I have an array with keys like so:
['asdf12','39342aa','12399','129asg',...]
and a collection which has these keys in each object like so:
[{guid: '39342aa', name: 'John'},{guid: '129asg', name: 'Mary'}, ... ]
Is there a fast way to sort the collection based on the order of keys in the first array?
var sortedCollection = _.sortBy(collection, function(item){
return firstArray.indexOf(item.guid)
});
a fast way to sort the collection in the question. But this is not a fast way, if you really meant the runtime performance.
_.sortBy(collection, (item) => { const index = firstArray.indexOf(item.guid); return index === -1 ? Infinity : index; });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;
});
Input:
var data1 = ['129asg', '39342aa'];
var data2 = [{
guid: '39342aa',
name: 'John'
}, {
guid: '129asg',
name: 'Mary'
}];
First create an index object, with _.reduce, like this
var indexObject = _.reduce(data2, function(result, currentObject) {
result[currentObject.guid] = currentObject;
return result;
}, {});
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.
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;
}
You can use indexBy(), and at() to sort your collection. The advantage being that concise code and performance. Using sortBy() here does the trick, but your external array is already sorted:
var ids = [ 'cbdbac14', 'cf3526e2', '189af064' ];
var collection = [
{ guid: '189af064', name: 'John' },
{ guid: 'cf3526e2', name: 'Julie' },
{ guid: 'cbdbac14', name: 'James' }
];
_(collection)
.indexBy('guid')
.at(ids)
.pluck('name')
.value();
// → [ 'James', 'Julie', 'John' ]
Using at(), you can iterate over the sorted external collection, building a new collection from the source collection. The source collection has been transformed into an object using indexBy(). You do this so at() has key-based access for each of it's ids.
