基于其他数组中的索引对对象数组进行排序

Question

I have the following array of unique IDs:

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

And I have another array of objects:

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

How can I sort the second array based on the index of the first array? Preferably using lodash, as the arrays can grow a bit larger.

So far, I have the following to get the indexes from the first array:

var sortArray = _.toPairs(idArray)

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

Trying different combinations of _.map() and _.sortBy() I can't seem to properly get the result I want which is:

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

Answer 1

This is possible to do without any library using 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) }); 

Answer 2

I believe the sort solution is very ineffective especially since you expect the arrays to grow bigger later. Sort is "at best" an O(2n) operation while you have two indexOf operations per cycle another O(2n^2). I propose the following which will outperform the sort method in large arrays.

 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); 

Only O(n^2)

Answer 3

Try this

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];});