArray.indexOf(a.id) - idArray.indexOf(b.id);
});

ordered.forEach( element =>{ console.log(element) });

我认为排序解决方案非常无效,尤其是因为您希望数组稍后会变大。排序是 "at best" 一个 O(2n) 操作,而每个周期有两个 indexOf 操作另一个 O(2n^2)。我提出以下建议,它将优于大型数组中的排序方法。

var stories = [{"title": 'Story2', id : "56f4cf96dd2ca7275feaf7b7"},
{"title": 'Story4', id : "56f4cf96dd2ca7275feaf7ac"},
{"title": 'Story1', id : "56f4cf96dd2ca7275feaf802"},
{"title": 'Story3', id : "56f4cf96dd2ca7275feaf805"}],

    idArray = ["56f4cf96dd2ca7275feaf802",
"56f4cf96dd2ca7275feaf7b7",
"56f4cf96dd2ca7275feaf805",
"56f4cf96dd2ca7275feaf7ac"],

ordered = idArray.reduce((p,c) => p.concat(stories.find(f => f.id == c)) ,[]);

console.log(ordered);

仅 O(n^2)