Lodash sort collection based on external array

This is the efficient & clean way:

(Import lodash identity and sortBy):

TS:

function sortByArray<T, U>({ source, by, sourceTransformer = identity }: { source: T[]; by: U[]; sourceTransformer?: (item: T) => U }) {
  const indexesByElements = new Map(by.map((item, idx) => [item, idx]));
  const orderedResult = sortBy(source, (p) => indexesByElements.get(sourceTransformer(p)));
  return orderedResult;
}

Or in JS:

function sortByArray({ source, by, sourceTransformer = _.identity }) {
    const indexesByElements = new Map(by.map((item, idx) => [item, idx]));
    const orderedResult = _.sortBy(source, (p) => indexesByElements.get(sourceTransformer(p)));
    return orderedResult;
}

Input:

var data1 = ['129asg', '39342aa'];
var data2 = [{
    guid: '39342aa',
    name: 'John'
}, {
    guid: '129asg',
    name: 'Mary'
}];

1. First create an index object, with _.reduce, like this

   var indexObject = _.reduce(data2, function(result, currentObject) {
       result[currentObject.guid] = currentObject;
       return result;
   }, {});
   

2. And then map the items of the first array with the objects from the indexObject, like this

   console.log(_.map(data1, function(currentGUID) {
       return indexObject[currentGUID]
   }));
   

Output

[ { guid: '129asg', name: 'Mary' },
  { guid: '39342aa', name: 'John' } ]

Note: This method will be very efficient if you want to sort so many objects, because it will reduce the linear look-up in the second array which would make the entire logic run in O(M * N) time complexity.

Here is just a simple add to the accepted answer in case you want to put the unmatched elements at the end of the sortedCollection and not at the beginning:

const last = collection.length;

var sortedCollection = _.sortBy(collection, function(item) {
  return firstArray.indexOf(item.guid) !== -1? firstArray.indexOf(item.guid) : last;
});

var sortedCollection = _.sortBy(collection, function(item){
  return firstArray.indexOf(item.guid)
});